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2610 MSE Introduction to the Mechanical Properties of Materials Fall 2008 Homework Solutions #2 DUE: 18 September 2008 1.) Sketch the planes and directions listed below. Be sure to label your axes. a.) Planes in a cubic crystal: (011), (111), (212) b.) Directions in a cubic crystal: [111] , [214], [103] c.) Planes in a hexagonal crystal: (1011), (1010) Not Graded d.) Directions in a hexagonal crystal: [1001], [0112] Not Graded 2.) a.) Sketch the simple, face-centered, and body-centered cubic unit cells Simple Cubic Body-Centered Cubic Face-Centered Cubic b.) Calculate the atomic packing factor of the three cubic Bravais lattices - Simple Cubic 2r = a 4 / 3r 3 4 / 3r 3 fSC = = = = 0.523 6 a3 (2r )3 - Body-Centered Cubic 4r = a 3 4 / 3r 3 4 / 3 r 3 3 = = = 0.680 fBCC = 2 3 3 8 a 4/ 3r -Face-Centered Cubic 4r = a 2 4 / 3r 3 4 / 3 r 3 2 = = = 0.740 fFCC = 4 3 3 6 a 4/ 2r ( ) ( ) c.) Are any cubic lattices close-packed? If so, which families of planes and directions are close-packed? - The FCC lattice is a close-packed structure. In FCC, the {111} family of planes are close- packed planes, and the <110> family of directions are close-packed directions. d.) Calculate the theoretical density of aluminum based on the crystal structure and molar mass. Use the table in Callister to find the necessary data. - physical constants from Callister: r = 0.143 nm m = 26.98 g / mol FCC (4 atoms / cell) - Volume of the unit cell: 4r = a 2 Vunit cell = a 3 = 4r / 2 ( ) 3 -Density (4 atoms)(26.98 g / mol) nm nm = = = 3 3 Vunit cellN A 4r / 2 N A 4(1.43x10 -8 cm) / 2 (6.022 x10 23 atoms / mol) ( ) ( ) = 2.71 g / cm 3 3.) There is an unoccupied interstitial site in the unit cell structure of halite at locations with coordinates , measured as fractions of the cell edge. a) How many nearest-neighbor ions surround this site? - 4 Cl- ions, 4 Na+ ions b) How many unoccupied interstitial sites are contained within one unit cell? - 8 interstitial sites c) What is the maximum radius of a sphere that could fit into such a site without producing any distortion of the structure? (Do this in terms of RNa+ and RCl-) s2 s3 s s = R Cl + R Na R Cl + 2r + R Na = s 3 R Cl + 2r + R Na = (R Cl + R Na ) 3 r= ( 3 -1 [R Cl + R Na ] 2 ) d) If young's Modulus, E, were measured for a single crystal of halite, along which family of directions <100>, <110>, or <111> would you expect E to be highest, lowest, why? - The <100> family of directions have the highest packing density in NaCl; therefore, the <100> directions would be expected to have the highest modulus. -The <111> directions have the lowest packing density; therefore, the <111> directions would expected be to have the lowest modulus. 4.) A schematic of lamellar composite with many layers is shown below. The thick sheets (1 mm) are aluminum and thin sheets (0.1 mm) are silicon carbide. z Al Al Al Al x y SiC SiC SiC The Young's modulus of aluminum and SiC are 70 GPa and 300 GPa, respectively. The Poisson ratio is 0.34 for aluminum and 0.17 for SiC. The laminate is under biaxial load, 20 MPa in both the z and the x directions. Find the Poisson strain in the y direction. SiC = E SiC SiC Al = E Al Al In the z and the x directions: SiC = Al A SiC A + Al Al A A A A C = E SiC SiC + E Al Al A A A A E C = E SiC SiC + E Al Al A A In the y direction: SiC = Al C = SiC SiC = C = C E SiC Al = C E Al Parallel C VSiC C VAl + E SiC V E Al V C 1 VSiC 1 VAl = + C E SiC V E Al V 1 1 VSiC 1 VAl Transverse = + E C E SiC V E Al V Because strain in each layer is not constrained in the transverse (y) direction, the Poisson's ratio of the composite in the transverse direction is a weighted average of the Poisson's ratio of the respective constituents: VAl V + L SiC SiC V V V V c = Al Al + SiC SiC V V V V C = SiC SiC + Al Al Transverse V V Lc = L Al y = y E transverse - C Eparallel [ x + z ] V V 2 x SiC SiC + Al Al 2(20 MPa)[(0.17)(0.091) + (0.34)(0.82)] V V =- y = - A SiC A Al [(300 GPa)(0.091) + (70 GPa)(0.82)] ESiC A + E Al A y = 1.39x10 - 4 5.) Which polymer of each set would be more likely to have a higher Tg? Why? a.) Poly(styrene) or poly(ethylene) - The poly (styrene) would have a higher glass transition temperature because it has a phenyl side group that inhibits chain sliding. b.) Poly(vinyl chloride) or poly(ethylene) - The poly (vinyl chloride) would have the higher glass transition temperature because it has chlorine bonded to the backbone of the polymer chain. The dipole moment caused by the strongly electronegative chlorine atom causes dipole-dipole bonding between chains and inhibits chain sliding. c.) Cis-poly(isoprene) or heavily cross-linked cis-poly(isoprene) - The heavily cross-linked cis-poly(isoprene) would have the higher glass transition temperature because the strong covalent bonds of the cross-linking molecules between separate polymer backbones prevent chain sliding. d.) Polystyrene (MW = 10,000 g/mol) or polystyrene (MW=100,000 g/mol) - The higher molecular weight polymer would have a higher glass transition temperature. The longer polymer chain has higher conformational entropy which makes it more entangled with neighboring chains, and the entanglement inhibits chain sliding. Cubic lattices for question 1: z y (011) x (111) (212) [111] [214] [103]

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