Research Papers

Using Topology Optimization to Numerically Improve Barriers to Reverse Engineering

[+] Author and Article Information
Devin D. LeBaron

Graduate Research Assistant
e-mail: dblebaron@gmail.com

Christopher A. Mattson

Associate Professor
e-mail: mattson@byu.edu
Department of Mechanical Engineering,
Brigham Young University,
Provo, UT 84602

1Corresponding author.

Contributed by the Design Automation Committee of ASME for publication in the JOURNAL OF MECHANICAL DESIGN. Manuscript received February 26, 2013; final manuscript received October 19, 2013; published online December 11, 2013. Assoc. Editor: Shinji Nishiwaki.

J. Mech. Des 136(2), 021007 (Dec 11, 2013) (8 pages) Paper No: MD-13-1100; doi: 10.1115/1.4025962 History: Received February 26, 2013; Revised October 19, 2013

Here explored is a method by which designers can use the tool of topology optimization to numerically improve barriers to reverse engineering. Recently developed metrics, which characterize the time (T) to reverse engineer a product, enable this optimization. A key parameter used in the calculation of T is information content (K). The method presented in this paper pursues traditional topology optimization objectives while simultaneously maximizing K, and thus T, in the resulting topology. New aspects of this paper include algorithms to (1) evaluate K for any topology, (2) increase K for a topology by manipulating macroscale geometry and microscale crystallographic information for each element, and (3) simultaneously maximize K and minimize structural compliance (a traditional topology optimization objective). These algorithms lead designers to desirable topologies with increased barriers to reverse engineering. The authors conclude that barriers to reverse engineering can indeed be increased without sacrificing the desirable structural characteristic of compliance. This has been shown through the example of a novel electrical contact for a consumer electronics product.

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Grahic Jump Location
Fig. 1

Design domain used for all the examples in Sec. 2. Dimensions are 60 units in the horizontal direction and 30 units in the vertical direction. The design domain corresponds to half the MBB-beam.

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Fig. 2

Representation of how method describes voids. (a) Benchmark A. This is the solution to the topology optimization problem without attempting to add information content. This will be used for comparison to other isotropic examples. C¯A = 100 and K¯A = 100. (b) Convex hull representation of voids in Fig. 2(a)

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Fig. 8

Design domain with one element highlighted. This elements microstructure orientation is determined by the value of Θ.

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Fig. 4

Use of voids to increase information content. (a) Design domain with 5 equally spaced voids in an attempt to complicate the geometry (b) Resulting topology using the design domain in Fig. 4(a). C¯A = 125 and K¯A = 234. Where C¯A = (C/CA*100) and K¯A = (K/KA*100). C and K are evaluated from this topology, and CA and KA represent the C and K of the benchmark topology in Fig. 2(a).

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Fig. 3

An example of using different microstructure orientations in a topology. Each unique orientation adds 1 piece of information content. In this example information content (K) has been increased by 7.

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Fig. 7

Resulting topology from using four voids and allowing them to move their location and size to find an optimum solution. C¯A = 134 and K¯A = 192.

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Fig. 6

Flow chart summarizing the algorithm that finds the optimal placement of the voids

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Fig. 5

Result of individually orienting the microstructure at each element. (a) Benchmark B. A topology optimization using a uniformly oriented anisotropic material. C¯B = 100 and K¯B = 100. Examples using an anisotropic material will be compared to this benchmark. (b) A topology optimization result using an anisotropic material that is optimally oriented at each element. C¯B = 78 and K¯B = 115

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Fig. 14

3D Model of the design in case 4

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Fig. 9

Design domain for the novel electrical contact

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Fig. 10

Case 1:(Benchmark C) Uniformly oriented Topology with no forced voids. C¯C = 100 and K¯C = 100.

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Fig. 11

Case 2: Uniformly oriented topology with 4 forced voids that have been optimally located and sized. C¯C = 109.5 and K¯C = 192.

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Fig. 12

Case 3: Optimally oriented topology with no forced voids. C¯C = 71 and K¯C = 132

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Fig. 13

Case 4: Optimally oriented topology with 4 forced voids that have been optimally located and sized. C¯C = 79 and K¯C = 250.




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