1. OPTIMIZATION OF CURVILINEAR FIBER PATHS FOR AN ORTHOTROPIC LAMINA
Pınar Acar
Avinkrishnan A. Vijayachandran
Veera Sundararaghavan
Anthony M. Waas
Good morning everybody, I am Pinar Acar. I am a PhD student and graduate research assistant at University of Michigan, Aerospace Engineering. Today I am gonna talk about our work which is interested in optimization of curvilinear fiber paths for an orthotropic lamina
15th Pan-American Congress of Applied Mechanics
21 May 2015, Urbana-Champaign, IL, USA

2. Outline Motivation Plate Problem Definition Algorithm and Methodology
Solution of Special Cases
Isotropic Plate
Orthotropic Unidirectional Lamina
Optimization of Fiber Path
Concluding Remarks and Future Work
I would like to start with the outline of the study. First, I will explain the motivation for our work. Then I will explain the plate problem that we have. After then, I will give information about the algorithm and the methodology that we used to solve the problem, and I will show you some verification results for some special cases which are isotropic lamina and orthotropic unidirectional lamina. After the solution of the plate lamina problem, I will give information about the optimization problem that we set up to find the fiber path. This section is the core section for our work. In this section, I will be talking about our objective, optimization variables and optimization method as well as the optimization results. And I will finish my talk with some concluding remarks and possible extensions for a future work

3. 1. Motivation Problem Definition
Tension Problem – An infinite plate with a circular cutout at the center is under the effect of uniform tension
Computation
Stress Intensity Factor (SIF) – It has very large values around the cutout and can cause matrix fracture
Solution
Optimization – Aims to find the best fiber path to reduce SIF and avoid matrix fracture
So, the problem of interest in this work is a single layer plate with a circular cutout at the center. We are basically solving a tension problem as the plate is under the effect of a uniform axial force acting only on the horizontal axis. I will show the figure that visualizes the problem very soon. Our key idea is to get rid of high stress concentrations in order to get rid of the matrix fracture. We can represent the stress concentrations with Stress Intensity Factor and it is very likely to see very large SIF values around the cutout in this problem. So, in order to achieve our objective, we tried to find an optimum fiber path distribution which can help us reduce the stress intensity factor and then avoid the matrix fracture

4. 2. Plate Problem Definition
One quarter of the plate is modeled
Symmetric boundary conditions are used to model quarter plate
Assumptions
Plane stress
Small strains
Cutout dimension (radius)
is much smaller than plate
dimensions (height and
width)
We can see our initial problem definition in this figure. So we have some applied axial forces in horizontal direction. For this problem, it is possible to solve only one quarter of the plate and get the stress concentrations for the quarter part. So, we preferred to do that since it is computationally cheaper. But to model the quarter plate, we had to use the symmetric boundary conditions that mean as if we have some roller at the locations we defined the symmetry conditions. This condition means that the axial displacement and the shear traction is zero. The other basic assumptions are plane stress assumption for the lamina and the second one is the cutout dimension, radius, is much smaller than the plate dimensions since this is an infinite plate problem

5. 3. Algorithm and Methodology
Fiber angle is the same in radial direction
Ti : Initial fiber angle
Tf : Final fiber angle
Fiber angle is changing
in tangential direction
Purpose: Ensure continuity
for manufacturability Ti
This figure is explaining the way we followed so as to mesh our problem. So, the basic assumption is that the fiber angle is only changing in tangential direction and it is constant along radial direction. The fiber angle is also constant along one finite element. So, we are starting with an initial fiber angle Ti which we defined in x is equal to zero axis and then we are changing the fiber angle dtheta amount for the next finite element. So, Ti plus dtheta is the initial fiber angle of the second finite element. We keep increasing the fiber angle with dtheta when we receive a new finite element. So, the only difference in fiber angles of two successive finite element in tangential direction is dtheta. And finally we are ending up with a final fiber angle of Tf with respect to y is equal to zero axis. We defined the variation of the fiber angles this way in order to provide a smooth distribution throughout the quarter plate. Because the main idea is to find an optimum fiber path that is also manufacturable. Tf

