Effect of Orientation on Stiffness of Aligned Fibre Composites

In this section we are going to look at the effect of loading a composite in a direction that is neither parallel nor perpendicular to the fibres. This section is not easy and will make extensive use of the tensor representation of stress and strain in both 2- and 3-dimensions as well as the use of matrix algebra. I'll start by defining the 3 special directions in aligned fibre composites.

The three mutually perpendicular orientations in space are arranged such that the "1" direction is paralle to the axis of the reinforcing fibres, the "2" direction is perpedicular to the fibres such that the plane defined by the "1" and "2" directions is the plane of the laminate or ply. Most structures manufactured from fibre re-inforced composites are thin sheets made up from several layers of fibre fabric layed up on top of each other. The "3" direction is perpendicular to the plane of the laminate or ply.


The stresses acting in 3-dimensions are shown in the figure above - note the designation of the indices defining the tensor form of each stress, the first index represents the direction of the perpendicular to the plane to which the stress is applied and the second index, the direction in which the stress acts. In laminates, it is normal to assume that the laminate is thin with respect to its size in the "1" and "2" directions and, as such, a state of plane stress exists in which there are no stresses acting perpendicular to the plane of the laminate, i.e.,


The stiffness matrix, Q, for plane stress is given by the matrix shown below, where is Poisson's ratio representing a strain in the "2" direction resulting from a stress applied in the "1" direction, i.e. ; similarly . Since stress is the product of stiffness and strain, then in tensor format, the stress is given by



where



However, we must remember that the composite is not isotropic, rather it is orthotropic and thus E11 and E22 are not the same, neither are the two poisson's ratios. We can also express strain as the product of compliance, S, which is the inverse of stiffness, Q. The compliance matrix S is much simpler than the stiffness matrix Q. Both tensors are symmetric about the diagonal.




In the previous lectures, we developed equations that described the principal elastic constants of the composite in terms of the volume fraction of fibres, namely the elastic modulus parallel (E11)and elastic modulus transverse (E22) to the fibres as well as the shear modulus and Poisson's ratio. Thus we can populate both the compliance, S, and stiffness, Q, tensors of the composite, defining both in terms of the volume fraction of fibres. Since the compliance matrix is algebraically the simlest to write down, this is the only tensor we shall define explicitly. The stiffness tensor is denoted simply as the inverse of compliance.

Next we must define the rotation from the more general "x-y" co-ordinate system that is aligned with the direction of loading, to the special "1-2" co-ordinate system that is aligned with the fibres, being the angle between the two.

In order to determine the elastic properties of the composite at any angle to the fibres we shall first define two transformation matrices. The matrix R, defines the conversion from tensor strain to engineering strain. You should recall the definition of tensor shear strain as being half the engineering shear strain. The matrix T defines the rotation of a second order tensor through an arbitrary angle .

The matrix T simply expresses mathematically the Mohr's circle of stress construction in two dimensions.

The stress tensor in the "1-2" co-ordinate system are obtained by rotating the stress tensor in the "x-y" system by T. Note the order of multiplication of the two matrices , they are not commutive, i.e. a.b is not the same, nor equal to b.a!

hence

Since we know the stiffness tensor in the "1-2" orientation we can substitute the product of the stiffness and strain tensors for the stress tensor, i.e.,

hence

Question.

What is the difference between engineering and tensor strain?

Answer.

An object subjected to a shear stress as shown in the uppermost diagram will distort in shear and the engineering shear strain is simply defined as tan(g). When the strains are small, as they typically are in elastic solids, tan(g)~g.

In a free body, the application of a shear stress against just one set of parallel surfaces results in a net torque on the body and for equilibrium, i.e. no rotation, an equal and opposite shear stress must exist. The two shear stresses result in equal shear in two directions. In order for the body to exhibit the same overall engineering shear strain, the actual shear strain on each of the two sides must be g/2!

So tensor shear strains have half the magnitude of the equivalent engineering shear strains. The T matrix can only be used to rotate a tensor hence the need to convert engineering shear strain to its tensor equivalent before rotation and back again after rotation.

The engineering strains in the "1-2" co-ordinate system must first be converted to tensor strains by

hence


The tensor strain in the "1-2" co-ordinate system is obtained by the same rotation as stress i.e.,

and thus


Finally, the tensor strain is converted back to engineering strain

hence


Since stress is just stiffness multiplied by strain, the stiffness of the composite at any angle and volume fraction f is given by



Return to review elastic properties of aligned fibre composites.
In the next section i'll show you how to use the tensor description of stiffness in MathCad to plot out the variation of the various elastic constants with orientation at any volume fraction of fibres.