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Modeling and Design of Complex Composite Structural Parts. Optimization M. Delfour J. Deteix M. Fortin G. Gendron. Introduction. ADS Composite supplies parts to Bombardier and Prévost (for buses, trains, recreational vehicles, …).
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Modeling and Design of Complex Composite Structural Parts Optimization M. Delfour J. Deteix M. Fortin G. Gendron
Introduction ADS Composite supplies parts to Bombardier and Prévost (for buses, trains, recreational vehicles, …). These companies need a tool to improve the design and to optimize the manufacturing process. Moreover those parts have almost no structural role. Making them participate in the structural behaviour (fuel efficiency) leads to design optimization. The GIREF (with MEF++) can provide a solution to those companies.
Overview of the project Damage and degradation analysis: development of a model of degradation for random short fibers composites. Structural analysis for static loading: automatic F.E. grid adaptation and development of a 3d shell element. Design optimization: development of an optimization process for a better structural behaviour. To this we add a graphical interface capable of creating meshes by extrusion and importing data of various format (CATIA, I-DEAS, etc).
Composite Parts • The resin and fibers are projected simultaneously on a mould by robotic projection. The speed and pattern of displacements of the robot allows the manufacturing of complex parts with a variable thickness. • Reinforcement can be added manually by use of unidirectional plies of fibers and/or stiffeners. • The parts are thin in large region (shell) but 2d model is not acceptable in some important places: holes, metallic inserts, reinforcing struts.
3D Shell Finite Element • WHY • Model is easier to formulate and the optimization problem is easier to construct. • Give a complete description of the stress. • If we choose a 2D model we will need to mix it with a 3D model in some regions. • HOW • Based on theoretical results (Delfour, Zolésio) consistent with classical model as the thickness goes to zero. • Prismatic element of at least degree 2 in ‘the thickness’ • Locking is prevented by reduced integration and stabilization to avoid singularities
Design problem The design problem is to optimize shells by acting upon their thickness, the presence and the orientationof patches of unidirectional fibers and the presence of stiffeners. We are working on various parts (seat, side panel,…) with various functionality so the mechanical requirements cannot be fixed for full reusability. In a first step we will work uniquely on the thickness of the shell. Leaving the process of adding stiffeners or patches of fibers as it is.
Ideal problem The quantities of interest are generally as follow: • weight / volume / thickness (geometric) • overall / punctual measure of displacement (stiffness) • overall / punctual measure of stress (Von Mises, …) • Ideally: thickness
Proposed Approach End T Step 1 Step 2 T Overkill lighter part lk=l0 Mec. Cond. Min J Geo. Cond. lk += eDk F Bad Design heavier part lk -= eDk k++ • Construction of a cost functional depending on weight and stiffness: • Design decision as a 2 steps loop: • Step1 Fix the multiplier and calculate a minimizer • Step2 Verify the others mechanical properties. J =l*Weight +Compliance
Remarks This process will give a sub-optimal solution relatively to the ‘ideal’ problem. However • The optimization problem is simple since it contains only geometrical constraints. • The choice of the compliance in the cost gives • a gradient which is easy (i.e. fast) to calculate, • a solution to Step 1 is of interest (it is the stiffest part for its weight). • In Step2 we can include any conceivable mechanical conditions. We can even add/eliminate conditions without supplementary calculations.
Explicit Formulation weight - stiffness Generally Step 1 is Denoting • hm, hMmin. / max. thickness • udisplacements • fv,fsloads (vol./surf.) • r density • h thickness • Wh, Gh part/surface
Numerical Aspects • To discretize the optimization problem: • use the F.E. method to evaluate u and J. • fix anapproximation of h if needed. • use the speed method to obtain the gradient of J Approximation of h: NURBS, linear, quadratic, piecewise constant on the F.E. grid or on a coarser grid. The corresponding discrete optimization problem is the minimization of a continuous function. We chose to use classical optimization methods. Several technique have been implemented. Generally we use a SQP method with feasible points (Herskovits).
Gradient of J The expression for the gradient depend of the nature of h. To have a reusable code: • Construct the derivatives of J with respect to the nodes of the F.E. grid. • Suppose that the relation between the nodes and the thickness is known (so the chain rule canbe defined) MEF++ user where h = (h1,…,hm) and Si are N the nodes of the F.E. grid.
