Ch.6: Analysis of Laminated Composites
Stacking of plies with differentThe transverse properties of unidirectional compositesangles for tailoring
(stiffness, thermal stability)
The transverse properties of unidirectional composites are unsatisfactory for most practical applications.
The goal of this chapter is to analyse the stacking sequencein order to achieve adequate anisotropic properties.
One plyL
T
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Stress and strain variation in a laminate
Kirchhoff plate theory:A line ABCD originally straight and normalA line ABCD originally straight and normalto the mid‐plane remains straight in the
deformed state: A’B’C’D’(no shear deformation)
Displacements of the midplane:
Slope of the laminate in the (x,z) plane:
Displacements at a point at aDisplacements at a point at a distance z from the midplane:
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Plate curvatures
Mid‐plane strains(membrane)( )
The strain varies linearly across the thicknessHowever, the stiffness properties are discontinuous from one layer to the next
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Every layer is characterized by its stiffness matrix
kk k
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Resultant forces and momentsResultant forces and moments
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Sum of the contributionsof the various layersof the various layers
For every layer:
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Do not depend on z
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A B
B D
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Extensional stiffness matrix
Coupling stiffness matrix(B=0 for symmetric stacking)
Bending stiffness matrix
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Example: Non‐symmetric two‐ply laminate(5mm at 0° and 3mm at 45°) Calculate the stiffness matrix
Stiffness matrix of one plyin principal material axes:in principal material axes:
Step 1: Compute the stiffness matrix for the ply at 45°[using formula (5.61) with =45°]
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Stiffness matrix in arbitrary axes
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Step 2: Global stiffness matrix
Opposite signs !
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Constitutive equation for the two‐ply laminate
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Symmetric laminatey
Contribution to B of symmetric layers:
No coupling between in‐plane forcesand out of plane deformations
(very important for thermal stability!)
Orthotropic in the plane
thicknessFor every ply with +, there should be another
14Odd function of
thickness ply with the same thickness oriented at ‐
Example: Four‐ply laminate
Each ply has a thickness of 3 mm
orthotropic
Because of symmetry:Coupling
Bending‐torsiong
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How to minimize the coupling between bending and torsion ?
Q16 and Q26 are odd functions of 16 26
Option 1: All layers oriented at 0° or 90°
Option 2: For every layer at + above the mid‐plane, there should be a layer with the same thickness and oriented at –, at the same distance below the midplane.B t thi i i tibl ith t !
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But this is incompatible with symmetry !
For a symmetric laminate, D16 and D26 cannot be zero. However, by stackingthe layers alternatively at + and –, D16 and D26 can be minimized, especiallyif the number of layers is largeif the number of layers is large.
45° ‐45°
unchanged
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Quasi‐isotropic laminate
Constitutive equations of an isotropic material:
A quasi‐isotropic laminate has the extensionalstiffness properties of an isotropic material:
Construction:•The total number n of layers must be 3 or more•Identical individual layers (Q and t)•The layers must be oriented at equal angles:•The layers must be oriented at equal angles:
/n between two layers
Examples:
Also: But not:
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Stresses and strains in the layers
Step 1: Invert the stiffness matrix to compute the mid plane strains and the curvatures:
Step 2: for every layer, one compute the stresses in global coordinates (x,y):
[ ][ ]k k
(linear over the thickness of the layer)
Step 3: before applying the failure criteria, one must transform the stresses in the (L,T) frame
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Example: three‐ply laminate [45°/0]S
In plane forcesIn‐plane forces:
A 1A‐1
The strains are identicalfor all layers
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Before applying a failure test, one needs to transform into the (L,T) frame
T(45°)T(45°)=
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Stresses and strains in (x,y) frameStresses and strains in (x,y) frame
Stresses and strains in (L,T) frame
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