Combining materials for composite-material cars

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Combining materials for composite-material cars Ford initiated research at a time when they took a look at making cars from composite materials. Graphite-epoxy is too expensive, glass-epoxy is not stiff enough. Grosset, L., Venkataraman, S., and Haftka, R.T., “Genetic optimization of two-material composite laminates,” Proceedings, 16th ASC Technical Meeting, Blacksburg, VA, September 2001

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Combining materials for composite-material cars. Ford initiated research at a time when they took a look at making cars from composite materials. Graphite-epoxy is too expensive, glass-epoxy is not stiff enough. - PowerPoint PPT Presentation

Transcript of Combining materials for composite-material cars

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Combining materials for composite-material carsFord initiated research at a time when they took a look at making cars from composite materials.Graphite-epoxy is too expensive, glass-epoxy is not stiff enough.Grosset, L., Venkataraman, S., and Haftka, R.T., Genetic optimization of two-material composite laminates, Proceedings, 16th ASC Technical Meeting, Blacksburg, VA, September 2001

With the pressure of having to reduce fuel consumption, automotive companies look from time to time at using composite materials instead of steel. This usually comes to nought, but it is probably right to examine it periodically. This research was initiated in 2000 when Ford was considering this option.Their basic dilemma was that graphite-epoxy, which is ideal for automotive applications from the standpoint of performance is too expensive, and glass-epoxy, which has much lower cost is not stiff or strong enough.So the forcus of this research was to devise a genetic algorithm that would design laminates made from a combination of the two materials.1Multi-material laminateMaterials: one material = 1 ply ( matrix or fiber materials)E.g.: glass-epoxy, graphite-epoxy, Kevlar-epoxyUse two materials in order to combine high efficiency (stiffness) and low costGraphite-epoxy: very stiff but expensive; glass-epoxy: less stiff, less expensiveObjective: use graphite-epoxy only where most efficient, use glass-epoxy for the remaining plies

The idea was to have some plies made from one material and some plies made from another material, and in this case the two materials were glass epoxy and graphite epoxy.

The goal is to use the expensive graphite epoxy only where it does most good and use glass epoxy elsewhere.Multi-criterion optimizationTwo competing objective functions: WEIGHT and COSTDesign variables: number of pliesply orientationsply materialsNo single design minimizes weight and cost simultaneously: A design is Pareto-optimal (non-dominated) if there is no design for which both Weight and Cost are lowerGoal: construct the trade-off curve between weight and cost (set of Pareto-optimal designs, also called Pareto front)

We cast the optimization problem as a bi-objective optimization, with the two objectives being weight and cost. The design variables were the number of plies, the ply orientation and the ply materials. The goal was to construct the Pareto front of non-dominated design, which would provide the tradeoff curve between weight and cost.3Material propertiesGraphite-epoxyLongitudinal modulus, E1: 20.01 106 psiTransverse modulus, E2: 1.30 106 psiShear modulus, G12: 1.03 106 psiPoissons ratio, 12: 0.3Ply thickness, t: 0.005 inDensity, : 5.8 10-2 lb/in3Ultimate shear strain, ult: 1.5 10-2Cost index: $8/lbGlass-epoxyLongitudinal modulus, E1: 6.30 106 psiTransverse modulus, E2: 1.29 106 psiShear modulus, G12: 6.60 105 psiPoissons ratio, 12: 0.27Ply thickness, t: 0.005 inDensity, : 7.2 10-2 lb/in3Ultimate shear strain, ult: 2.5 10-2Cost index: $1/lb

This slides shows most of the relevant material properties of the two materials. It is seen that graphite epoxy is more than three times stiffer in tension, while it does not have much of an advantage in transverse stiffness or shear stiffness. The latter is not too important, because both transverse and shear stiffness for a laminate are obtained by having plies in different directions. For example, the maximum shear stiffness is achieved from a +-45 laminate.The graphite epoxy also has the advantage that it is about 20% lighter than glass epoxy. However, at the time it was 8 times more expensive.4Optimization problem Minimize weight and cost of a 30x36 plateBy changing ply orientations and material misubject to a frequency constraint:

Since Ford would not release the actual constraints, we agreed on design for a stand-in constraint of minimum frequency of 25Hz.

The frequency is found from classical lamination theory and the square of the frequencies is proportional to the D terms in the bending matrix. For fixed ply angle these are proportional to the cube of the thickness.5Constructing the Pareto trade-off curveSimple method: weighting method. A composite function is constructed by combining the 2 objectives:

W: weightC: cost: weighting parameter (01)A succession of optimizations with varying from 0 to 1 is solved. The set of optimum designs builds up the Pareto trade-off curveThis is not a fool-proof approach (Why?), and it is time consuming. Genetic algorithms provide a shortcut.

