Midterm Examination
Thursday, February 14,, 8:00 – 9:15 AM
Place all answers in a 8.5" x 11" Bluebook
Allowed: 8.5" x 11" page of notes – must return with exam – and a calculator
Required: ID for entrance to the exam.
Note: 5 Problems for this exam (both side of this page).
1. Miller Indices (20 points)
(a) For a cubic system (FCC or BCC), find the plane that has both the [1 10] and [0 2 1]
directions within the plane. In a drawing, show the two directions listed above and
show the plane. Find a 3rd direction in this plane.
(b) For this plane, write the representation for the family of equivalent planes and write
out all equivalent planes.
ANSWER:
This is a vector normal to the plane.
Since a direction [xyz] is normal to a plane (xyz) with the same indices, we have the
final plane that contains both directions as: (112)
An easy way to find another direction that’ss in the plane is to simply add the two
directions given. (though this direction is not shown in the drawing)
b. 112 121
112 121 211 112
12 121 211 112 121 211 112
1 211
2 12
1 21
1 112
121
211
112 121 211 112
121
211
11
12
121
2. Atomic Bonding (20 points)
(a) Calculate the planar atom density for the (100) and (111) plane for FCC - for area,
use the lattice constant a (unit cell dimension) as the reference.
(b) Calculate the unit cell body diagonal length (i.e., in the [111] direction) for Pt
(platinum), given that Pt has a FCC structure, a density of 21.4 g/cm3 and an atomic
weight of 195.09 g/mol.
(c) How many moles of Pt are in 1 cm3 of material.
ANSWERS:
2. a. Planar Atom Density (PD) = (# atoms centered on plane)/(area of plane)
(100) plane:
4R
a
a
There are 2 atoms on plane (one in the middle and 1/4 on each of the 4 corners)
Area of plane = A = a2
PD = 2/(a2)
(111) plane:
Call red line
‘h’
a
h
All sides of triangle
a/2*¦ 2
are of length a¦ 2
There are 2 atoms on plane (3*1/2 atoms in middle of each line + 3*1/6 atoms at corners)
Find h through Pythagorean Theorem, using purple and blue lines shown on unit cell:
h2 = (a/2√2)2+(a)2 h = a√1.5
Area of plane = A = 1/2* a√2* a√1.5 A = a2/2* √3
PD = (2atoms)/( a2/2* √3) PD = 4/(a2*√3)
b.
d
a
a¦ 2
Pythagorean Theorem: d2 = (a)2+(a√2)2
d = a√3
nA
VC N A
• ρ is the density and is given as 21.4 g/cm3
• n is the number atoms in the unit cell. For fcc, n = 4 (1/2 atom on each of
the 6 faces + 1/8 atom at each of the 8 corners)
• A is the atomic weight and is given as 195.09 g/mol
• Vc is the unit cell volume; Vc =a3.
• NA is Avogadro’s number, 6.023*1023 atoms/mol.
ρ=
Å
Vc = a 3 =
4(195.09
g
)
mol
g
atoms
(21.4 3 )(6.022 *10 23
)
cm
mol
d = a 3 → d = 6.80 *10 −8 cm = 6.8Å
3
c. 1cm (21.4
a = 3.93*10−8 cm
g
mole
)(
) = 0.1097moles
3
cm 195.09g
3. Point Defects (20 points)
(a) Calculate the energy for vacancy formation for gold (Au) (FCC packing) given that
the equivalent number of vacancies at 800 °C is 1.64 x 1017 cm-3. The density and
AW for Au at 800 °C are 19.3 g/cm3 and 196.96 g/mol, respectively.
(b) What is the vacancy concentration in Au at room temperature?
(c) What is the average spacing between vacancies at room temperature and 800 °C?
