MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Midterm Examination
Thursday, February 21, 11:00 AM - 12:15 PM
1.
Polymers
a) Draw syndiotactic polystyrene. The mer unit of polystyrene is shown below. (5 pts)
Solution:
b) Name three traits (regarding chemical structure or processing) that characteristically lead
to highly crystalline polymers. (7 pts)
Simple mer structures, linear polymers, regular polymers (e.g. isotactic and
syndiotactic), and/or slow cooling.
c) Assume polydimethylsiloxane (PDMS) exhibits waxy behavior below 10,000 g/mol, and
solid above. How many mer units long would a PDMS chain need to be to meet this
minimum criterion for a solid? (Note, Atomic weights: H: 1 g/mol; C: 12 g/mol; O: 16
g/mol; Si: 28 g/mol.) (8 pts)
Structural formula of PDMS is -[SiO(CH3)2]-. It has a molecular weight of
28+16+2*(12+3) = 74g/mol. 10,000/74 = 135.1  need a minimum of 136 mer units.
MATERIALS 101
2.
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Miller Indices
a) Consider a cubic crystal structure. Find the plane (hkl) that contains both the [ ̅ 10] and
[213] directions. Draw the plane and the vectors in a cubic unit cell. Make sure the
vectors are drawn in the plane. Note: The vectors and plane do not need to be drawn from
the same origin. (10 pts)
The cross product of two vectors gives a third vector orthogonal to the first two. The
cross product can be determined by the determinant of the matrix featured below:
̂
Det ̅
̂
̂
= ̂ (1*3 - 1*0) - ̂ (-1*3 - 2*0) + ̂ (-1*1 – 1*2) = [33 ̅ ] →
[11 ̅ ]
For cubic systems, the vector [hkl] is normal to the plane (hkl)
[
̅]
(11 ̅ ) Plane
(5pts: 2 for determinant, 2 for correct calculation, 1 for reducing [33 ̅ ] to [11 ̅ ])
̂
̂
̂
[110]
[213]
(111) plane
(5pts: 1 for cube, 2 for correct plane, 1 for each vector)
b) A stress of
is applied along the [010] direction, resulting in slip in the
above plane. The direction of the slip is [ ̅ 10]. Calculate the resolved shear stress. (8 pts)
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
The resolved shear stress is given by
where the angle φ is
between the vector normal to the slip plane and tension direction and the angle λ is
between the slip direction and tension direction
[
The cosine of an angle is related to the dot product by
Where the angle
[
̅] [
[̅
√
][
√
√
][
]
√
is between the vectors [abc] and [def].
]
]
=
=
√ √
√
√
(2pts)
(2pts)
= 81.65MPa
(2pts)
c) Write down the slip system of this crystal. What sort of cubic structure is this? SC, BCC
or FCC? (2 pts)
Slip system: {111}<110> (1pt)
FCC (1pt)
3.
Point Defects
A theoretical material called Herbium has a density and atomic weight of 20 g/cm 3 and 205
g/mol. Boltzmann’s constant is 8.62x10-5eV atom-1 K-1 and the gas constant is 5.189x1018eV
mol-1 K-1.
a) Analysis of Positron Annihilation Spectroscopy data indicates that the average distance
between vacancies in Herbium at 300 K is about 0.15 cm. Assuming that the vacancies
are uniformly distributed throughout the material, calculate the activation energy of
vacancies in Herbium at 300 K. State your answer in eV. (10 pts)
(1 vacancy per 0.15 cms)3 = 296.30 vacancies cm-3 = Nv
(
)(
)
Nv = N*exp[-Qv/(kbT)]
Ln[5.047x10-21]=-Qv/(8.62x10-5*ev atom-1 Kelvin-1 *300 Kelvin)
Qv = 1.21eV
b) Another element called Skendzium has a density of 10 g/cm3 and atomic weight of 151
g/mol. The two may join together to form an ionic ceramic called 2HS, whereby Herbium
is the anion and Skendzium is the cation. This ceramic has a density of 15 g/cm3.
Theoretical calculations indicate that the formation energy of a Schottky defect in 2HS is
3.2 eV. Calculate the Schottky defect concentration in 2HS at 1000K. State the answer in
defects per cm3. (10 pts)
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Ns = N*exp[-Qv/(2kbT)]
Ns = 2.54x1022 cm-3*exp[-3.2/(2*8.62x10-5*ev atom-1 Kelvin-1 *1000 Kelvin)]
= 2.54x1022 cm-3 *8.68x10-9
= 2.21x1014 defects cm-3.
4.
Shear Stress and Slip
α-Fe crystallizes in the BCC structure and has an atomic radius of 0.1241nm and atomic weight
of 55.85g/mol.
a) In a slip system, the slip will occur in the {110} close packed family of planes of the
BCC structure. Draw the (110) plane in the unit cell and a 2D projection of one of these
planes with atoms on it. (2 pts)
b) Slip occurs along the closest packed family of directions in the slip plane and describes
the most probable direction of dislocation motion. What is the closest packed family of
directions in the (110) plane? Use standard notation. Calculate the linear density along
this direction. (3 pts)
<111> is the family of close packed directions in the BCC crystal system.
c) Find the unit cell edge length of α-Fe. Then, calculate the theoretical density of α-Fe. (5
pts)
√
√
√
√
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
(
√
)
(
√
)
(
WINTER 2012
)
d) A single crystal of α-Fe is loaded uniaxially with a stress of 50 MPa in the [001]
direction. Will this single crystal have a resolved shear stress in the (110) plane? If not,
name one of the most likely planes for slip? (3 pts)
The direction perpendicular to this plane is the [110] direction. Since the stress is being
applied to the [001] direction, these directions are perpendicular and there will be no
resolved shear stress in the (110) plane. A plane that will feel a resolved shear stress is
the (011), (101), (011) or (101) planes.
e) Given that the single crystal of α-Fe is isotropic and has an elastic modulus of 211GPa
and Poisson ratio is 0.29, calculate the shear modulus. Calculate the transverse strain
([100] and [010] directions) for uniaxial loading of 50 MPa in the [001] direction. (2 pts)
81.8GPa
f) You are given a different, very carefully grown, perfect cube of an isotropic single crystal
of α-Fe that has an edge length of 5 mm. Unfortunately, while sailing through the Pacific
Ocean directly over the Marianas Trench, you drop your single crystal into the trench!
The hydrostatic pressure at the bottom of the Marianas Trench is ~100 MPa. What will
the change in the volume of the single crystal be once it is sitting at the bottom of the
trench? (5 pts)
=
5.
Mechanical Properties
You have been given a sample of a special unknown alloy. The sample is a square bar 100 mm
long with sides 15mm wide. Along with the sample you have been given preliminary testing
results from another sample of the same alloy in the form of a stress-strain curve. Use a 0.002
offset if required.
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
Stress-Strain Behavior of Unknown Alloy
1400
1200
1000
800
σ (MPa)
600
400
200
0
0
0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016
ε
a) Determine the modulus of elasticity from the stress-strain curve above. (4 pts) From the
curve provided, estimate the maximum load which can be safely applied to the bar? (2
pts)
Stress-Strain Behavior of Unknown Alloy
1400
1200
1000
800
σ (MPa)
600
Stress
400
.002 offset
200
0
0 0.0020.0040.0060.008 0.01 0.0120.0140.0160.018 0.02
ε
The yield strength  y  1220MPa
The elastic modulus is the slope of the linear portion of the curve which is:
   2 800MPa  0
E 1

