MIME 260 Midterm Winter 2016
MID TERM TEST, Feb. 25th, 2016
DURATION: 2 HRS
Student No. Name
McGill University Dept. of Mining and Materials Engineering
MATERIALS SCIENCE AND ENGINEERING MIME-260 MIDTERM Examiner: Prof. Jun Song
Date: Feb. 25, 2016
Instructions 1. 2. 3. 4. 5. 6. 7. 8. 9.
Read the instructions carefully. Put your name and ID# on this page and every page after (in the page header) except the formula page. Read each question all the way through before starting it. The parts may be connected. Write your solutions in the space provided below the relevant question. No additional pages! Use the blank sides of this examination book for rough work. You are permitted TRANSLATION dictionaries ONLY. STANDARD CALCULATOR permitted ONLY. This is a CLOSED BOOK examination. No documentation allowed, see formula sheet. The value of each question is respectively indicated under the section title. For instructor’s use only
Question Points
1
2
3
4
5
6
7
Total
15
12
15
14
14
12
18
100
1
Student No.
Name:
Question 1 (2.5 × 6 = 15 pts) A. Determine the Miller indices of the following planes within a cubic unit cell. (A.1)
Solution: (102)
(A.2)
Solution: (2 1 2)
B. Determine the Miller-Bravais indices of the following planes within a hexagonal unit cell. (B.1)
Solution:
(1 2 1 0)
(B.2)
Solution: (0 1 1 1)
2
Student No.
Name:
C. Determine the Miller-Bravais indices of the following directions within a hexagonal unit cell. (C.1) Note: M is the face center of the shaded plane.
M
Solution: [101 1]
(C.2)
Solution: [1210]
3
Student No.
Name:
Question 2 (12 pts) For an ionic pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to
EA
1.5 , r
ER
6 106 r9
For these expressions, energies are expressed in electron volts (eV), and r is the distance in nanometers (nm). The net energy EN is just the sum of the two terms, EA and ER, expressions above. Determine the equilibrium spacing r0 between two ions, and the magnitude of the bonding energy E0 between the two ions. Solution:
dE 0 r0 dr dE 1.5 5.4 105 0 0.278 nm dr r 2 r10 The bonding energy is
Er r0
1.5 6 106 4. 79 eV 0.278 0.2789
4
Student No.
Name:
Question 3 (3 + 6 + 6 = 15 pts) (A) ZnS has a cubic unit cell in which Zn atoms (white) occupy the corner and face-center sites while S atoms (black) are enclosed inside the cell. Each S is neighboring 4 Zn with the nearest Zn-S distance being 0.23 nm. How many Zn atoms and how many S atoms are there in each unit cell (3 pts)? Solution: # of Zn: 4
# of S: 4
(B) Consider a crystalline material that has a hexagonal close-packed (HCP) structure, with lattice constant a = 3.6 ×10−10 m and c/a = 1.633. This material has an atomic weight of 30 g/mol. Answer the following questions (assuming hard-sphere model): (B.1) Compute the density of the material. (6 pts) Solution: # of atoms per unit cell: 6
Volume of the unit cell:
3 3a 2c 2
Thus the density is
6 30 g/mol
3 3 3 1.633 3.6 1010 m 6.02 1023 mol 1 2 1.51 106 g/m 3
5
Student No.
Name:
(B.2) Suppose that the material can undergo a phase transformation from HCP to Face-Centered Cubic (FCC) structure. Denote the FCC lattice constant as a1. If the material’s density increases by 25% after the HCPFCC transformation, what is the relation between a and a1? (6 pts)
Solution: For HCP: The unit cell volume is:
3 3a 2c 2 There are 6 of atoms per cell, thus the volume per atom is:
3 3a 2c 3a 2c 2 6 4 For FCC: The unit cell volume is:
a13 There are 4 of atoms per cell, thus the volume per atom is:
a13 4 Since the density increases by 25%, we have 1 1.25 a13 4 3a 2c / 5 3 2 a1 / 4 3a c / 4 1/3
a1 4 3a 2c / 5
1.31a
or a 0.763a1
6
Student No.
Name:
Question 4 (7 + 7 = 14 pts) Consider a material, being a single crystal, has a HCP crystal structure and a melting temperature of 1500oC. Assuming the energy for vacancy formation is 0.8 eV/atom, and the energy for self-interstitial formation is 1.8 eV/atom. Answer the following questions: (A) Calculate the number of atom sites that are vacant per unit cell at temperature = 1100 oC (7 pts).