Note: Boltzmann’s constant = 8.6174x10-5 eV K-1
Note: Avogrado’s # = 6.0221 x 1023/mol
Au: thermal expansion coefficient: 14x10-6 K-1
Au: volume expansion coefficient: 42x10-6 K-1
ANSWERS:
25℃ .× /!"#
=0.113 cm
/%
$
4. Polymers (20 points)
The weight average molecular weight of a polydimethylsiloxane (PDMS) is 500,000
g/mol. PDMS is the most widely used silicon-based organic polymer, and is
particularly known for its unusual rheological (or flow) properties. Its applications
range from contact lenses and medical devices to elastomers, caulking, lubricating
oils and heat resistant tiles.
[Note: the polymer backbone has silicon-oxygen bonds]
(a) Compute the weight average degree of polymerization. See the mer unit for PDMS
below:
(b) Is PDMS a thermoplastic or thermoset? Explain.
(c) Draw several repeat units of PDMS.
AW: Note AWs: H: 1; C: 12.01; O: 15.994; Si: 28.08 (all g/mol)
ANSWER:
4. a. m = 2(12.01) + 1(28.08) + 1(15.994) + 6(1) = 74.094 g/mol
4,6/"78
6748
&'(&& )* +),-.&(/012/)3 9.: 6/"78
b. (This problem was not graded due to a misstatement in last year’s solution set and
an unrealistic number for the molecular weight of PDMS) PDMS is a thermoset because
it is an elastomer. Elastomer means that the polymer is crosslinked which allows it so the
polymer does not melt and just decomposes at a high temperature.
c.
5. Mechanical Properties (20 points)
(a) A cylindrical rod, 150 mm long, 15 mm diameter, is loaded in uniaxial tension to a
load (force) of 45,000 N. It must not experience either plastic deformation or a
diameter reduction of more than 1.2x10-2 mm. Of the materials listed below, which
are possible candidates:
Material
Young’s
Modulus
Yield Strength
Poisson’s Ratio
(MPa)
(GPa)
Aluminum alloy
70
250
0.33
Titanium alloy
105
850
0.36
Steel alloy
205
550
0.27
Magnesium alloy
45
170
0.20
σyield (MPa)
250
850
550
170
ν
0.33
0.36
0.27
0.2
(b) For aluminum, find the bulk modulus.
ANSWER:
Material
E (GPa)
70
Aluminum Alloy
105
Titanium Alloy
205
Steel Alloy
45
Magnesium Alloy
z
15mm
x
∆d=-1.2x10-2mm
υ =−
εx
εz
∆d −0.012mm
=
= −0.0008
d
15mm
F
45000N
σz =
=
2 = 254.648MN
Ao π (15mm) 2 1m
4
1000mm
εx (max) =
150mm
(
)
Now that the stress along the z axis is known,
you can compute the strain along z using
Young’s Modulus.
εz =
σz
E
Steel :
45000N
εz =
254.648MPa
= 0.0012
205GPa *1000MPa /GPa
Repeat this for the remaining three candidate materials.
Using Poisson’s Ratio, we can then calculate the strain in x, which is limited by the value
calculated above for maximum strain:
εx = −εzυ = −0.0012 * 0.27 = −0.000335
−0.000335 < −0.0008
Since this is true, steel would be an appropriate candidate for this application.
If you repeat this procedure for the remaining three alloys, none of the others will
work. The results are summarized in the table below:
E (GPa)
Aluminum Alloy
Titanium Alloy
Steel Alloy
Magnesium Alloy
70
105
205
45
syield (MPa)
250
850
550
170
PR
z stress (MPa)
0.33
0.36
0.27
0.2
254.648
254.648
254.648
254.648
(b) The bulk modulus for aluminum is given by:
E
70
K=
=
= 68.627GPa
3(1− 2υ ) 3(1− 2 * 0.33)
Strain
in z
0.0036
0.0024
0.0012
0.0057
Strain in x
W/F
-0.001200
-0.000873
-0.000335
-0.001132
Fail
Fail
Work
Fail