Plastics Engineering 3E Episode 7 pot

198 Mechanical Behaviour of Composites ii Loading off Fibre Axis For the situation where the loading is applied off the fibre axis, then the above approach involving the Plate Constitu

194 Mechanical Behaviour of Composites when the stresses a ay and txy are applied Calculate also the strains in the global X - Y directions

[SI = [a]-' where [a] = [To]-' [ e ] [T,]

The local compliance matrix is

Mechanical Behaviour of Composites 195 Directly by matrix manipulation

E, = 1.318 x lod3 cy = -1.509 x yxy = 8.626 x

or by multiplying out the terms

E, = [(s11) (a,) + (512) (ay)] + (516) ( t x y ) E, = 1.318 x

and similarly for the other two strains

3.8 General Deformation Behaviour of a Single Ply

The previous section has considered the in-plane deformations of a single ply

In practice, real engineering components are likely to be subjected to this type

of loading plus (or as an alternative) bending deformations It is convenient

at this stage to consider the flexural loading of a single ply because this will develop the method of solution for multi-ply laminates

(i) Loading on Fibre Axis

Consider a unidirectional sheet of material with the fibres aligned in the x-

direction and subjected to a stress, u, If the sheet has thickness, h, as shown

in Fig 3.15 and we consider unit width, then the normal axial force N, is given by

196 Mechanical Behaviour of Composites

Or more generally for all the force components N,, N, and N,,

where [Q] is the stiffness matrix as defined earlier

Now, from equation (3.27), the forces [N] are given by

where [A] is the Extensional Stiffness matrix (= [Qlh) and [ B ] is a Coupling Matrix It may be observed that in the above analysis [B] is in fact zero for

Mechanical Behaviour of Composites 197

this simple single ply situation However, its identity has been retained as it has relevance in laminate analysis, to be studied later

Returning to equation (3.28), the moments may be written as

where [ D ] is the Bending Stiffness Matrix (= [Q]h3/12)

Equations (3.29) and (3.30) may be grouped into the following form

It should be noted that it is only possible to utilise [a] = [AI-' and [dl =

[DI-' for the special case where [B] = 0 In other cases, the terms in the [a]

and [ d ] matrices have to be determined from

A B -'

[i : I = [ , D ] (3.34)

198 Mechanical Behaviour of Composites

(ii) Loading off Fibre Axis

For the situation where the loading is applied off the fibre axis, then the

above approach involving the Plate Constitutive Equations can be used but it

is necessary to use the transformed stiffness matrix terms 0

Hence, in the above analysis

[AI = @I * h

103 = @I h3/12

The use of the Plate Constitutive Equations is illustrated in the following

Examples

Example 3.9 For the 2 mm thick unidirectional carbon fibre/PEEK

composite described in Example 3.6, calculate the values of the moduli,

Poisson’s Ratio and strains in the global direction when a stress of a =

50 MN/m2 is applied You should use

(i) the lamina stiffness and compliance matrix approach and

(ii) the Plate Constitutive Equation approach

Mechanical Behaviour of Composites 199

2.79 x io4 5.51 x io3 1.61 x io4

a = A-' and d = D-' (since B = 0)

-3.12 2.06 x 1 0 - ~ -1.29 8.89 x x 1 0 - ~ -3.12 7.16 x x 1 0 - ~ 1

200 Mechanical Behaviour of Composites

This is because in composites we can often get coupling between the different

modes of deformation This will also be seen later where coupling between

axial and flexural deformations can occur in unsymmetric laminates Fig 3.17

illustrates why the shear strains arise in uniaxially stressed single ply in this

Example

Orlginal unidirectional composite

Deformed shape Fig 3.17 Coupling effects between extension and shear

Mechanical Behaviour of Composites 20 1 Example 3.10 If a moment of M y = 100 Nm/m is applied to the unidirec- tional composite described in the previous Example, calculate the curvatures which will occur Determine also the stress and strain distributions in the global

( x - y ) and local (1-2) directions

Solution Using the D and d matrices from the previous Example, then for the applied moment My = 100 N:

This enables the curvatures to be determined as

202 Mechanical Behaviour of Composites

[":I = T e a [ :]

Y 1 2 Y X Y

so

~1 = -5.14 x ~2 = 8 x y 1 2 = 0.014

These stresses and strains are illustrated in Fig 3.18

Fig 3.18 Stresses and strains, Example 3.10

Note that if both plane stresses and moments are applied then the total stresses will be the algebraic sum of the individual stresses

3.9 Deformation Behaviour of Laminates

(i) Laminates Made from Unidirectional Plies

The previous analysis has shown that the properties of unidirectional fibre composites are highly anisotropic To alleviate this problem, it is common to build up laminates consisting of stacks of unidirectional lamina arranged at different orientations Clearly many permutations are possible in terms of the numbers of layers (or plies) and the relative orientation of the fibres in each

