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Exam 1 - MatS 2001
Introduction to the Science of Engineering Materials
Fall 2013 – October 4, 2013
Name (Print)
ID
EXAM RULES:
All notes, books, notebooks, note cards, cell phone, etc. must be off the desk
and inside closed and zippered backpacks. Students should be separated by at
least one empty chair. Only simple scientific calculators (cannot store
formulae or text) are allowed.
SHOW YOUR WORK!!
Numerical answers with incorrect concepts, or solution procedures or no work shown will
receive no credit, even if the numerical answer is correct.
GOT QUESTIONS? ASK!!
Equations you may need and the periodic table are on the last page. If you need extra paper to
work a problem use the back of the page but please indicate this.
Constants
Avogadro's number N A
= 6.023 x 10
1eV = 1.602 x 10
J
atomic mass unit 1 amu = 1.66 x 10
kg speed of light c = 3.0 x 10
m/sec
Bohr radius a o
= 0.529 Å electron charge e = 1.602 x 10
Boltzman constant k = 8.617 x 10
eV/K Planck constant h = 6.626 x 10
J sec
proton mass m p
= 1.672 x 10
kg neutron mass m n
= 1.675 x 10
kg
electron mass m e
= 9.11 x 10
kg gas constant, R = 1.9872 cal mol
K
gas constant, R = 8.314 J mol
K
1 Pa = 1.45 x 10
psi
Question 1 (15 points)
Circle True or False
This time, question 1 was graded as 1 point when right, 0 when wrong or not answered. (In the past, it
was graded with 1 point when right, 0 when not answered, -1 when wrong.) The average estimated would
be 8 or 9. There were students guessing the answer by choosing true for all of them.
a) T F The strain hardening exponent in the relationship
s T
= Ke T
n
is equal tois equal to
the true strain at the point where necking begins.
b) T F A stiff material has a low modulus of elasticity.
c) T F Poisson’s ratio is based on constant volume in the elastic region
d) T F If internal and surface cracks are initially the same size, the surface crack
will cause fatigue failure before the internal crack.
e) T F The Larson – Miller Parameter directly relates creep failure time to
applied stress.
f) T F Typical yield stresses for metals are in the range of 10-300 MPa.
g) T F The most important factor in determining a material’s resistance to fracture is
the number of cracks in it.
h) T F Crack growth rate in fatigue increases as the crack length increases.
i) T F Ceramics have higher KIC values than metals.
j) T F KIC is small for metals because they have some ductility.
k) T F Fracture can happen before the yield stress is reached.
l) T F Polymers are covalently bonded.
m) T F Elements on the left side of the periodic table tend to form ionic bonds with
elements from the right side.
n) T F Melting temperature increases as bond energy increases.
o) T F In metals the valence electrons are free to drift through the entire metal.
(a) (10 pts) Calculate the elastic modulus, (in GPa), the yield stress, yield strain,
tensile strength, and strain at necking. Show your work on the graph.
(2 points for each of the five items, elastic modulus, the yield stress, yield strain,
tensile strength, and strain at necking)
..
200 MPa 0 MPa
E
0 0
80
025 0 0
GPa
(If the method shown on the graph is wrong and the answer
is not close to 80 GPa, 0 point; if the method shown on the graph is right, but answer is not
close to 80GPa, 1 point. )
Yield stress is found using a line with slope = E and 0.2% offset. (If did not use the 0.
method, 0 point)
As shown on the graph: y = 250 MPa
Yield strain can be read directly from the graph: y = 0.
As an alternative,
.
,
. y
250 MPa
0 002
80 000 M a
0
P
005
Tensile strength is the maximum of the curve = 450 MPa
Strain at necking is strain at tensile strength = 0.
Most people got 10 points for this part.
(b) (15 Pts) A 2 m long, 2mm diameter rod of this alloy is loaded with 128 kg. What
are the resulting stress, strain, length, and diameter of the rod while
loaded? When the load is removed, what is the length of the rod?
(3 points for each of the five items, stress, strain, length, diameter, length after load
is removed)
With 128 kg.,
2 2
(128 kg)(9.81 m)(4)
399.3 MPa
(2 mm) s
From the graph: total = 0.125 (Most people used Hooke’s law to do this, which is
not right. At this stress the material is plastically deformed. If instead Hooke’s
law was used, the strain, 0.005, is much too small (just check the plot) and 0
points were given for the strain part. For the length and diameter, if the 0.
strain was used and the calculation done correctly, 3 points for each correct
answer, but the total score of part (b) should not be more than 10 points. If the
answer is not correct, 0, 1, or 2 points were given for the answers, depending on
quality of the equations and calculations.)
0
l l ( 1 ) ( 2 0 m)(1+0.125) = 2.25 m.
2
2
0 0
0 0
2
mm 2 m
, or , or , D=1.89 mm
4 2.25 m
or
1 399.3 MPa 1.125 450 MPA
A 2.79 mm , D 1.88 mm
i
i i i
i
T
i i
T
A l D
Al A l A
l
F
plastic 0 plastic
plastic total elastic plastic
400 MPa
80 GPa
l l (1 ) (2.0 m)(1 0.120) 2.24 m
or
Most people got 4-10 points for part b.
2
1/
1/
0.0026 m
600 MPa 0.0026 54.2 MPa-m
54.2 MPa-m
967 MPa
IC
c
IC
IC
c
K
a
Y
K
K
Y a
Question 4 (10 points)
The hook shown to the right will support a container of molten steel and must carry a
load of 40,000 lb. The hook operates continuously at a temperature of 560 C. The
relationship between stress and the Larson – Miller parameter is shown on the graph
below:
where, for this material:
3
L-M T 38.316 1.796 log 10
day
t x
T is in Kelvin and rupture time is in days.
If the useful life of the hook, with a safety factor of 1.6 for the load, is to be 7 years of
continuous operation, what must be the diameter (in inches) of the indicated portion of
the hook?
Most students got the L-M value but many students could not calculate the stress or forgot to consider the
safety factor
L M (^) 560 (^273) [ 38. 316 1. 796 log( 365 x 7 yrs )] x 10
3
From graph, Stress» 5 , 200 psi. Applying a safety factor of 1. 6 , the working stress
is 2800 psi.
F
A , or
40 , 000 lb
lb
in
2
22. 86 in
2
, d
2
29. 1 in
2
d 5. 39 in.
0
F
A
T
i
F
A
0
l
l
ln
f
T
i
l
l
E
E 2 G 1
G (^ )
n
T T
K
1/
C
E
a
K^ Y^^ ^ a
da
dN
A K
m
2 /2 2 /
0
/
m m
c
f (^) m m m
a a
N
m AY
1
exp
n C
d Q
K
dN RT
3
( log ) 10 r
LM T C t
% 1 exp 0. 25 ( E )
2