The bi-objective optimization was performed by repeated optimization of a convex combinations of the two objectives, with the weight assigned to the first one being changed from 0 to 1, in steps of 0.1.

This is not a fool-proof way of getting the front (does not work when it is not convex), and it is also time consuming. Multi-objective genetic algorithms provide a shortcut of obtaining the entire Pareto front in a single optimization.Multi-material Genetic AlgorithmTwo variables for each ply:Fiber orientationMaterialEach laminate is represented by 2 strings:Orientation stringMaterial stringExample:[45/0/30/0/90] is represented by:Orientation:45-0-30-0-90Material: 2-2-1-2-1GA maximizes fitness: Fitness = -F1: graphite-epoxy2: glass-epoxy

The coding had each ply represented by two numbers, ply orientation and material number. The two numbers, though were put into two different strings, or chromosomes.7Simple vibrating plate problemMinimize the weight (W) and cost (C) of a 36x30 rectangular laminated plate19 possible ply angles from 0 to 90 in 5-degree stepConstraints:Balanced laminate (for each + ply, there must be a - ply in the laminate)first natural frequency > 25 HzFrequency calculated using Classical Lamination Theory

The additional information in this slide is that we limit ourselves to 19 possible ply angles, primarily in order to reduce the design space, and we also insist on balanced laminate, so that for every plus angle except for 90-deg, there will be also a ply with the negative angle.8How constraints are enforcedBalance constraint hard coded in the strings: stacks of are usedExample: (45-0-30-25-90) represents [45/0/30/25/90]sFrequency constraint is incorporated into the objective function by a penalty, which is proportional to the constraint violation

The balance constraint was enforced by making each angle in the string that was different from 0 or 90 represent a double ply with plus and minus angles. The frequency constraint was enforced with a penalty parameter when it was violated, including a 10% penalty for any violation at all. When the constraint was satisfied there was a small bonus (0.01g).9Genetic operatorsRoulette wheel selection based on rankTwo-point crossoverMutation and permutationPly deletion and additionOperators apply to each chromosome individually.Best individual passed to the next generation (elitist)

Selection was based on roulette wheel allocating pie sectors based on rank. Two-point crossover, mutation and permutation were used to create children as well as ply deletion and addition. The crossover, mutation, and permutation operators were applied to each chromosome individually. Finally the best individual was passed to the next generation.

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Pareto Trade-off curve

($)(lb)A (16.3,16.3)B (6.9,55.1)Cpoint C64% lighter than A; 17% more expensive53% cheaper than B; 25% heavier

The figure shows many of the designs that were visited on the way to the optimum for each value of the weight multiplier alpha. For alpha=1, where the weight is the only thing that count, we get an all-graphite epoxy design, with a weight of 6.9lb and a cost of $55.1. For alpha=0, when only cost counts we get a design with all glass-epoxy, with weight of 16.3 lb, and cost of $16.3.All the points on the Pareto front are non-dominated, and point C may be a good compromise between the two objectives. It may also be the closest point to the utopia point where both objectives are equal to their best values (details in the next slide)Optimum laminatesCost minimization: [5010/0]s, cost = 16.33, weight = 16.33Weight minimization: [505/0]s, cost = 55.12, weight = 6.89Intermediate design: [502/505]s, cost = 27.82, weight = 10.28Glass-epoxy in the core layersto increase the thicknessGraphite-epoxy as outer pliesfor a high frequencyMidplaneIntermediate optimum laminates:sandwich-type laminates

In retrospect the solution found by the genetic algorithm is simple enough that we probably could have found it without a genetic algorithm. The best angle for this plate aspect ratio and material properties appears to be near 50 degrees. The only place where another angle appears is for the cost minimization, because using one zero-degree ply allows us a single ply instead of two.As expected, when we use both materials, the graphite epoxy is on the outside, because the problem is controlled by bending properties (the vibration is normal to the plane of the laminate).Finally, the compromise laminate is close to the geometric average of the extreme laminates in terms of weight and cost.12Problems two-material laminateCheck the reasonableness of the weight ratio of the two extreme laminates (16.3/6.9) with simple calculation. SolutionIllustrate the difference between crossover of two laminates when (a) the materials and angles are segregated into two chromosomes and (b) there is one string with the two numbers for each ply being together. Solution.Why does the figure of the Pareto front not show values of alpha between 0 and 0.7? Solution13