6. Methodology Find fiber angle of each finite element
where m is the current and n is the maximum
element number in tangential direction
Find optimum fiber angle path to minimize the maximum energy fraction
Find material matrix of each finite element
So, how are we solving this problem. First, we need to find the fiber angles for each finite element. Basically we have two variables, the initial fiber angle Ti, the final fiber angle Tf and the increment or decrement of the angle at each finite element, dtheta. I said both increment and decrement for dtheta because we did not constraint dtheta as a positive or negative value, it can be both. Either we have decreasing or increasing fiber angles, they will be still continuous with this definition. One other point is that the fiber angle is the same along one finite element and that means we can directly find the material matrix for that particular element using that fiber angle. Since we are solving the problem with finite element approach we need to apply the traditional finite element procedures to solve for the stresses. Since this problem is a tension problem, we did not have to use the plate elements for solution, instead we used quadrilateral finite elements to solve the problem. So, our approach to find the stress intensity factor requires computing the maximum energy fraction value. So, we computed the energy faction values for each finite element and the maximum of them is an indicator for the maximum stress location. So, in order to decrease the high stress concentration around that location, we decided to minimize the maximum of the energy fraction values. We performed a global optimization to find an optimum fiber path that minimizes the maximum energy fraction value.
Solve for stress distribution
Compute maximum energy fraction

7. Stress-Strain Formulation
“Stress-Strain Relation” for each finite element
Q material matrix defining the relation between stress and strain in global coordinate frame should be transformed to material (fiber) frame.
So, after finding the fiber angle for the particular finite element, we can define the fundamental stress-strain relation with that fiber angle. The regular Q matrix is defined in global frame so in order to define the stress-strain relation in material frame or in fiber frame we need to transform the global Q matrix to material axis using the particular finite element. As a note, for this work, we used the lamina properties for the given material.
Material matrix of finite element formulation
* IM7/ lamina properties were used to find Q matrix in this work.

8. Stress-Strain Formulation
“Strain Energy Density” for each finite element
Total Strain Energy Density
“Energy Fraction” for each finite element
(where m=1,2,6 corresponds to xx, yy, xy directions)
(nel is the total number of elements)
After setting up and solving the finite element problem for stresses, we compute the strain energy density value for each finite element by using the stress and strain values for that element. Our aim was to get the energy fraction value, so we computed the energy fraction for each finite element as the ratio of strain energy density of that particular element and the total strain energy density.

9. 4. Solution of Special Cases
Isotropic Plate
The finite element model
delivers matching results
since SIF is obtained
as 2.47
(Theoretical value: SIF=3)
Material Properties:
E = 70 GPa
ν = 0.3
Now, I am gonna show you some results that we found with our finite element code. The first case is including an isotropic lamina. So, we are applying an axial stress of sigma in the horizontal direction. And the maximum stresses are seen around the cutout as expected. And the value of stress intensity factor at x is zero and y is one point was found as In literature, the theoretical value was given as 3 sigma. So, we can say we are close.

10. Solution of Special Cases
Orthotropic Unidirectional Lamina (IM7/8551-7)
Fibers are aligned at an
angle 0o.
The FE model delivers
a very high SIF of 11.2
Material Properties:
The second application is including an orthotropic unidirectional lamina which has fibers aligned at an angle of zero degree. So, we computed the maximum stress intensity factor of this problem as According to general literature, for this problem, the maximum stress intensity factor should be in the range from 8 to 12. So, it shows that our solution is in the range.

11. 5. Optimization of Fiber Path
Objective – Minimize Stress Concentrations
Maximum stress concentrations are seen around cutout
Energy density is a representative value for stress concentrations
Lower energy density values are corresponding to lower stress concentrations
Representative Objective – Minimize Ef,max
Minimum “maximum energy density” relates to minimum SIF (small strain assumption)
Ef is always a positive value by definition. Thus it is easier to minimize Ef than σx which can be negative too.
Since we computed a very high stress intensity factor for the orthotropic unidirectional lamina, now we will be trying to decrease it. So, our fundamental objective in this section is to minimize the stress concentrations. We can use the energy density or fraction value as a representative for stress concentrations. This means that lower energy density values indicate the lower stress concentrations. Since it is a representative quantity, we selected our representative objective as the minimization of the maximum energy fraction value. It is better to use only the maximum value since if we can minimize the maximum value then it means the other stress concentrations are already lower than that value. And also, since we are using only one parameter as the objective, we have only one objective functions and we will notbe dealing with a multi-objective optimization problem which is more costly.

