Structural scales and types of analysis in composite materials

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Structural scales and
types of analysis in
composite materials
Daniel & Ishai: Engineering Mechanics of
Composite Materials
• Micromechanics
- which fibre?
- how much fibre?
- arrangement of fibres?
>>> LAYER PROPERTIES
(strength, stiffness)
• Laminate Theory
- which layers?
- how many layers?
- how thick?
>>> LAMINATE PROPERTIES
» LAMINATE PROPERTIES
>>> BEHAVIOUR UNDER LOADS
(strains, stresses, curvature, failure mode…)
Polymer composites are usually
laminated from several individual layers of
material. Layers can be ‘different’ in the
sense of:
•
•
•
•
•
different type of reinforcement
different geometrical arrangement
different orientation of reinforcement
different amount of reinforcement
different matrix
Typical laminate
configurations
for storage
tanks to BS4994
Eckold (1994)
E2
fibre
direction
E1
The unidirectional ply (or lamina) has
maximum stiffness anisotropy - E1»E2
90o
0o
We could remove the in-plane
anisotropy by constructing a ‘cross-ply’
laminate, with UD plies oriented at 0
and 90o. Now E1 = E2.
But under the action of an in-plane
load, the strain in the relatively stiff 0o
layer is less than that in the 90o layer.
Direct stress thus results in bending:
This is analogous to a metal laminate consisting of
one sheet of steel (modulus ~ 210 GPa) bonded to
one of aluminium (modulus ~ 70 GPa):
P Powell: Engineering with
Fibre-Polymer Laminates
Note the small anticlastic bending due to the
different Poisson’s ratio of steel and aluminium.
In this laminate, direct stress and
bending are said to be coupled.
Thermal and
moisture effects also
result in coupling in
certain laminates consider the familiar
bi-metallic strip:
A single ‘angle-ply’ UD lamina (ie fibre
orientation q  0o or 90o) will shear
under direct stress:
q
In a 2-ply laminate (q, -q), the shear
deformations cancel out, but result in
tension-twist coupling:
To avoid coupling effects, the cross-ply
laminate must be symmetric - each ply must
be mirrored (in terms of thickness and
orientation) about the centre.
Possible symmetric arrangements would be:
0o
90o
0/90/90/0
[0,90]s
90/0/0/90
[90,0]s
Both these laminates have the same inplane stiffness.
How do the flexural stiffnesses compare?
0o
90o
0/90/90/0
[0,90]s
90/0/0/90
[90,0]s
• The two laminates [0,90]s and [90,0]s
have the same in-plane stiffness, but
different flexural stiffnesses
• Ply orientations determine in-plane
properties.
• Stacking sequence determines flexural
properties.
• The [0,90]s laminate becomes [90,0]s if
rotated. So this cross-ply laminate has
flexural properties which depend on
how the load is applied!
HAWT (2004)
VAWT (1987)
• To avoid all coupling effects, a laminate
containing an angle ply must be balanced
as well as symmetric - for every ply at angle
q, the laminate must contain another at -q.
• Balance and symmetry are not the same:
0/30/-30/30/0 - symmetric but not balanced
= direct stress/shear strain coupling.
30/30/-30/-30 - balanced but not symmetric
= direct stress/twist coupling.
ODE/BMT: FRP Design Guide
• The [0,90] cross-ply
laminate (WR) has
equal properties at 0o
and 90o, but is not
isotropic in plane.
• A ‘quasi-isotropic’
laminate must contain
at least 3 different
equally-spaced
orientations:
0,60,-60;
0,90,+45,-45; etc.
UD (0o)
laminate
proportion of
plies at 90o
proportion of
plies at 0o
proportion of
plies at 45o
Carpet plot for tensile modulus of
glass/epoxy laminate
UD (90o)
laminate
0/90 (cross-ply)
E = 29 GPa
0/90/±45 (quasi-isotropic)
E = 22 GPa
Classical Plate Analysis
• Plane stress (through-thickness and
interlaminar shear ignored).
• ‘Thin’ laminates; ‘small’ out-of plane
deflections
• Plate loading described by equivalent
force and moment resultants.
Nx  
h/ 2
h/ 2
σ(z).dz
Mx  
h/ 2
h/ 2
σ(z).z.dz
• If stress is constant through thickness h,
Nx = h sx, etc.
Classical Plate Analysis
• Plate bending is described by
curvatures kx, ky, kxy.
• The ‘curvature’ is equal to 1 / radius of
curvature.
• Total plate strain results from in-plane
loads and curvature according to:
[ ( z)]  [  ]  z[ k ]
o
where z is distance from centre of plate
Classical Plate Analysis
Stress = stiffness x strain:
s( z)  [Q][  ]  z[Q][ k ]
o
Giving:
[N ] 
h/2
h/2
h/2
h / 2
h / 2
h / 2
o
o
[
Q
][

