Homework_2_solution - Introduction To Materials Science...

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Unformatted text preview: Introduction To Materials Science, Chapter 1, Introduction Composites Thin crystalline platelets grown fromCr,lut-on ,- Mo ihs voldhis crystal structure so !i Fe cha na f e t back and forth: chain-folded model Spring 2005 Engineering Material Polyethylene Polymer composite materials: reinforcing glass fibers in a polymer matrix. University of Virginia, Dept. of Materials Science and Engineering Homework #2 18 (Solution) The average chain length is much greater than the thickness of the crystallite University of Virginia, Dept. of Materials Science and Engiineeriing of Virgi24ia, Dept. of Materials Science and Engineering U n v e r s ty n 9 Reading Assignments Chapters 1, 2 Chapter 3 sections 1-14 Chapter 12 sections 1-4 Chater 14 sections 1-4 Problem 3 a. We need to consider two limiting cases when calculating the interstitial site radius, first that it fits in plane (001) and the second that it first in plane (002). Case of plane (001) In this case the condition is 2 rinterstitial + 2 R = 2a noting that for a bcc lattice a= 4R 3 we get after some manipulation: ⎛2 2 ⎞ rinterstitial = ⎜ − 1⎟ R ⎜ ⎟ ⎝3 ⎠ Spring 2005 Engineering Material Case of plane (002) In this case the geometric condition becomes 2 rinterstitial + 2 R = a which yields Energy x A B rinterstitial = 1 ⎛ 4 R⎞ ⎜ ⎟−R 2 ⎝ 3⎠ ⎛2 ⎞ =⎜ − 1⎟ R ⎝3 ⎠ which is obviously the more constraining condition, thus ⎛2 ⎞ rinterstitial = ⎜ − 1⎟ R ⎝3 ⎠ b. c. Let NV1 = Nv at 1120˚K and NV2 = Nv at 1000˚K Engineering Material ⎜−⎟ − ⎜ R ⎜T T ⎟⎟ N V1 Ne = 5= = e⎝ ⎝ 1 2 ⎠ ⎠ ⎛Q⎞ ⎜− V ⎟ NV 2 ⎜ ⎟ Ne⎝ RT2 ⎠ R ln 5 (8.31 J / mol˚ K ) ln 5 ⇒ QV = − 1 1 = − = 116300 J / mol 1 1 − − T1 T2 1120 1000 ⎛Q ⎜− V ⎜ RT ⎝ 1 ⎞ ⎟ ⎟ ⎠ ⎛Q⎛ 1 V ⎜ ⎞ 1 ⎞⎟ Problem 4 a. ⎛2 ⎞ rinterstital = ⎜ − 1⎟ R ⎝3 ⎠ with R = 0.124 nm we get rinterstital=4 .0 1 = 0.077nm ⇒ rC a r b o n 0.077 = rinterstital 0.0192 this explains the low solubility of Carbon in α-Fe b. c. 4R = 2 R + 2rinterstitial 2 2R ⇒ rinterstitial = − R = ( 2 − 1) R = 0.0526nm 2 r 0.77 ⇒ Carbon = = 1.46 rinterstitial 0.0526 a= ...
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