230 f14 Hw2 Sols
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1 MSE 230 Assignment 2 Solutions Fall 2014 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 r 0 (therefore dE/dr equals zero when r= r 0 ). Second, the force vs. interatomic separation curve equals zero when r= r 0 . Using this information allows one to calculate the a) E A = -A/r, E R = B/r n E t = -A/r + B/r n (E min is @ r o ∴ dE dr = 0@r o ) 0 = dE dr = A/r 2 − nB / r n +1 A = 2.0x10 -28 J.m r=r o = 4 x10 −10 m n = 9 B = 1.456 x 10 -104 J.m 9 b) When plotting Force vs. r, (see Fig. 2.8a), slope @ r=r o → bond stiffness or modulus. Young’s Modulus α dF dr = d 2 E dr 2 = − 4x10 −28 J.m r 3 + 1.31x10 −102 r 11 J ⋅ m 9 Strain = 0.001 means that the bond is stretched to 4.004 Å. If we find d 2 E dr 2 (or dF dr ) at 4Å and 4.004 Å we can calculate the per cent change in Young’s modulus. dF dr (4Å) = 25 J/m 2 dF dr (4.004 Å) = 24.677 J/m 2 25 − 24.677 25 = 0.0129 or ≈1.3% decrease in Young’s modulus. Please note that the units of J/m 2 (or N/m) are not the same as those for Young’s modulus (Pa or N/m 2 ). This is why the problem states that “Young’s modulus is proportional to the slope…”. We would need to include some proportionality constant to get the units and numerical values the same as those for Young’s modulus.
Transcript of 230 f14 Hw2 Sols
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MSE 230 Assignment 2 Solutions Fall 2014 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 (Emin is @ ro ∴dEdr
= 0@ro )
0 =dEdr
= A / r2 − nB / rn +1 A = 2.0x10-28 J.mr = ro = 4x10−10m
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 α dFdr
=d2Edr2
= −4x10−28J.m
r3+1.31x10−102
r11J ⋅m9
Strain = 0.001 means that the bond is stretched to 4.004 Å. If we find d2Edr2
(or dFdr
) at 4Å
and 4.004 Å we can calculate the per cent change in Young’s modulus.
dFdr
(4Å) = 25 J/m2
dFdr
(4.004 Å) = 24.677 J/m2
25 − 24.67725
= 0.0129 or ≈1.3% decrease in Young’s modulus.
Please note that the units of J/m2 (or N/m) are not the same as those for Young’s modulus (Pa or N/m2). This is why the problem states that “Young’s modulus is proportional to the slope…”. We would need to include some proportionality constant to get the units and numerical values the same as those for Young’s modulus.
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2.* This problem calls for a demonstration of the relationship a = 4R 3 for BCC. Consider the
BCC unit cell shown below
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 = 4R
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3. This problem calls for a computation of the density of iron. According to Equation (3.8)
ρ = nAFe
VCNA
For BCC, n = 2 atoms/unit cell, and
VC = ⎝⎜⎛
⎠⎟⎞4R
3 3
Thus, ρ =
(2 atoms/unit cell)(55.9 g/mol)
[ ](4)(0.124 x 10-7 cm)3/ 3 3/(unit cell)(6.023 x 1023 atoms/mol)
= 7.90 g/cm3
The value given inside the front cover is 7.87 g/cm3.
a
a
NO
P
Q
a
3
4.* Callister 3.16: For each of these three alloys we need to, by trial and error, calculate the density using Equation (3.8), and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of n are 1, 2, and 4, whereas the expressions for a (since VC = a3) are 2R, 4R/ 3 and 2R 2 .
For alloy A, assume a body centered cubic crystal structure.
ρ = nAA
VCNA
= (2atoms / unit cell)(43.1g /mol)
[(4 3)(1.22x10−8cm)"# ]3 / unit cell(6.023x1023atoms /mol)
= 6.4 g/cm3
Therefore, its crystal structure is BCC.
For alloy B, assume a simple cubic crystal structure.
ρ =
€
(1atom/unitcell)(184.4g /mol)[(2)(1.46x10−8cm)[ ]3 /unitcell(6.023x1023atoms/mol)
= 12.30 g/cm3
Therefore, its crystal structure is simple cubic.
For alloy C, assume a BCC crystal structure.
ρ =
€
(2atoms/unitcell)(91.6g /mol)[(4 3)(1.37x10−8cm)[ ]3 /unitcell(6.023x1023atoms/mol)
= 9.60 g/cm3
Therefore, its crystal structure is BCC.
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5) See Figure below