Solution: For each HCP unit cell, there are 6 atoms. Thus the # of vacancies per unit cell is
0.8 6 exp 6. 96 103 5 (1100 273) 8.62 10
(B) At what temperature the population of vacancies is 12,000 times the population of self-interstitials within the material? (7 pts) Solution: The population of vacancies and self-interstitials within the material are (denote the # of lattice atoms as N0):
0.8 N v N 0 exp k BT 1.8 N SI N 0 exp k BT With
N SI 104 N v We have
ln N 0
0.8 1.8 ln12000 ln N 0 k BT k BT
1 1 ln12000 T 1235 K 5 k BT 8.62 10 ln 12000
7
Student No.
Name:
Question 5 (8+6=14 pts) A steel alloy specimen having a circular cross section of radius r = 5 mm, has an elastic modulus E = 200 GPa. If this specimen is subjected to a tensile force of 40,000 N that produces a total strain of 0.01, (A) Determine the elastic and plastic strain values. (8pts) Solution: The stress is
5 103
The elastic strain is
e The plastic strain is then
40000
E
2
509 MPa
0.509 0.002545 200
p 0.01 e 0.007455
8
Student No.
Name:
(B) If its original length is 480 mm, what will be its final length after the load of 40,000 N is applied and then released? (6 pts) Solution: The elastic strain will be recovered after unloading. Thus the final length is
480 1 0.007545 483. 62 mm
9
Student No.
Name:
Question 6: DEFINITION and SHORT-ANSWER (4 × 3 = 12 pts) 6.1. Explain why the modulus of elasticity of polymers, such as polyethylene and polystyrene, is expected to be very low compared with that of metals and ceramics. (3 pts). Solution: Van der Waals interactions in polymers
6.2.
Sketch the typical stress-strain curve for a material that exhibits discontinuous yielding. (3 pts)
Solution: See lecture notes (Lecture 5)
6.3. In an alloy, what characteristics decide if the solute atoms will form a solid solution or a second phase in the solvent? (3 pts) Solution: Physical concentration of the solute atoms Difference in atomic radius (compared to the host atom) Difference in electronegativity (compared to the host atom)
6.4.
Explain why metallic materials tend to be densely packed (3 pts)
Solution: Atomic radii of metal atoms are similar Metallic bond is non-directional Electron “sea” shields ion-ion interactions
10
Student No.
Name:
Question 7: TRUE OR FALSE (PLEASE EXPLAIN IF STATEMENT IF FALSE) 3 pts each question. In the case when explanation is required, the explanation weights 1 out of 3 pts. 7.1.
A HCP structure with the ideal c/a ratio (i.e., c / a 2 6 / 3 1.633 ) has a lower Atomic Packing Factor (APF) than the Face-Centered Cubic (FCC) structure. False. The APF are the same
7.2.
Dislocation is a two-dimensional planar defect.
False. Dislocation is a 1D defect
7.3.
The material SiC is more likely to have a directional bonding than NiAl.
True. (SiC is covalent).
7.4.
The hardness number is an indication of the material’s resistance to elastic deformation. False. Plastic deformation
7.5.
The van der Waals bonding forces arise from atomic or molecular dipoles.
True
7.6.
The bonding in diamond is primarily ionic in nature.
False. The bonding is primarily covalent.
11
Student No.
Name:
Do not write below this line EQUATIONS [u ' v ' w '] [u v t w] u (2u ' v ') / 3, t (u v ),
w w'
F AO
,
E 2G 1 , K
APF
li lo lo
,
i ( h k )
F A0
1 U r y y 2 F Ai
T KTn
G
,
X Y Z Z
E 3 1 2
total sphere volume VS Qv , NV N exp , total unit cell volume VC k BT
l f lo % EL 100 lo
T
(h k l ) (h k i l )
v (2v ' u ') / 3,
A AF % RA 0 A0
100
Q N SI N exp SI kBT y
U r d 0
2 1 1 y y U r y y y 2 2 E 2E
T ln
li lo
T 1
T ln 1
TS MPa 3.45HB
Avogadro's number = 6.02 1023 mol1
T ( K ) T ( o C) 273
Boltzmann constant = 8.62 105 eV K 1 1.38 1023 J K 1 MPa = 106 Pa, GPa = 109 Pa, Psi = 6895 Pa
12
DURATION: 2 HRS
Student No. Name
McGill University Dept. of Mining and Materials Engineering
MATERIALS SCIENCE AND ENGINEERING MIME-260 MIDTERM Examiner: Prof. Jun Song
Date: Feb. 25, 2016
Instructions 1. 2. 3. 4. 5. 6. 7. 8. 9.
