The Islamic University of Gaza
Industrial Engineering Department
Engineering Materials, EIND 3301
Final Exam
Instructor: Dr. Mohammad Abuhaiba, P.E.
Exam date: 20/01/2011
Fall 2010
Final Exam
Pages 2 &3: Closed book
Exam Duration: 2 hours
Pages 4 to 12: Open Book
Name: _______________________________ ID: ________________________
Question
Grade
1
Maximum Grade
10
5
5
9
9
6
16
2.5
2.5
12
6
4
10
3
100
2
3
4
5
6
7
8
9
10
11
12
13
14
Total
Page 1 of 14
Question #1: (10 points) Fill in the blanks
1. The two fundamental types of dislocations are ____Edge_ and _Screw_.
2. Plastic deformation corresponds to the deformation of large number of _Dislocation_.
3. For a particular crystal structure, the slip plane is the plane that has the _greatest
planner density_.
4. The smallest repetitive volume which contains the complete lattice pattern of a crystal
is called _Unit Cell_.
5. The 4 Point Defects are:
 __ Vacancy
 __ Self-interstitial atom
 __Substitional impurity atom
 __ Intersitial impurity atom
6. For the iron-iron carbide phase diagram:
a. For pure iron, at room temperature the stable form of solid solution called ferrite
(or  iron) has _BCC_ crystal structure with maximum soluibilty of _0.022 at a
temperature of _727 o C_.
b. Ferrite experiences a polymorphic transformation to austenite (or  iron) at 912o C
which has _FCC_ crystal structure with maximum solubility _2.14%_ at temperature
_1147 o C_.
c. Austenite persists to 1394o C at which it reverts back to a BCC phase known as 
ferrite which finally melts at _1538o C.
d. Composition axis extends only to 6.7 wt% C at which _Cementite_ is formed along
the right vertical axis.
e. One _Eutectoid_ point at 0.76 wt % C and 727o C; at which  solid phase
transforms into  iron and cementite.
f. Ferrite is soft and __Ductile_ and cementite is __Hard__ and brittle.
Page 2 of 14
Question #2: (5 points) True or False
1. The ratio of the shear stress to the shear strain in the elastic range is known as the
modulus of elasticity. (False)
2. Alloying metals with impurity atoms (substitutional or interstitial) into solid solution is
another technique to strengthen and harden metals (True)
3. For a particular crystal structure, the slip direction is that direction in the slip plane
having the lowest linear density. (False)
4. A fine-grained material (one that has small grains) is softer than one that is coarse
grained. (False)
5. High-purity metals are almost always softer and weaker than alloys composed of the
same base metal. (True)
Question #3: (5 points) Select the most correct answer
1. If either stress or temperature is increased, which of the following combinations of
effects will result?
a. The steady-state creep rate decreases, the instantaneous strain at the time of stress
application decreases, and the rupture lifetime decreases.
b. The steady-state creep rate decreases, the instantaneous strain at the time of stress
application decreases, and the rupture lifetime increases.
c. The steady-state creep rate increases, the instantaneous strain at the time of stress
application decreases, and the rupture lifetime decreases.
d. The steady-state creep rate increases, the instantaneous strain at the time of stress
application increases, and the rupture lifetime decreases.
2. Once a system is at a state of equilibrium, a shift from equilibrium may result by
alteration of which of the following?
a. Pressure
b. Composition.
c. Temperature.
d. All of the above.
3. A eutectoid reaction involves which of the following phases?
a. One liquid and one solid
b. One liquid and two solid
c. Two liquids and one solid
d. Three solid
4. A phase diagram, shows the relationships among:
a. The temperature, the composition, and pressure in a particular alloy system under
equilibrium conditions.
b. the temperature, the composition, and maximum tensile stress of a particular alloy
system under equilibrium conditions
c. The temperature, the composition, and the phases present in a particular alloy
system under equilibrium conditions.
5. Metals such as lead and tin do not strain harden at room temperature because their recrystallization temperatures lie
a. Below room temperature
b. Above room temperature
c. None of the above.