Derivatives of J • Apply the speed method to obtain a derivative. • Choose specific velocities to obtain the desired derivatives. Let ji be the shape function (linear): ji(Sj) = dijand
Final Remarks • MEF++ gave us: • tensorial calculus (simplify the opt. and the F.E.) • manipulation of algebraic expression (à la MAPLE) • elements related to shape opt. are embedded in the library. • At the moment we work on • final validation and automatic stabilization of the 3d shell • validation and first application on ‘real life’ problem • What is next • thickness/orientation, thickness/fibers/orientation problems • Possibly topological (should be easy in MEF++) • Economical/manufacturing considerations in the problem (variables,cost,conditions)
Numerical Results • We are presently working on the preparation of practical problems. For our first ‘real life’ problem we chose the design of a seat (specs comes from the New York metro). • In that case real life is: • a mesh of one layer of 6648 elements (6820 nodes) • adapted to maximize accuracy of the F.E. solution. • the design must satisfy 3 sets of loads and mechanical • and geometrical conditions. • We will present only simple ones: plates,hemispheric shells, etc. • The density being constant weight is in fact equivalent to volume.
Plate 1 • Square 100 cm x 100 cm • Thickness 3 – 7 mm • Mesh: one layer of elements • Design variables: nodes of the top surface. • Constraints: • 0.3 £ zi£ 0.7 . . y . . x z bdy conditions: clamped at the 4 ‘corners lines’ uniform pressure on part of the top surface E = 7.e+9 n = 0.3 fs= (0,0,-1000)
Various Designs P1 l: 0.25 Vol.: 6415 Wmax: -0.0236 Compl: 2351 l: 0.30 Vol.: 6007 Wmax: -0.0244 Compl: 2437 l: 0.40 Vol.: 5512 Wmax: -0.0259 Compl: 2593 l: 1.0 Vol.: 4582 Wmax: -0.0310 Compl: 3103
Plate 2 • Square 100 cm x 100 cm • Thickness 3 – 7 mm • Mesh: two layers of elements • Initial Volume: 5000 cm3 • Design variables: nodes of the top surface. • Constraints: • 0.1 £ zi£ 0.5 z = 0 y x bdy conditions: clamped on 4 surfaces at z= 0 uniform pressure on the top surface E = 7.e+9 n = 0.3 fs= (0,0,-1)
Various Designs P2 vol.: 5222 vol.: 5019 vol.: 4667 vol.: 4370
Shell 1 z • ‘Square’ 100 cm x 100 cm • Thickness 3 – 7 mm • Mesh: one layer of elements • Design variables: nodes of the top surface. . . . y . x • Constraints: • R + 0.3 £√(xi2 + yi2 + zi2 ) £ R + 0.7 bdy conditions: clamped at the 4 corner lines uniform pressure on the top E = 7.e+9 n = 0.3 fs = (0,0,-1)
Various Designs S1 l: 0.008 Vol.: 7585 Wmax: -0.0030 Compl: 184 l: 0.035 Vol.: 6453 Wmax: -0.0033 Compl: 199 l: 0.07 Vol.: 5831 Wmax: -0.0041 Compl: 229 l: 0.14 Vol.: 4993 Wmax: -0.0061 Compl: 307
Shell 2 z • Radius 10 m • Thickness 6 – 10 cm • Design variables: nodes of the interior surface. . . . . y x Constraints: 9.90 £√(xi2 + yi2 + zi2 ) £ 9.94 bdy conditions: clamped at the 4 ‘corners’ non-uniform / radial pressure E = 2.e+6 n = 0. fs = (0,0, –10/(x2 + y2 + 0.4))
Various Designs S2 l: 0.035 Wmax: -0.1094 l: 0.045 Wmax: -0.1098 l: 0.05 Wmax: -0.1102 l: 0.20 Wmax: -0.1165
Cylinder z • Height 100 cm • Diameter 16 – 26 cm • Design variables: nodes of the lateral surface. • Constraints: • 8 £√(xi2 + yi2 )£ 13 y We chosel= 125 (representative of all l). x bdy conditions: clamped on the bottom at z= 0 shearing forces on the top end E = 2.e+6 n = 0. fs = (250,0,0)