Student Name ………………………… Student Number …………………
DEPARTMENT OF MECHANICAL ENGINEERING
MECH 473 Ferrous and Non-Ferrous Materials
1 hour and 15 minutes Mid-term Test No.1 Monday June 5, 2006
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1. Calculate the theoretical density of NaCl given that the atomic mass of Na is 22.99
g/mole and Cl is 35.45 g/mole and that the Na ion radius is 1.26 A and Cl ion radius is
1.81 A and Avogadro’s no. is 6.02x10exp(23) atoms/mol.
ao  2rNa 1  2rCl 1  21.26  21.81  6.14 A
o

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4  22.99  4  35.45  1.68 g/cm 3
mass

3
volume
6.14 10 8 6.02 10 23



2. How is the modulus of elasticity related to the melting point of a material? What is the
long range attractive energy between the positive and negative charges of the atoms?
What is the relationship between the coefficient of linear thermal expansion and the
elastic modulus? What is the relationship between the coefficient of linear thermal
expansion and the absolute melting temperature? State whether the interatomic binding
energy of the four atomic bonding mechanisms is strong, medium or weak.
The modulus of elasticity, E, is linearly related to the absolute melting point, Tm, of a
material by
E
100 kTm

where  is the atomic volume. (1/2)
The long range attractive energy between the positive and negative charges of the atom is
Coulombic. (1/2)
There is an inverse relationship between the modulus of elasticity and the melting
temperature with the coefficient of linear thermal expansion given by

 G  Cv

G
3E
100TM
where  is the density of the material, Cv is the volumetric specific heat and G is the
Gruneisen’s constant. The larger E and Tm, the smaller is the linear thermal expansion.
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Metal – medium (1/2)
Ionic – strong (1/2)
Covalent – strong (1/2)
Van der Waals – weak (1/2)
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3. Sketch the following planes and directions within the hexagonal unit cells.
a) 2110, b) 1121, c) 10 1 0, d) (1210), e) (0002), f) (01 1 1)
Directions
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Plane
4. What are the Miller indices of the four directions of a form <111> that lie in the (1 01)
plane of a cubic cell? If each <111> on a {110} represents a slip system of a dislocation,
how many slip systems will exist in a material having a BCC structure? Show if it is
possible, or not, for two dislocations to interact to form one dislocation in order to work
harden the material that has a BCC crystal structure?
(3 marks – one for (1 01) plane and 2 for
four directions)
There are six {110} planes having four <111> directions giving 24 slip systems. (1
mark)
A BCC material cannot work harden because the dislocations, which have <111>{110},
do not interact in an energetically favorable manner, e.g.,
111  1 1 1  202
111  1 1 1   0
111  1 1 1   020
These interactions never produce a dislocation having <111>, which is the close-packed
direction in BCC materials. (2 marks)
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5. As a rule of thumb, at what temperature range should a material be used in order to
neglect a) creep effects and b) general environmental effects on its mechanical properties.
For creep effects only, the temperature should be less than T/Tm = 0.4 to 0.5
For general environmental effects (fatigue, creep, corrosion, etc), the temperature should
be less than T/Tm = 0.3
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6. What would happen to the thermal conductivity of a material if the mean free path of
the electron inside the material should increase? What effect does solute atoms have in
materials on the thermal conductivity?
The thermal conductivity would increase if the electron’s mean free path should increase.
Solute atoms (and defects) scatter electrons, which reduces their mean free path and thus
the thermal conductivity of the material.
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7. In metals, heat is conducted by electrons. What conducts heat in the other types of
materials such as ceramics, semiconductors, glasses and polymers? What is an
approximate difference in heat conduction between the metals and the other materials
and state why is there this difference?
Phonons conduct heat in these materials, which is about 2 orders of magnitude slower
than in the metals because the elastic wave speed (~10exp(3) m/s) of the phonon is less
than the electron velocity in metals (~10exp(5) m/s).
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8. What is the failure mechanism(s) for a) metals, b) ceramics and glasses and c)
polymers (2 possible).
a) metals – dislocations (1/2)
b) ceramics and glasses – cracks (1/2)
c) polymers – cracks (hard polymers) and glide of polymer chains (soft polymers) (1)
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9. List five (5) mechanisms that we can use to strengthen or increase the hardness of a
material and give a brief description of each. Indicate two mechanisms that are also
effective in reducing high-temperature creep.
1) Strain hardening occurs by cold working a material to introduce dislocations, which
act as barrier to the movement of other dislocations. (1)
2) Solid solution hardening, which involves adding solutes, alloying addition or foreign
atoms to the material, which block the movement of dislocations, increase its strength
and its resistance to high temperature creep. (1.5)
3) Secondary phases involves adding an alloying addition to induce multiple phases such
as the alpha and beta phases in brass, which block the movement of dislocations at their
phase boundaries increasing the materials strength. (1)
4) Dispersion hardening (aging) involves the formation of precipitates within a grain,
block the movement of dislocations, increase its strength and its resistance to high
temperature creep. (1.5)
5) Grain refinement involves reducing the size of the grains by rapidly cooling the
material from a high temperature or by recrystallization during annealing after cold work.
The grain boundaries block the movement of dislocations, which increases its strength
but not the material resistance against high temperature creep. (1)
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10. A ceramic matrix composite contains internal flaws as large as 0.001 cm in length.
The plain strain fracture toughness of the composite is 45 MPa sqrt(m) and the tensile
strength is 550 MPa. Will the flaw cause the composite to fail before the tensile strength
is reached? Assume f = 1.
Since the crack is internal, 2a = 0.001 cm = 0.00001 m. Therefore a = 0.000005 m.
K Ic  f a or   K Ic / f a


   MPa m /(1)  .005 m)  11,354 MPa
The applied stress required for the crack to cause failure is much larger than the tensile
strength of 550 MPa. Failure of the ceramic should be expected due to the massive
overload and not because of the presence of the flaws.
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11. For a material component that is exposed to cyclical stress, if the nominal stress is
equal to the ultimate stress, how soon will the component fail? At a nominal stress is
below the yield point and above the endurance limit, will the component fail? In ironbased materials, what is the design parameter used by engineers below which a nominal
stress will not fatigue the material component. In FCC materials such as Al, Cu, and Ni
how many cycles are often used to specify the design limit for the fatigue strength.
The component will fail after the first half cycle.
Yes, the component will eventually fail although it may be after a large number of cycles.
The fatigue limit or the endurance limit.
107 cycles
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12. Bonus – What are four material phenomena that plasma crystals grown in space can
be useful to study?
Plasma crystals give insight in physical phenomena, which is otherwise unobservable on earth,
such as (five given)
• Study the transition between solid – liquid – gas – plasma states of matter.
• Study the formation of defects
• Study the phase separation (different crystal structures formed by small and large atoms)
• Study the convection of fluids at interfaces
• Study the flow of heat through a crystal via phonons (lattice vibrations)