Page 3 of 14
Question #4: (9 points)
Below is shown the lead-tin phase diagram. Find out the phase(s) present, their
composition(s), and their amount(s) for:
a. An alloy of composition 30 wt% Sn-70 wt% Pb at 200°C.
b. An alloy of composition 80 wt% Sn -20 wt% Pb at 250°C
Solution:
30 wt% Sn-70 wt% Pb at 200°C
80 wt% Sn -20 wt% Pb at 250°C
Phase
+L
L
Composition
 wt% Sn + L (56.8 wt% Sn)
L (80 wt% Sn)
= (56.8-30)/(56.8-17.7) = 68.5%
Amount
100% L
L = 31.5%
Page 4 of 14
Question #5: (9 points)
For a bronze alloy, the stress at which plastic deformation begins is 267MPa and the modulus
of elasticity is 115GPa.
a. What is the maximum load (in N) that may be applied to a specimen having a crosssectional area of 300mm2 without plastic deformation?
b. If the original specimen length is 137mm, what is the maximum length (in mm) to which
it may be stretched without causing plastic deformation?
Solution:
(a) The maximum stress that may be applied without plastic deformation taking place is
the yield strength, σy;
The load at yielding (or maximum load), Fy, is
= 80,100 N
(b) Combining Hooke's law (at the stress corresponding to the yield strength)
,
= 137.3 mm
Page 5 of 14
Question #6: (6 points)
The yield strength for an alloy that has an average grain diameter of 1.6 x 10 -2 mm is 320MPa.
At a grain diameter 3.9 x 10-3 mm, the yield strength increases to 480MPa. At what grain
diameter (in mm) will the yield strength be 441MPa?
Solution:
It is necessary to set up two simultaneous equations of the Hall-Petch form,
Now, incorporating values of σy and d, these equations become
And, from these equations it is possible to solve for values of σ 0 and ky; thus
σ0 = 164 MPa,
ky = 19.7 MPa-mm1/2
Finally, the grain diameter required to give a yield strength of 441 MPa may be determined by
rearrangement of the first equation--that is
= 0.00506 mm
Page 6 of 14
Question #7: (16 points)
Refer to the Iron-Iron carbide phase diagram shown below. Find:
a.
b.
c.
d.
e.
f.
g.
h.
i.
Show all liquidus and solidus lines on the diagram. (3)
The eutectic temperature and composition. (1)
The eutectoid temperature and composition. (1)
the main phases present for an alloy of 1% C at 1600 oC, 1400 oC, 1200 oC, 740 oC,
and 600oC. (5)
The chemical composition of each phase present for an alloy of 1% C at 1600 oC and
1400 oC. (1.5)
Amount of each phase present for an alloy of 1% C at 1200 oC and 600oC.(1.5)
Amount of pearlite for an alloy of 1% C at 1200 oC and 600oC. (1)
Locate the hypo-eutectoid and hyper-eutectoid steels in terms of carbon content
on the diagram. (1)
What is the maximum solubility limit of C in ferrite, and at which temperature it
occurs? (1)
Liquidus
Solidus
Solidus
Solution:
0.014
0.72
1.74
See the labels on the figure
The eutectic temperature and composition: 1147°C, 4.30 wt%C (1.0 point)
The eutectoid temperature and composition: 727°C, 0.76 wt% C(1.0 point)
the main phases present for an alloy of 1% C at: (5.0 point)
 1600 oC: liquid
 1400 oC:  + L
 1200 oC: 
 740 oC:  + Fe3C
 600oC:  + Fe3C
e. The chemical composition of each phase present for an alloy of 1% C at: (1.5 point)
 1600 oC: 1.0 wt% C
 1400 oC:  (0.72 wt%C+ L (1.74 wt%C)
a.
b.
c.
d.
Page 7 of 14
f. Amount of each phase present for an alloy of 1% C at: (1.5 point)
 1200 oC: 100% 
 600oC: W =100*(6.7-1)/(6.7-0.014) = 85%, WFe3C = 15%
g. Amount of pearlite for an alloy of 1% C at: 1.0 point)
 1200 oC: 0%
 600oC: Wpearlite = 100*(6.7-1)/(6.7-0.76) = 96%
h. hypo-eutectoid steels: %wt C < 0.76wt% (1.0 point)
 hyper-eutectoid steels:2.14wt%< %wt C > 0.76wt%
i. Maximum solubility limit of C in ferrite = 0.022wt%C at which 727°C (1.0 point)
Page 8 of 14
Question #8: (2.5 points)
What are the Miller indices for the plane shown below?
Z'
Solution:
X'
Shift xyz coordinate system to the right as shown.