 200000MPa  200GPa
1   2
0.004  0
b) Poisson’s ratio describes strain perpendicular to the load direction due to strain in the
load direction. A load applied in the z-direction (parallel to the 100 mm long edge)
causing compressive strain is usually accompanied by tensile strain in the x and y
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
directions (parallel to one of the 15mm edges) for positive values of Poisson’s ratio. (14
pts)
i. What is the expression for Poisson’s ratio relating strain in the z-direction to strain in
the x-direction?

  x
z
ii. If the alloy sample is compressed by 0.05 mm in the z-direction, what would the xdirection width of the sample be for a Poisson ratio of 0.30?
zo  100mm
xo  15mm
  0.30
z f  zo  0.05mm
z 
z f  zo
zo

100mm  0.05mm   100mm  0.05mm  0.0005
100mm
100mm
x
 0.30
z
 x  0.30 z  0.30  0.0005  0.00015
 
x 
x f  xo
xo
0.00015 
x f  15mm
15mm
x f   0.0001515mm   15mm  15.00225mm
iii. What force must be applied along the z-axis for the width in the x-direction to be
increased by 0.0045mm?
x f  xo  0.0045mm
0.0045mm
 0.0003
15mm

0.0003
z   x  
 0.001

0.30
 z  E z   200GPa  0.001  0.2GPa  200MPa
x 
z 
Fz
Ao
Fz   z Ao   200MPa 15mm 15mm   45000 N
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2012
a) You are characterizing this and many other materials in order to choose the safest for use
as a bridge support material. Given the two stress-strain curves below, which material
would you choose to be the safest and why? (2 pts)
Material A
σ
Material B
1400
1400
1200
1200
1000
1000
800
800
σ
600
600
400
400
200
200
0
0
0
0.004
0.008
ε
0.012
0.016
0
0.004
0.008
0.012
0.016
ε
Material B would be the better choice for safety judging from the stress-strain curves.
Both curves are the same until material A fractures at a strain of ~0.009 while material B
will continue to plastically deform until more work is done; fracturing at a strain of
~0.015.