T.C.
DOKUZ EYLÜL ÜNİVERSİTESİ
MÜHENDİSLİK FAKÜLTESİ
METALURJİ VE MALZEME MÜHENDİSLİĞİ BÖLÜMÜ
Name:
Learning outcomes
questions
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MME 2001
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MIDTERM II
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3.12.2014
90 minutes
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4
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13
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Total grade
100
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1.
The correct order of the coordination number in SC, BCC, FCC and HCP unit cells is
(A) 12, 8, 12, 6.
(B) 6, 8, 12, 12.
(C) 8, 6, 12, 12.
(D) 6, 12, 12, 8.
2.
Frenkel and Schottky imperfections are
(A) dislocations in ionic crystals.
(B) Grain boundaries in covalent crystals.
(C) Vacancies in ionic crystals.
(D) Vacancies in covalent crystals.
3.
Metallic bond is not characterized by
(A) ductility.
(B) high conductivity.
4.
(C) directionality.
15
9
(D) opacity.
z
The direction indices for the crystal direction indicated by the vector is
(3 points)
__
[661]
1/3
1/2
y
x
5.
The miller indices for the marked plane is,
(3 points)
_
(324)
z
1/2 y
x
6.
2/3
When monochromatic x-radiation having a wavelength of 0.1028 nm is focused on a metal with an FCC
structure, the angle of diffraction (2θ) for the (311) set of planes in this metal occurs at 71.2 degrees (for the
first order reflection n=1).
a.Calculate the interplanar spacing for this set of planes.
b.Calculate the lattice parameter for this metal.
7.
Niobium has an atomic radius of 0.1430 nm and a density of 8.57 g/cm3. Determine whether it has an FCC or a
BCC crystal structure. Show your work.
Cannot be solved without the atomic weight information!
8.
An Fe-C alloy that has 0.20 wt% carbon is to be treated at 950C. If the concentration of Carbon at the surface is
suddenly brought to and maintained at 1.0 wt%, how long will it take to achieve a carbon content of 0.78 wt%
0.2mm below the surface? The diffusion coefficient for carbon in Fe at this temperature is 1.6x10-11m2/s.
assume that the steel piece is semi-infinite.
Non steady state diffusion problem
C
=
0.20 wt% C
0
C
=
1.00 wt% C
Cx
=
0.78 wt% C
x
=
0.5mm = 5.10 m
D
=
1.6. 10 m /s
s
-11
-4
2
2
0.78 - 0.20
1.00 - 0.20
25
0.275
erf (z) = 0.275 olursa; z= 0.25
0.25 =
9.
25
t = 12345 s = 3.43 h
Cite the primary differences between elastic, anelastic, and plastic deformation behaviors.
Elastic deformation is time-independent and nonpermanent.
anelastic deformation is time-dependent and nonpermanent.
plastic deformation is permanent.
10.
estimate from the stress-strain plot below a) the yield strength, b) the tensile strength, c) the fracture stress,
d) the Young’s modulus, e) elastic deformation and f) the plastic deformation.
a)
b)
c)
d)
e)
f)
yield strength:  540 MPa
Tensile strength:  700 MPa
Fracture stress:  550 MPa
Young’s Modulus:  50 GPa
Elastic deformation:  0.012
Plastic deformation: 0.16-0.012=0.148
11.
A cylindrical specimen of brass having a diameter of 19 mm and a length of 200 mm is deformed elastically in
tension with a force of 48800 N. E= 97 GPa. Poisson’s ratio=0.34. Compute how much this sample will elongate
in the direction of the applied stress; and the change in the diameter of the specimen.
= F/A0
 = l/l0
l = 4 F l0 / E  d02
= E
 l = [4(48800)(200x10-3)]/[ (19x10-3)2(97x109)]
= 3.55 x 10-4 = 0.36 mm
the change in the diameter of the specimen.
 = - x/z = - (d/d0) / (l/l0)
d =-(.l. d0)/l0 =-(0.33)(0.36)(19)/200 =-0.011 mm
12.
stress-strain curves of three different metals are shown below: (a) Which is the most and which is the least
ductile of these three metals. (b) Which has the highest toughness?

(a) The most ductile: C
The least ductile: A
(b) The highest toughness: B
A
B
C

13.
(a) Which will experience the greatest percent reduction in area? Why?
(b) Which is the strongest? Why?
(c) Which is the stiffest? Why?
(d) which is the most resilient? Why?
material
Young’s Modulus
(GPa)
Fracture
(MPa)
strength
Tensile
(MPa)
strength
Yield
(MPa)
strength
Strain at fracture
A
350
650
Fractures before yielding
B
150
105
120
100
0.40
C
310
500
550
415
0.15
D
210
265
340
310
0.23
E
210
720
850
700
0.14
(a) Material B will experience the greatest percent area reduction since it has the highest strain at fracture, and,
therefore is most ductile.
(b) Material E is the strongest because it has the highest yield and tensile strengths.
(c) Material A is the stiffest because it has the highest elastic modulus.
(d) Material C is the most resilient because it has the highest yield strength and the Young’s Modulus.
14.
A true stress of 415 MPa produces in an alloy a true strain of 0.10; (a) What is the strain hardening exponent of
this alloy? (b) This alloy undergoes necking starting at a true strain of 0.15. Calculate its true tensile stress?
(K=1035 MPa.)
= 1035 (0.15)0.4 = 484.6 MPa
A cylindrical specimen of metal A with an original diameter of 15 mm was tested in tension and fractured with a
final diameter of 12 mm. Its engineering stress-strain curve is shown below. Estimate the true yield and tensile
strengths, true yield and tensile strains and the Resilience Module of this metal.
True Yield strength & strain
YT = (1+) = 450 (1+0.005) = 452.25 MPa
YT = ln(1+) = ln (1+0.005) = 0.00499
600
500
True tensile strength and strain
T = (1+) = 480 (1+0.010) = 485 MPa
T = ln(1+) = ln (1+0.010) = 0.00995
Resilience Module:
Ur = Y2/2E = (452x106N/m2)2/[2(100x109N/m2)]
= 1.02x105 J/m3
stress (MPa)
15.
400
300
200
100
0
0,000
0,005
strain
T = K Tn
Ur = Y2/2E
T =  (1 + )
n  = 2 d sin
d = a / (h2+k2+l2)
 = - x/z
T = ln(1+)
0,010
0,015