Slides (part 5)

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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