Mechanical Behaviour of Composites 203 layer At first glance it might appear that the best means of achieving a more isotropic behaviour would be to have two layers with the unidirectional fibres arranged perpendicular to each other For example, two layers arranged at 0" and 90" to the global x-direction or at +45" and -45" to the x-direction might appear to offer more balanced properties in all directions In fact the lack of symmetry about the centre plane of the laminate causes very complex behaviour

in such cases

In general it is best to aim for symmetry about the centre plane A lami- nate in which the layers above the centre plane are a mirror image of those

below it is described as symmetric Thus a four stack laminate with fibres

oriented at 0", 90", 90" and 0" is symmetric The convention is to denote this

as [oo/900/900/o"]T or [0", 90;, Oo]T or [0"/90"], In general terms any laminate

of the type [e, -8, -8, e]T is symmetric and there may of course be any even number of layers or plies They do not all have to be the same thickness but symmetry must be maintained In the case of a symmetric laminate where the central ply is not repeated, this can be denoted by the use of an overbar Thus the laminate [45/ - 45/0/90/0/ - 45/45]T can be written as [f45,0, %lS

In-plane Behaviour of a Symmetric Laminate

The in-plane stiffness behaviour of symmetric laminates may be analysed as follows The plies in a laminate are all securely bonded together so that when the laminate is subjected to a force in the plane of the laminate, all the plies deform by the same amount Hence, the strain is the same in every ply but because the modulus of each ply is different, the stresses are not the same This

is illustrated in Fig 3.19

Fig 3.19 Stresses and strains in a symmetric laminate

When external forces are applied in the global x - y direction, they will equate

to the summation of all the forces in the individual plies Thus, for unit width

where h is the thickness of the laminate a N, are the overall stresses or forces and (a)f is the stress in the ply 'f' (see Fig 3.20)

f th layer

hl2

L -

Fig 3.20 Ply f in the laminate

In matrix form we can write

Mechanical Behaviour of Composites 205

As the strains are independent of 2 they can be taken outside the integral:

where, for example,

Having obtained all the terms for the extensional stiffness matrix [A], this

may then be inverted to give the compliance matrix [ a ]

where h is the thickness of the laminate

V y x = -

206 Mechanical Behaviour of Composites

3.10 Summary of Steps to Predict Stiffness of Symmetric Laminates

1 The Stiffness matrix [GI is obtained as earlier each individual ply in the laminate

2 The Stiffness matrix [A] for the laminate is determined by adding the product of thickness and [GI for each ply

3 The Compliance matrix [a] for the laminate is determined by inverting

[A] ie [ a ] = [AI-'

4 The stresses and strains in the laminate are then determined from

{ ~ } = ~ a ~ { : } h

Example 3.11 A series of individual plies with the properties listed below

are laid in the following sequence to make a laminate

Determine the moduli for the laminate in the global X-Y directions and the

strains in the laminate when stresses of a , = 10 MN/m2, cy = - 14 MNlm2 and tXy = -5 MN/m2 are applied The thickness of each is 1 mm

E1 = 125000 MN/m2 E2 = 7800 MN/m2 G12 = 4400 MN/m2

~ 1 2 = 0.34

Solution The behaviour of each ply when subjected to loading at 13 degrees

off the fibre axis is determined using Matrix manipulation as follows:

Stress Transformation Matrix Strain Transformation Matrix

-sc sc (2 - s2)

s2 c2 -2sc ]

Mechanical Behaviour of Composites 207 Overall Stiffness Matrix Overall Compliance Matrix

-

6.27 io4 2.71 io4 3.66 io4 2.71 x IO4 2.21 x IO4 1.87 x IO4

3.66 x io4 1.87 io4 2.88 io4

Hence, the Extension Stiffness matrix is given by

208 Mechanical Behaviour of Composites 3.1 1 General Deformation Behaviour of Laminates

The previous section has illustrated a simple convenient means of analysing in-plane loading of symmetric laminates Many laminates are of this type and

so this approach is justified However, there are also many situations where other types of loading (including bending) are applied to laminates which may

be symmetric or non-symmetric In order to deal with these situations it is necessary to adopt a more general type of analysis

Convention for defining thicknesses and positions of plies

In this more general analysis it is essential to be able to define the position and thickness of each ply within a laminate The convention is that the geometrical mid-plane is taken as the datum The top and bottom of each ply are then defined relative to this Those above the mid-plane will have negative co-ordinates and those below will be positive The bottom surface of the fth ply has address

hf and the top surface of this ply has address hf-1 Hence the thickness of the

fth ply is given by

h ( f ) = hf - hf-1

For the 6 ply laminate shown in Fig 3.21, the thickness of ply 5 is given by

h(5) = hs -h4 = 3 - 1 = 2 mm

Fig 3.21 Six ply laminate

The thickness of ply 1 is given by

h(1) = hl - h~ = (-3) - (-6) = 3 mm

Mechanical Behaviour of Composites 209

This is called the Extensional StifSness Matrix and the similarity with that

derived earlier for the single ply should be noted

Also, the Coupling Matrix, [B] is given by

(3.38)

