Tests

Material Science
Fall 2008
E317: TEST 1 – Fall 2008
Name: ___________________________________________
1) (6%) List the six major categories of materials. Give an example of each.
Metals: Copper
Ceramics: Oxides
Polymers: Plastic bottles
Elastomers: Rubber
Glass: Glass
Hybrid: Composites
2) (2%) Modes of loading include Tension/Compression, Bending, Hydrostatic
Pressure and Shear.
3) (2%) (True, False) If a material is good in one mode of loading then it is good
in all of the modes of loading.
4) (10%) Sketch the stress-strain curves for the following materials. Label the
yield strength (σy) and the tensile strength (σts).
5) (6%) An Aluminum rod (E = 69 GPa, A = 80 mm2) is loaded in tension by a 4
kN force. Calculate the strain.
σ = F/A = 4000 N/(80x(10-3)2 m2) = 50 MPa
σ = Eε
ε = 50x106 / 69x109 = 7.25x10-4 mm/mm
6) (4%) Poisson’s ratio (ν) is the ratio of transverse strain over axial strain.
Material Science
Fall 2008
7) (2%) Give the electron configuration for the following elements and their ions:
Br, Fe, Br-, and Fe2+.
Br: 1s22s22p63s23p63d 104s24p5
Br-: 1s22s22p63s23p63d 104s24p6
Fe: 1s22s22p63s23p63d 64s2
Fe2+: 1s22s22p63s23p63d 6
8) (2%) Given the following electron configurations classify the element as
reactive or non-reactive.
a) 1s22s22p6 : Non-reactive
b) 1s22s1: Reactive
9) (8%) Explain the mechanism of an ionic, covalent, metallic and Van Der
Waals bonds.
Ionic Bond: Occurs only between metallic and nonmetallic elements.
The metallic elements gives up its valance electrons to the nonmetallic
element. In the process all the atoms acquire stable configurations and
become ions and an electric charge which are attracted to each other
through Coulombic Forces.
Covalent Bond: In covalent bonding, stable electron configurations are
assumed by the sharing of electrons between adjacent (similar or
dissimilar) atoms.
Metallic Bond: These valance electrons of metals are bonded relatively
loosely to the nucleus. Therefore when a number of atoms are put
together, the outer electrons leave individual atoms to enter a common
“electron cloud”. The remaining non-valence electrons and atomic
nuclei form what are called “ion cores”, which possess a net positive
charge.
Van Der Waals Bond: Secondary bonding that exists between virtually
all atoms or molecules. It arises from atomic or molecular dipoles and
the attraction between the oppositely charged ends of these dipoles.
10) (6%) What type of bond would you expect for each of the following materials.
a) Gold : Metallic
b) Rubber : Covalent / Hydrogen
c) Methane (CH4)
C: Nonmetal
H: Nonmetal
Covalent Bond
Material Science
Fall 2008
11) (4%) Why are metals electrically conductive?
High electrical and thermal conductivity, because the free electrons in
the electron cloud can move readily giving a current.
12) (2%) True or False? The bonds in most ionically and covalently bonded
materials are purely ionic or purely covalent.
13) (6%) Calculate the % Ionic Character (if appropriate).
% Ionic Character = {1-Exp[-.25(XA – XB)2]}*100
a) Iron Oxide (FeO)
XFe = 1.8
XO = 3.5
%IC = 51.4% ⇒ Ionic Bond
b) Brass (CuZn)
Metallic Bond
14) (6%) A hypothetical metal has a BCC crystal structure and a density of 8.22
g/cm3 and an atomic weight of 77.4 g/mol. What is the volume of its unit cell.
ρ = nA / VC N A
NA = Avogadro’s Number = 6.0223x1023 atoms/mol
n=2
ρ = nA / VC N A = 2 atoms (77.4 g/mol) / (VC 6.0223x1023 atoms/mol) = 8.22 g/cm3
Vuc = 31.27x10-24 cm3 = 31.27 Å3
15) (2%) (True, False) Glass is a crystalline solid.
16) (2%) (True, False) Polymers are made up of long organic chains.
17) (2%) Thermoplastics have (no, some, many) cross-links.
18) (2%) Thermosets have (no, some, many) cross-links.
19) (2%) Elastomers have (no, some, many) cross-links.
20) (4%)Why did people turn to iron if it is softer than bronze? It is more
plentiful.
21) (4%) Steel is basically an alloy between Iron and Carbon.
22) (4%) A super fluid has zero viscosity at absolute zero.
