Science Applied 278
Midterm Examination February 25, 2008
THE UNIVERSITY OF BRITISH COLUMBIA
Department of Materials Engineering
APSC 278 Engineering Materials
MID-TERM EXAMINATION, February 25, 2008
This is a closed book examination.
Use of relevant material stored in programmable calculators prohibited.
Answer all questions. Show all work and units.
The complete exam is 8 pages in length
Name:
Student #:
1
Applied Science 278
Midterm Examination February 25, 2008
1. Stress-strain Curves (8p)
A standard tensile sample (meeting ASTM standards) with a gauge length of 60 mm was originally
used as a demonstration for APSC 279 laboratory. For this sample, the change in length for at the
necking point was measured to be 25 mm and the change in length between the necking point and
failure was found to be 10 mm.
i. calculate
a. the percent uniform elongation
b. percent total elongation
Solution
=
=
25mm
x100%
60mm
35mm
x100%
60mm
= 41.667%
= 58.333%
ii. later in the term, a lack of material led to redesign of the sample such that the gauge length
was reduced to 30 mm. Assuming the material was exactly the same, predict the percent
total elongation for this test.
Solution
The change in length up to necking is proportional to the gauge length and the
post-necking change in length is independent of lo. Thus Total Elongation (%) =
=
12.5 + 10
x100% = 75%
30
For a load of 50 kN, a cylindrical tensile sample with initial diameter of 10 mm was observed to
have a diameter under load of 9.94 mm. Calculate the modulus of elasticity. Given: Poissons
ratio = 0.3.
Solution
= F/A = (50000N)/[(5mm)2] = 636.62 MPa
d/d = (9.94-10)/10 = -0.006
= -d/A
Strain in loading direction A = -(-0.006)/0.3 = 0.02
E = / = 636.62 MPa / 0.02 = 31.83 GPa
2
Applied Science 278
Midterm Examination February 25, 2008
2. Energy (force) vs. interatomic separations/bonding (8p)
Sketch the force vs. atomic spacing diagram for an ionic bond such as NaCl. Clearly identify, the
equilibrium bond spacing and describe the relationship between this diagram and the relevant
macroscopic mechanical properties.
Solution
- Modulus of elasticity (E) is related to the slope, dF/dr at the point of equilibrium
bond spacing.
- Theoretical fracture stress is related to Fmax.
3
Applied Science 278
Midterm Examination February 25, 2008
3. Crystallography (8p)
Calculate the radius of a tantalum atom, given that Ta has an BCC crystal structure, a density of
16.6 g/cm3, and an atomic weight of 180.9 g/mol.
Solution
For BCC, n = 2 atoms/unit cell
Solving for R:
4
Applied Science 278
Midterm Examination February 25, 2008
For the plane designated below, the determine Miller Indices. Show all steps in the identification
process.
Solution
2/3
4/7
4/7
Coordinate
X
Y
Z
Intercepts
4a/7
b
2c/3
Intercepts
4/7
1
2/3
Reciprocal
7/4
1
3/2
Enclosure
(7 4 6)
5
Applied Science 278
Midterm Examination February 25, 2008
4. Recrystallization / strengthening (4p)
Two Al-Mg alloys with grain sizes of 5 m and 100 m were tested in a standard tensile test. It
was determined that the yield stress of the two samples was 130 MPa and 70 MPa, respectively. If
a yield stress of 90 MPa is required for a certain component in an automobile, determine the
required grain size.
Solution
y
= o + ky d-0.5
130
= o + ky (5x10-6m)-0.5
(1)
70
= o + ky (100x10-6m)-0.5
(2)
(1) - (2):
ky
= 0.1728 MPa m0.5
o
= 52.72 MPa
When y = 90 MPa,
90
= 52.72 + 0.1728 d-0.5
d
= 21.49 m
6
Applied Science 278
Midterm Examination February 25, 2008
5. Creep (4p)
Steady-state creep data taken for an unknown material are given in the table below.
Steady state creep rate
(h-1)
5x10-5
4x10-3
Stress (MPa)
Temperature (oC)
50
50
300
400
Compute the steady state creep rate at 350 oC for a stress of 50 MPa.
350
(350oC)
At 350oC,
5.30x10
5.30x10-4 h-1
7
Applied Science 278
Midterm Examination February 25, 2008
6. Fracture Mechanics (8p)
Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture
toughness of 40 MPa m0.5. It has been determined that fracture results at a stress of 260 MPa when
the maximum (or critical) internal crack length is 6.0 mm. For this same component and alloy, will
fracture occur at a stress level of 300 MPa when the maximum internal crack length is 4.0 mm?
Why or why not?
Solution
Determine Y:
Y=
K Ic
a
=
40MPa m
= 1.58
260MPa ( )(0.006m / 2)
Compute Y a for the other set of conditions:
Y a = 1.58(300MPa ) (0.004m / 2) = 37.7 MPa m
0.5
Therefore, fracture will NOT occur since this value (37.7 MPa m ) is smaller
than the KIc of the material, 40MPa m .
8
Applied Science 278
Midterm Examination February 25, 2008
Useful Equations
=
y
x
=
z
z
Q
RT
&
s = K 2 n exp
E = 2G(1 + )
Qv
N v = N exp
RT
y
1
Ur = y =
2
2E
2
toughness
y +u
2
theoretical
f
T = ln(1 + )
=
R = cos cos
a
m = o 1 + 2
t
T = (1 + )
T =
E
10
a
m = 2 o
t
n
K T
nA / N A
Vc
2 E s
c =
a
1
1
1
2
2
2
n = 2d sin
d hkl =
a
(
h2 + k 2 + l 2
rate = Ae Q
)
1
2
RT
t recrx = Ae Q RT
= o + kyd
1
2
&
s = K 1 n
R = 8.314 J/K-mol
2 E ( s + p )
c =
a
y =
K
2r
1
2
f y ( )
K = Y a
K
B 2.5 Ic
y
2
min
max
Na =6.02x1023 atoms/mol
R=
9