The Coupling Matrix will be zero for a symmetrical laminate

(ii) Moment Equilibrium: As in the case of the forces, the moments may be summed across F plies to give

[MIL = C [ M I f = / [UlfdZ

f = I f = I

210 Mechanical Behaviour of Composites and once again using the expressions from the analysis of a single ply,

[MIL = J ([QI[&lZ + [al[Klz2)dz

hf-1

f = 1

where [ B ] is as defined above and

This equation may be utilised to give elastic properties, strains, curvatures, etc

It is much more general than the approach in the previous section and can accommodate bending as well as plane stresses Its use is illustrated in the following Examples

Example 3.12 For the laminate [0/352/ - 3521, determine the elastic constants in the global directions using the Plate Constitutive Equation When stresses of a = 10 MN/m2, u - -14 MN/m2 and txy = -5 MN/m2

y -

are applied, calculate the stresses and strams in each ply in the local and global directions If a moment of M, = lo00 N m/m is added, determine the new stresses, strains and curvatures in the laminate The plies are each 1 mm thick

El = 125 GN/m2, E2 = 7.8 GN/m2, G12 = 4.4 GN/m2, u12 = 0.34 Solution The locations of each ply are illustrated in Fig 3.22

Using the definitions given above, and the values for each ply, we may determine the matrices A, B and D from

10

A = C i Z f ( h f -hf-l),

f = 1

Mechanical Behaviour of Composites 21 1

Fig 3.22 Ten ply laminate

Then

EX N X

[;;I = a I [;; = a [3 * h

where h = full laminate thickness = 10 mm, and a = A-’ since [B] = 0

This matrix equation gives the global strains as

212 Mechanical Behaviour of Composites when f = 3 , f = 4 , f = 7 and f = 8

a, = -10.8 MNIm’, a, = -19.2 MNlm2, txy = -11.8 MNlm2 when f = 5 a n d 6

a, = 62.6 MNlm2, a,, = -9 MNlm2, txy = -0.9 MNIm’

Note that in the original question, the applied force per unit width in the x-direction was 100 Nlmm (ie a, (10)) As each ply is 1 mm thick, then the above stresses are also equal to the forces per unit width for each ply If we add the above values for all 10 plies, then it will be seen that the answer is

100 Nlmm as it should be for equilibrium Similarly, if we add N, and N,, for each ply, these come to -140 Nlmm and -50 Nlmm which also agree with the applied forces in these directions

In the local (1 -2) directions we can obtain the stresses and strains by using

the transformation matrices Hence, for the tf’th ply

Mechanical Behaviour of Composites 213 where d = D-' since [B] = 0

So that

a, = -47 MN/m2, uy = 5.7 MN/m2, rxy = 15.4 MN/m2 The local stresses and strains are then obtained from the stress and strain transformations

u2 = 2.8 MN/m2,

u1 = -44.1 MN/m2, ti2 = -19.5 MN/m2

= -3.6 E2 = 4.7 x y 1 2 = -4.44 x 1 0 - ~ For the next interface, z = -4 mm, the new values of E ~ , and yxy can be calculated and hence the stresses in the global and local co-ordinates f = 1 and f = 2 need to be analysed for this interface but there will be continuity across the interface because the orientation of the plies is the same in both cases However, at z = -3 mm there will be a discontinuity of stresses in the global direction and discontinuity of stresses and strains in the local directions due to the difference in fibre orientation in plies 2 and 3

The overall distribution of stresses and strains in the local and global direc- tions is shown in Fig 3.23 If both the normal stress and the bending are applied together then it is necessary to add the effects of each separate condition That

is, direct superposition can be used to determine the overall stresses

214 Mechanical Behaviour of Composites

Global strains Global stresses

Fig 3.23 Stresses and strains, Example 3.12: (a) global; (b) local

Note, to assist the reader the values of the terms in the matrices are

2.24 x lo6 1.84 x lo6 -9.042 x lo5 Nmm,

5.25 x lo6 2.24 x lo6 -1.75 x lo6

Mechanical Behaviour of Composites 215

the matrix manipulation - although this is not difficult, it is quite time- consuming if it has to be done manually

The following Example compares the behaviour of a single ply and a laminate

Example 3.13 A single ply of carbon fibre/epoxy composite has the following properties:

El = 175 GN/m2, E2 = 12 GN/m2, G12 = 5 GN/m2, u12 = 0.3

Plot the variations of E,, E,, G,, and uxy for values of 8 in the range 0 to 90" for (i) a single ply 0.4 mm thick and (ii) a laminate with the stacking sequence [4=8], This laminate has four plies, each 0.1 mm thick Discuss the meaning

uxy = - - 0.495 and a12

a22

u,, = - = 0.205

Fig 3.24 shows the variation of these elastic constants for all values of 8 between 0 and 90"