Intercepts: (1, -1, 2/3),
reciprocals: (1, -1, 3/2),
Clear fractions: (2, -2, 3)
Plan:
Question #9: (2.5 points)
Determine the Miller indices of the direction shown in the figure.
Solution:
Tail: (2/3,2/3,0),
head: (1/2,0,1/2),
Head – Tail = (-1/6, -2/3, 1/2),
Clear Fractions: (-1, -4, 3)
Direction:
Page 9 of 14
Question #10: (12 points)
Determine the carburizing time (in s) necessary to achieve a carbon concentration of 0.44 wt%
at a position 1.9 mm into an iron-carbon alloy that initially contains 0.031 wt% C. The surface
concentration is to be maintained at 1.2 wt% C, and the treatment is to be conducted at
1020°C. Assume that D0 = 5.1 x 10-5 m2/s and Qd =154 kJ/mol.
Solution:
The time required to achieve the stipulated carburization time may be determined by solving
Fick's second law:
For the given boundary conditions, the solution is as follows:
Incorporation of composition values provided in the problem statement leads to
The value of z for which the error function is 0.6500 may be determined by using data in the
table provided in the problem statement and by linear interpolation as follows:
which leads to z = 0.6612. Thus
Or
Now, solving for the diffusion coefficient at this temperature, using data provided in the
problem statement
Incorporation of this value into the above equation, the time is equal to
= 67,900 s
Page 10 of 14
Question #11: (6 points)
A fatigue test was conducted in which the mean stress was 46.2MPa and the stress amplitude
was 219MPa.
a.
b.
c.
d.
Calculate the maximum stress level (in MPa).
Calculate the minimum stress level (in MPa).
Calculate the stress ratio.
Calculate the magnitude of the stress range (in MPa).
Solution:
Solving these two expressions simultaneously for σmax and σmin (and incorporating values of σm
and σa given in the problem statement) yields
(c) The stress ratio, R, is defined as
(d) The stress range σr is equal to
Page 11 of 14
Question #12: (4 points)
Estimate the rupture lifetime (in h) for some cylindrical component originally 10.5mm in
diameter and 496mm long that is fabricated from a low carbon-nickel alloy; assume that it is to
be exposed to a tensile load of 6300N at 538°C. The logarithm stress versus logarithm rupture
lifetime plot for this alloy is given below.
Solution:
For a cylindrical specimen that is loaded in tension, load and stress are related by the equation
And, incorporating values of F and d0 provided in the problem statement, the stress is equal to
From the 538°C line in the plot provided in the problem statement, the rupture lifetime
corresponding to 72.8MPa is about 104 h.
Page 12 of 14
Question #13: (10 points)
For some metal alloy it is known that the kinetics of re-crystallization obey the Avrami
equation, and that the value of k in the exponential is 2.73 x 10-6, for time in seconds. If, at
some temperature, the rate of re-crystallization is 0.0014 s-1, what total time (in s) is required
for the re-crystallization reaction to go to 90% completion?
Solution:
Since the rate of transformation r is defined as:
using the specified rate value it is possible to solve for t0.5 ( the time required for the
transformation to go to 50% completion), as
Using the Avrami equation--i.e.,
at the point of 50% transformation completion, this equation takes the form
And, solving for n leads to
which, upon entering the value t0.5 computed above as well as the value of k provided in the
problem statement yields the following value for n:
Finally, using this n value, the time t required for the second degree of re-crystallization may
be determined using another rearranged form of the original Avrami equation as
Page 13 of 14
Question #14: (3 points)
Using the continuous cooling transformation diagram shown below for an iron-carbon alloy of
eutectoid composition, name the micro-structural products of specimens having this eutectoid
composition that are first completely transformed to austenite, then cooled to room
temperature at the following rates:
a. 175°C/s
b. 90°C/s
c. 15°C/s
Solution:
a. At a rate of 175°C/s, only martensite forms since this rate is greater than the critical
rate of 140°C.
b. At a rate of 90°C/s, both martensite and pearlite form since this rate is less than the
critical rate (140°C) yet greater the maximum rate for formation of a totally pearlitic
structure (35°C).
c. At a rate of 15C°C/s, only pearlite forms since this rate is less than the maximum rate
for the formation of a totally pearlitic structure (35°C).
Page 14 of 14