12. Optimization Workflow
Optimization Variables
Set all input variables: Ti and Tf
Define lower and upper limits
Sampling
Optimization Algorithm
Determines initial points
Design space should
be covered
Genetic Algorithm
for global optimum
Constraints
So, what are we doing in optimization? First, we have to define the optimization variables as well as their upper and lower limits. Since in this work we performed a global optimization, we need sampling as the second step. What sampling does is that it determines the initial design points. So, a good sampling algorithm should give information about the whole design space, so it should cover the design space very well. Then, we should set the optimization method using the initial design points. In his work, we used a genetic algorithm. Then, as the next step, the constraints should be specified to find the feasible solutions. Then, in the final step, we can select the optimum solution among all feasible solutions.
Optimum Solution
Check for feasible designs

13. Optimization Method
Optimization Variables
Latin Hypercube Sampling (LHS) was used as sampling method. LHS covers all design space equally
Sampling
In this problem, we have basically three variables: Ti, Tf and dtheta. But we can define dtheta using Ti and Tf, that means we have 2 independent optimization variables. We set the given limits for these angles. We used one hundred seventy nine degrees for the upper limit since one hundred eighty degrees would be the same with zero degree. For the second step, we used Latin Hypercube Sampling as the sampling algorithm to find the initial design points to initiate the optimization. The advantage of using Latin Hypercube Sampling is that it generates random design points in a kind of controlled way and so it covers all design space equally. We used one of the famous genetic optimization algorithm, which is known as Non-Dominated Sorting Genetic Algorithm as the optimization method. This optimization method is powerful since it guarantees the global optimum solution even in multi-objective problems. So, in case of a future extension for this work, for example if we solve this problem for tension and buckling at the same time, it means we will have multiple objective functions but we will still be able to use this algorithm so it was efficient in a way to implement this algorithm to the solution of this problem.
Non-Dominated Sorting Genetic Algorithm (NSGA-II) is one of the fastest genetic algorithms, it guarantees the global optimum even in multi-objective problems
Optimization Algorithm

14. Optimization Results Optimization Information Number of DOE: 10
Optimization Algorithm: NSGA-II
Number of Designs: 100 total feasible solutions
Optimization Results (for IM7/8551-7)
Ti = 55.94o
Tf = 33.56o
Ef,max=0.0033
Maximum SIF around cutout is 3.94
Here, the general information about the optimization is shown. At the end of the optimization, we ended up with 100 total feasible designs. Since we didn’t specify any design constraints particularly, all of the final optimization results were also the feasible designs. Among these designs, we selected the design that minimized the maximum energy fraction value. The values of the optimization variables in this design are given here. We were able to minimize the maximum energy fraction value with a corresponding maximum stress intensity factor value of Compared to the stress intensity factor value that we computed for the orthotropic unidirectional lamina, we were able to get a much lower stress intensity factor value by optimizing the fiber path. So, if I summarize the results. If the fibers are aligned at zero degree, which means we have an orthotropic unidirectional lamina, we have a maximum stress intensity factor of But if we have a fiber distribution that I showed the definition before with this initial fiber angle and this final fiber angle, then we can reduce the 11.2 value to 3.94 which is a very significant reduction.

15. Optimum Result
Optimum design reduces the stress concentrations significantly
Effect of cutout is nullified
by tailoring the curved
fiber path
SIF of optimum lamina,
3.94, is almost one- third
compared to the SIF of
unidirectional orthotropic
lamina, 11.2
(for IM7/8551-7)
So, this figure is showing the stress concentrations for our optimum design. Again, we see the higher stress concentrations around the cutout but still they are much lower with the optimization.

16. Optimum Fiber Path Fiber path of optimum design (for IM7/8551-7)
Fiber angles are plotted based
on the definition of Ti and Tf
by using discrete dθ to smooth
out the curves.
And for our quarter plate modeling, this plot shows the fiber path of our optimum design. We are starting with Ti angle with respect to x is equal to zero axis, and we are ending up with Tf angle with respect to y is equal to zero axis. And each finite element holds the same fiber angle.

17. 6. Concluding Remarks and Future Work
Present Work
The present work discusses solution and optimization of an one-layer plate tension problem using quadrilateral elements in FEM, with imposed constraints through meshing and fiber angle
Future Work
The present work can be extended to solve and optimize tension and/or buckling problems of composite laminates.
So, in this work, we solved a one-layer orthotropic plate tension problem with finite element approach using the constraints through meshing and fiber angle and then optimized the same problem to find the best fiber path. The possible extensions for this work could be solve the same problem for also buckling and again optimize it with multiple objectives including both tension and buckling. But for buckling we need to use plate finite elements this time. This work can also be extended in a way to include multiple layers in the solution.

18. THANK YOU! acarp@umich.edu avinav@umich.edu
So, these are all I wanted to say about our work today. Thank you very much for your attention.