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MIT 3 032 - Problem Set 4

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Metal wires (fibers) θ = 40° direction xMPa 3.032 Problem Set 4 Fall 2007 Due: Start of Lecture, 10.19.07 1. A microelectronic sensor is to be made of conductive wires deposited on a thin Si wafer. During design considerations, it was decided that only metals having cubic symmetry were to be used as fibers (metallic wires) in this composite sensor. Another design criterion for the sensor is that the fibers of the sensor do not undergo any elastic (true) normal or shear strains > 0.1% and 0.01%, respectively, when subjected to a given stress state. (We will assume plane stress since the device is very thin.) y Fiber ⎡80 45 ⎤σij = ⎣⎢45 130⎦⎥ Si wafer Figure 1: Orientation of composite sensor with respect to original loading conditions (a) The fibers in the sensor are subjected to an initial plane stress state, given below, with respect to the orientation seen in Fig. 1. Determine the stress state on the fibers by resolving the original stress state onto an axis-set aligned with the fibers (instead of aligned with the applied stress state σij as shown in Fig. 1). (b) Given the elastic compliance constants for various cubic crystals (see Hosford Table 2.2. below), calculate the complete strain tensor εij for the stress state you determined in part (a) for all of the metallic and organic materials from which we might make these fibers (writing a small program would be ideal here!). Identify the metals that satisfy the design criterion that the fibers do not undergo elastic normal strains (εii) equal or greater than 0.1% or true shear strains (εij) equal to or greater than 0.01%. (Note: The units of sij are given in (TPa)-1 = 1x10-12 Pa-1).2 MaterialS11S12S44CrFeMoNbTaWAgAlCuNiPbPdPtCGeSiMgOMnOLiFKClNaClZnSInPGaAs3.10 -0.46 10.107.56 -2.78 8.592.90 -0.816 8.216.50 -2.23 35.446.89 -2.57 12.112.45 -0.69 06.2222.46 -9.48 22.0315.82 -5.73 35.3415.25 -6.39 13.237.75 -2.98 8.0594.57 -43.56 67.1113.63 -5.95 13.947.34 -3.08 13.071.10 -1.51 1.929.80 -2.68 15.007.67 -2.14 12.544.05 -0.94 6.607.19 -2.52 12.6611.65 -3.43 15.7126.00 -2.85 158.622.80 -4.66 78.6218.77 -7.24 21.6516.48 -5.94 21.7411.72 -3.65 16.82Elastic compliances (TPa)-1 for various cubic crystalsMechanical Behavior of MaterialsTable by MIT OpenCourseWare.2. Atomic interactions can be modeled using a variety of potential energy approximations. One very common potential form is the Lennard-Jones 6:12 potential: ⎡σ12 σ6 ⎤⎛⎞ ⎛⎞ Ur() = 4ε−⎢⎜⎟ ⎜⎟⎥ ⎢⎣r r ⎥⎦⎝⎠ ⎝⎠ where σ and ε are constants specific to a given material (note: they are NOT equivalent to stress and strain, but this is the standard notation for the L-J parameters). The parameters σ and ε are related to the equilibrium bond length and the bond strength, respectively. Here, r is the interatomic spacing given in units of Angstroms (1 Angstrom = 10 nm = 10-10 m), and U(r) is given in units of eV/atom. A molecular dynamics simulation was performed by Zhang et al. [1] to study the properties of Al thin films in which the authors proposed a Lennard-Jones potential of the form above to model Al-Al interactions. The values used for the material parameters were: ε = 0.368 and σ = 2.548 (we’ve rounded off the values in the paper for your pset). (a) What are the assumed units of σ and ε in Zhang et al’s potential for aluminum? (b) Using the given material parameters and the form of the interatomic potential energy curve, plot U(r) for aluminum from r = 0 to 3.5 Angstroms in increments of <0.25 Angstroms. Tips: i. We strongly suggest you use a scientific programming language such as Mathematica, Matlab, or Maple here to graph and take derivatives of U(r), as this will be a big help in prep for Lab 2. If you didn’t take 3.016 or use this in another class, now is your chance to learn – e.g., by working with a classmate who did. Regardless of what program 3you use to do this, you must include your own program commands and output in the pset, of course, not a duplicate of someone else’s program. ii. Since r =0 will give you an error, use a very low value for your starting point instead, i.e., r=0.001 A. iii. In order to make your graph more readable, you may want to adjust the scale of your axes so that it only emphasizes the points around the equilibrium interatomic spacing, ro. (c) Determine the equation for and graph the interatomic forces F as a function of interatomic separation r for Al over the same range of r used in part (a), indicating units of F(r). Also, analytically and graphically determine the equilibrium interatomic spacing, ro. Mark this point on both the graphs produced in parts (a) and (b). (d) Compare this equilibrium interatomic spacing to the literature value of atomic radius for aluminum, and from that comparison explain what you think Zhang et al. assumed in choosing the constants σ and ε that made ro come out this way. (e) Figure 2 below shows interatomic energy curves [V(r) = our U(r)] for Mg that were calculated by Chavarria [2] (squares) and McMahan et al. [3] (dotted and solid curve). These data are shown in “arbitrary units” or “a.u.” which is typical of computational / experimental results that have funny units, but r turns out to be expressed in ~2 x Angstroms (i.e., 9 a.u. = 4.5 Angstroms). Comparing these curves with that calculated in part (a) for Al, explain whether you would expect magnesium to have a lower or higher elastic modulus than aluminum? Is this confirmed by the literature values of elastic properties and physical properties of Al and Mg? (Hint: One can compare the shape or curvature of the Al and Mg potential energy curves by normalizing energies by the minimum or binding energy Ub or (U/Ub) and distances by ro or (r/ro).) Figure 2: Interatomic potential for magnesium as calculated by Chavarria [2] (squares) and McMahan et al. [3] (dotted and solid curves). (e) Again, considering the relationship between the Young’s modulus and the equilibrium interatomic spacing and U(r) curvature, what effect do you think temperature has on the measured Young’s modulus? Provide a conceptual explanation of your answer. (Hint: Think about what happens to atoms inside of a material as you heat it up.) 43. Poly(dimethylsiloxane) is a crosslinked elastomer that is commonly used to build microscale devices via soft lithography (borrowing from microelectronics processing approaches to make flexible, patterned surfaces). The glass


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