]
dz

[
Q
][
k
]
zdz

[
Q
][

]


[M ]  [Q][ o ]
h/2

h / 2
h/2
zdz  [Q][k ]  z 2 dz
h / 2

h/2
dz  [Q][k ]  zdz
h / 2
In simpler terms:
[ N ]  [ A][  o ]  [ B ][ k ]
[ M ]  [ B ][  o ]  [ D][ k ]
[A] is a matrix defining the in-plate stiffness. For an isotropic
sheet, it is equal to the reduced stiffness multiplied by thickness
(units force/distance).
[B] is a coupling matrix, which relates curvature to in-plane
forces. For an isotropic sheet, it is identically zero.
[D] is the bending stiffness matrix. For a single isotropic sheet,
[D] = [Q] h3/12, so that D11=Eh3/12(1-n2), etc.
Classical Laminate Analysis
• Combines the principles of thin plate
theory with those of stress
transformation.
• Mathematically, integration is performed
over a single layer and summed over all
the layers in the laminate.
h
h


o
[ N ]   [Q ][ ]  dz  [Q ][ k ]  zdz 
j 1
h
h


n
j
j
j 1
j 1
Classical Laminate Analysis
• The result is a so-called constitutive
equation, which describes the
relationship between the applied loads
and laminate deformations.
[ N ]  [ A][ B]  [ ]


  
.
[M ] [ B][ D] [k ] 
o
[A], [B] and [D] are all 3x3 matrices.
Classical Laminate Analysis
• Matrix inversion gives strains resulting
from applied loads:
[ ] [a][b]  [ N ] 

  
.

[k ]  [c][d ] [ M ]
o
where:
1
[ A][ B] 
[a][b] 

  

[ B][ D]
[c][d ]
Effective Elastic Properties of the
Laminate (thickness h)
a 21
a12
1
1
1
Ex 
; Ey 
; G xy 
;n xy 
;n yx 
ha11
ha 22
ha 33
a11
a 22
Bending stiffness from the inverted D
matrix:
E
flex
x
12
12
flex
 3 ; Ey  3
h d11
h d 22
1
1
( EI ) x 
; ( EI ) y 
d11
d 22
Classical Laminate Analysis - assumptions
1 Layers in the laminate are perfectly bonded to each
other – strain is continuous at the interface between
plies.
2 The laminate is thin, and is in a state of plane stress.
This means that there can be no interlaminar shear or
through-thickness stresses (tyz = tzx = sz = 0).
3 Each ply of the laminate is assumed to be
homogeneous, with orthotropic properties.
4 Displacements are small compared to the thickness of
the laminate.
5 The constituent materials have linear elastic properties.
6 The strain associated with bending is proportional to the
distance from the neutral axis.
Steps in Classical Laminate Analysis
1. Define the laminate – number of layers, thickness, elastic
and strength properties and orientation of each layer.
2. Define the applied loads – any combination of force and
moment resultants.
3. Calculate terms in the constitutive equation matrices [A], [B] and [D].
4. Invert the property matrices – [a] = [A]-1, etc.
5. Calculate effective engineering properties.
6. Calculate mid-plane strains and curvatures.
7. Calculate strains in each layer.
8. Calculate stresses in each layer from strains, moments
and elastic properties.
9. Evaluate stresses and/or strains against failure criteria.
Use of LAP software to
calculate effect of cooling
from cure temperature
(non-symmetric laminate).

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