Mathcad Exercise 7. Calculation of ABD laminate stiffness tensor & elastic modulus

You should download the MATHCAD document that contains this design exercise and open it using version 11 or higher of MATHCAD in order to follow along. 

There are some exercises for you to try at the end of this section.

In this design tool, we first define the mechanical properties of the fibre and matrix materials, then the properties of the elastic and strength properties of the aligned fibre composite, parallel and perpendicular to the fibre axis, just as we did in the previous exercise. Secondly, we define the stiffness and compliance tensors for the composite as a function of volume fraction, and then the full stiffness and compliance tensors of the composite as a function of fibre volume fraction and orientation. At this point the design tool is exactly the same as that developed in the previous section.

In order to make the design tool as generic as possible we shall define two vectors, one containing the thickness of each ply in the laminate and the other the angle of the fibres in the ply. We shall assume that the fibre fraction is the same in each ply, though it need not be! To change the lay up of the laminate just replace the two vectors with ones of your choosing, making sure that the number of elements in each are the same.

In the above example we have a [0/60/120]s laminate, the "s" indicating that the laminate is symmetric about the centreline. In order to calculate the A,B and D matrices we need to establish the position of each interface between the plies relative to the centreline.

The matrices are written in terms of the volume fraction, f, and an arbitrary rotation that will enable the design tool to calculate each of the 3 matrices for any laminate orientation.

In order to determine stresses and strains for any imposed loads/moments/deflections, we will need to combine the 3x3 matrices into a larger 6x6 matrix "AB/BD". Here we encounter a problem because MathCAD requires that all terms within a matrix have the same units, the individual A,B and D matrices do not! A has units of stress * distance, B has units of stress * distance2, while D has units of stress * distance3. In order to combine the 3 matrices we shall divide B by metres (m) and use the AUGMENT() function in MathCAD to create a 6x3 AB matrix. We can do the same to combine B and D, remembering to first divide D by metres, to create the BD matrix. To create the full 6x6 ABD matrix we can use the STACK() function, but again we need to divide the BD matrix (whose units are stress * distance2) by metres to get the units the same as the AB matrix. The ABD matrix allows us to determine the force resultants and moment resultants given a set of imposed strains and curvatures. To calculate the curvatures and strains that result from a set of imposed forces and bending moments we will need the inverse of ABD, which by convention is deemed F. Note that both ABD and F are functions of the volume fraction of fibres and an arbitrary rotation.

We have already demonstrated that the tensile modulus can be obtained from the A matrix - below we can see that the [0/+60/-60]s laminate has a constant tensile modulus regardless of the orientation of the applied stress relative to the laminate - thus this symmetric laminate is isotropic in the x-y plane.

Question.

How can the stiffness of a random fibre composite be determined? This is a composite in which the volume fraction of fibres at any orientation is a constant.

Answer.

There are two possible ways to solve this problem. The first is to make up a laminate with any number of plies, n, each of which is offset from teh next by the same angle. Thus for two plies the angles would be 0 and 90, for 3 plies, 0, 60 and 120, for 4 plies 0,45,90 and 135, etc . etc.. The program block below shows how to generate the orientations automatically for any value of n and we'll assume all the layers have unit thickness - the A matrix is then the sum of the Q matrices for each orientation. The tensile modulus is obtained from A as before.

It's pretty obvious that after 3 equally oriented layers, the stiffness remains constant and independent of the number of orientations selected.

The second method just integrates each term in the stiffness matrix of a single layer over all orientations from 0 to 180o and then divides by pi to obtain the average. Each average is then loaded into its correct position in a new stiffness tensor, the tensor inverted to obtain compliance and the tensile modulus extracted as before...

The answers, of course, are the same!