MSE 230 Assignment 2 Solutions Fall 2013 1 To solve this problem there are two things you need to notice about plots in Fig 2 8 First the potential energy vs interatomic separation curve is at a minimum at the equilibrium spacing r0 therefore dE dr equals zero when r r0 Second the force vs interatomic separation curve equals zero when r r0 Using this information allows one to calculate the a EA A r ER B rn Et A r B rn 0 Emin is ro dE A r 2 nB rn 1 dr dE 0 ro dr A 2 0x10 28 J m r ro 4x10 10 m n 9 B 1 456 x 10 104 J m9 b When plotting Force vs r see Fig 2 8a slope r ro bond stiffness or modulus Young s Modulus dF d 2 E 4x10 28 J m 1 31x10 102 2 J m9 3 11 dr dr r r dF d 2E Strain 0 001 means that the bond is stretched to 4 004 If we find at 4 2 or dr dr and 4 004 we can calculate the per cent change in Young s modulus dF 4 25 J m2 dr dF 4 004 24 677 J m2 dr 25 24 677 0 0129 or 1 3 decrease in Young s modulus 25 2 Callister Problem 3 3 This problem calls for a demonstration of the relationship a 4R 3 for BCC Consider the BCC unit cell shown below 1 a Q a P N O a Using the triangle NOP NP 2 a2 a2 2a2 And then for triangle NPQ NQ 2 QP 2 NP 2 But NQ 4R R being the atomic radius Also QP a Therefore 4R 2 a2 2a2 or a 3 4R 3 Callister Problem 3 7 This problem calls for a computation of the density of iron According to Equation 3 5 nAFe r VCNA For BCC n 2 atoms unit cell and 4R 3 VC 3 Thus 2 atoms unit cell 55 9 g mol 7 3 3 23 4 0 124 x 10 cm 3 unit cell 6 023 x 10 atoms mol 7 90 g cm 3 The value given inside the front cover is 7 87 g cm3 2 4 3 14 For each of these three alloys we need to by trial and error calculate the density using Equation 3 5 and compare it to the value cited in the problem For SC BCC and FCC crystal 3 structures the respective values of n are 1 2 and 4 whereas the expressions for a since V a C are 2R 4R 3 and 2R 2 For alloy A assume a body centered cubic crystal structure nA A V N C A 2 atoms unit cell 43 1g mol 4 3 1 22x10 8 cm 3 unit cell 6 023x10 23 atoms mol 6 4 g cm 3 Therefore its crystal structure is BCC For alloy B assume a simple cubic crystal structure 1atom unitcell 184 4g mol 2 1 46x10 8 cm 3 unitcell 6 023x1023atoms mol 12 30 g cm 3 its crystal structure is simple cubic Therefore For alloy C assume a BCC crystal structure 2atoms unitcell 91 6g mol 4 3 1 37x10 8 cm 3 unitcell 6 023x10 23 atoms mol 9 60 g cm 3 its crystal structure is BCC Therefore 3 5 See Figure below For BCC a 4 r 3 Separation distance x a 2r 4 r 2r 3 x 0 31r Simple Cubic FCC BCC x a r Problem 5 Figure 4 Problem 6 Figure Origin z 1 2 1 1 1 1 y x 1 0 1 2 0 1 2 0 2 1 2 1 5 7 a Use the cross product 3 1 2 11 1 1 5 4 For cubic structures 1 5 4 is perpendicular to 1 5 4 You can check your answer using the dot product 3 1 2 1 5 4 3 5 8 0 1 1 1 1 5 4 1 5 4 0 b Cross product again two planes intersect in a line i j k 1 11 112 1 1 1 1i 3j 2k 132 1 1 2 c Dot product u v cos 3 1 2 11 1 0 14 3 cos cos 0 90 6
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