Read the instructions carefully. Put your name and ID# on this page and every page after (in the page header) except the formula page. Read each question all the way through before starting it. The parts may be connected. Write your solutions in the space provided below the relevant question. No additional pages! Use the blank sides of this examination book for rough work. You are permitted TRANSLATION dictionaries ONLY. STANDARD CALCULATOR permitted ONLY. This is a CLOSED BOOK examination. No documentation allowed, see formula sheet. The value of each question is respectively indicated under the section title. For instructor’s use only
Question Points
1
2
3
4
5
6
7
Total
15
12
15
14
14
12
18
100
1
Student No.
Name:
Question 1 (2.5 × 6 = 15 pts) A. Determine the Miller indices of the following planes within a cubic unit cell. (A.1)
Solution: (102)
(A.2)
Solution: (2 1 2)
B. Determine the Miller-Bravais indices of the following planes within a hexagonal unit cell. (B.1)
Solution:
(1 2 1 0)
(B.2)
Solution: (0 1 1 1)
2
Student No.
Name:
C. Determine the Miller-Bravais indices of the following directions within a hexagonal unit cell. (C.1) Note: M is the face center of the shaded plane.
M
Solution: [101 1]
(C.2)
Solution: [1210]
3
Student No.
Name:
Question 2 (12 pts) For an ionic pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to
EA
1.5 , r
ER
6 106 r9
For these expressions, energies are expressed in electron volts (eV), and r is the distance in nanometers (nm). The net energy EN is just the sum of the two terms, EA and ER, expressions above. Determine the equilibrium spacing r0 between two ions, and the magnitude of the bonding energy E0 between the two ions. Solution:
dE 0 r0 dr dE 1.5 5.4 105 0 0.278 nm dr r 2 r10 The bonding energy is
Er r0
1.5 6 106 4. 79 eV 0.278 0.2789
4
Student No.
Name:
Question 3 (3 + 6 + 6 = 15 pts) (A) ZnS has a cubic unit cell in which Zn atoms (white) occupy the corner and face-center sites while S atoms (black) are enclosed inside the cell. Each S is neighboring 4 Zn with the nearest Zn-S distance being 0.23 nm. How many Zn atoms and how many S atoms are there in each unit cell (3 pts)? Solution: # of Zn: 4
# of S: 4
(B) Consider a crystalline material that has a hexagonal close-packed (HCP) structure, with lattice constant a = 3.6 ×10−10 m and c/a = 1.633. This material has an atomic weight of 30 g/mol. Answer the following questions (assuming hard-sphere model): (B.1) Compute the density of the material. (6 pts) Solution: # of atoms per unit cell: 6
Volume of the unit cell:
3 3a 2c 2
Thus the density is
6 30 g/mol
3 3 3 1.633 3.6 1010 m 6.02 1023 mol 1 2 1.51 106 g/m 3
5
Student No.
Name:
(B.2) Suppose that the material can undergo a phase transformation from HCP to Face-Centered Cubic (FCC) structure. Denote the FCC lattice constant as a1. If the material’s density increases by 25% after the HCPFCC transformation, what is the relation between a and a1? (6 pts)
Solution: For HCP: The unit cell volume is:
3 3a 2c 2 There are 6 of atoms per cell, thus the volume per atom is:
3 3a 2c 3a 2c 2 6 4 For FCC: The unit cell volume is:
a13 There are 4 of atoms per cell, thus the volume per atom is:
a13 4 Since the density increases by 25%, we have 1 1.25 a13 4 3a 2c / 5 3 2 a1 / 4 3a c / 4 1/3
a1 4 3a 2c / 5
1.31a
or a 0.763a1
6
Student No.
Name:
Question 4 (7 + 7 = 14 pts) Consider a material, being a single crystal, has a HCP crystal structure and a melting temperature of 1500oC. Assuming the energy for vacancy formation is 0.8 eV/atom, and the energy for self-interstitial formation is 1.8 eV/atom. Answer the following questions: (A) Calculate the number of atom sites that are vacant per unit cell at temperature = 1100 oC (7 pts).
Solution: For each HCP unit cell, there are 6 atoms. Thus the # of vacancies per unit cell is
0.8 6 exp 6. 96 103 5 (1100 273) 8.62 10
(B) At what temperature the population of vacancies is 12,000 times the population of self-interstitials within the material? (7 pts) Solution: The population of vacancies and self-interstitials within the material are (denote the # of lattice atoms as N0):
0.8 N v N 0 exp k BT 1.8 N SI N 0 exp k BT With
N SI 104 N v We have
ln N 0
0.8 1.8 ln12000 ln N 0 k BT k BT
1 1 ln12000 T 1235 K 5 k BT 8.62 10 ln 12000
7
Student No.
Name:
Question 5 (8+6=14 pts) A steel alloy specimen having a circular cross section of radius r = 5 mm, has an elastic modulus E = 200 GPa. If this specimen is subjected to a tensile force of 40,000 N that produces a total strain of 0.01, (A) Determine the elastic and plastic strain values. (8pts) Solution: The stress is
5 103
The elastic strain is
e The plastic strain is then
40000
E
2
509 MPa
0.509 0.002545 200
p 0.01 e 0.007455
8
Student No.
Name:
(B) If its original length is 480 mm, what will be its final length after the load of 40,000 N is applied and then released? (6 pts) Solution: The elastic strain will be recovered after unloading. Thus the final length is
480 1 0.007545 483. 62 mm
9
Student No.
Name:
Question 6: DEFINITION and SHORT-ANSWER (4 × 3 = 12 pts) 6.1. Explain why the modulus of elasticity of polymers, such as polyethylene and polystyrene, is expected to be very low compared with that of metals and ceramics. (3 pts). Solution: Van der Waals interactions in polymers
6.2.
Sketch the typical stress-strain curve for a material that exhibits discontinuous yielding. (3 pts)
Solution: See lecture notes (Lecture 5)
6.3. In an alloy, what characteristics decide if the solute atoms will form a solid solution or a second phase in the solvent? (3 pts) Solution: Physical concentration of the solute atoms Difference in atomic radius (compared to the host atom) Difference in electronegativity (compared to the host atom)
6.4.
Explain why metallic materials tend to be densely packed (3 pts)
Solution: Atomic radii of metal atoms are similar Metallic bond is non-directional Electron “sea” shields ion-ion interactions
10
Student No.
Name:
Question 7: TRUE OR FALSE (PLEASE EXPLAIN IF STATEMENT IF FALSE) 3 pts each question. In the case when explanation is required, the explanation weights 1 out of 3 pts. 7.1.
A HCP structure with the ideal c/a ratio (i.e., c / a 2 6 / 3 1.633 ) has a lower Atomic Packing Factor (APF) than the Face-Centered Cubic (FCC) structure. False. The APF are the same
7.2.
Dislocation is a two-dimensional planar defect.
False. Dislocation is a 1D defect
7.3.
The material SiC is more likely to have a directional bonding than NiAl.
True. (SiC is covalent).
7.4.
The hardness number is an indication of the material’s resistance to elastic deformation. False. Plastic deformation
7.5.
The van der Waals bonding forces arise from atomic or molecular dipoles.
True
7.6.
The bonding in diamond is primarily ionic in nature.
False. The bonding is primarily covalent.
11
Student No.
Name:
Do not write below this line EQUATIONS [u ' v ' w '] [u v t w] u (2u ' v ') / 3, t (u v ),
w w'
F AO
,
E 2G 1 , K
APF
li lo lo
,
i ( h k )
F A0
1 U r y y 2 F Ai
T KTn
G
,
X Y Z Z
E 3 1 2
total sphere volume VS Qv , NV N exp , total unit cell volume VC k BT
l f lo % EL 100 lo
T
(h k l ) (h k i l )
v (2v ' u ') / 3,
A AF % RA 0 A0
100
Q N SI N exp SI kBT y
U r d 0
2 1 1 y y U r y y y 2 2 E 2E
T ln
li lo
T 1
T ln 1
TS MPa 3.45HB
Avogadro's number = 6.02 1023 mol1
T ( K ) T ( o C) 273
Boltzmann constant = 8.62 105 eV K 1 1.38 1023 J K 1 MPa = 106 Pa, GPa = 109 Pa, Psi = 6895 Pa
12