Material Science
Fall 2008
23) (4%) The north pole is the source of a magnet and the south pole is the sink
of the magnet.
24) (4%) EVA is a polymer that is used in mid sole of shoes.
25) (4%) Choice of material is the most important design consideration for
artificial joints.
E317: TEST#2 (Fall 2008)
Problem 1: (8%) Draw the stress distribution for each of the loading conditions.
Problem 2: (8%) For a bronze alloy, the stress at which plastic deformation begins is
275 MPa and the modulus of elasticity is 115 GPa.
a) (3%) What is the maximum load that may be applied to a specimen with a crosssectional area of 325 mm2 without plastic deformation?
σ=
F
A
275 = F / 325
F = 89.4 kN
b) (3%) If the original specimen length is 115 mm, what is the maximum length to
which it may be stretched without causing plastic deformation?
275 x 106 = 115x109 ε
σ = Eε
ε=
li − l o Δ l
=
lo
lo
.00239 = (lf – 115)/ 115
ε = .00239
lf = 115.27 mm
Problem 3: (8%) Derive the material index for a
pined-pined square beam of length L. We want to
minimize weight without sacrificing stiffness.
Consider a beam of length L that carries a force F,
deflects by a distance δ and operates in the elastic
range. The area A is free to vary and the material
properties (ρ, E) are free to vary.
Minimize mass:
Stiffness:
m = ALoρ
What will decrease the mass? m decreases as A or ρ decreases.
S = F/δ
δ= L3F/48EI (δ depends on the boundary conditions)
S = 48EI/L3
What will increase stiffness? S increases as I = A2/12 or E increases.
A conflicts with our two goals. Let’s eliminate it from the equation.
1/ 2
m ⎛ 12 SL3 ⎞
=⎜
⎟
Lρ ⎝ 48 E ⎠
1/ 2
Lρ ⎛ SL3 ⎞
m=
⎜
⎟
2 ⎝ E ⎠
=
S 1/ 2 L5/2 ⎛ ρ ⎞
⎜
⎟
2 ⎝ E1/ 2 ⎠
The stiffness and length are not free to vary. Therefore, we want to decrease (ρ/E1/2) or
increase (E1/2/ρ).
Material index for bending beam: Mb = E1/2/ρ
Problem 4: (16%) The following stress-strain graph is the result of a tensile test
performed on a 0.505 in diameter aluminum alloy test bar.
a) What is the yield strength? 35000 psi
b) What is the tensile strength? 40000 psi
c) What is the modulus of elasticity? E = 35000/0.0035 = 10000 ksi
d) What is the 0.2% offset proof stress? 35500 psi
e) What is the total strain right before fracture? 0.103
f) What is the plastic strain after fracture? εplas = εtotal - εelastic = 0.103 – 0.0035 =
0.0995
Stress ‐ Strain
45000
40000
35000
Stress (psi)
30000
25000
20000
15000
10000
5000
0
0
0.02
0.04
0.06
0.08
0.1
Strain (ε)
Problem 5: (3%) If you are testing the hardness of 2 materials, the depth of the
indentation of the harder material will be (less, more) than the indentation of the softer
one. Circle the correct answer.
Problem 6: (3%) To make crystalline materials stronger, we need to make it harder for
dislocations to move.
Problem 7: (5%) Match the following strengthening mechanisms with their definitions.
Grain Boundaries
Addition of foreign atoms that go into
either substitutional or interstitial sites.
Solid Solution Strengthening
Both metals are molten and upon
solidification the solute forms small
particles.
Dispersion Strengthening
Where crystal meet with different
orientations.
Precipitate Strengthening
Particles with high melting temperatures
are added to the liquid metal.
Problem 8: (5%) When a metal is cold worked, the strength increases and the
ductility decreases.
Problem 9: (5%) Two previously undeformed cylindrical specimens of an alloy are to
be strain hardened by reducing their cross-sectional areas. For one specimen, the initial
and deformed radii are 16 mm and 11 mm, respectively. The second specimen has an
initial radius of 12 mm. If the two specimens must have the same deformed hardness
compute the second specimen’s deformed radius.
%CW =
( Ao − A f )
Ao
First Specimen:
*100%
%CW =
Second Specimen: %CW =
(16 2 − 112 )
* 100% = 52.73%
16 2
(12 2 − r f2 )
12 2
* 100% = 52.73%
rf = 8.25 mm
Problem 10: (5%) Annealing increases ductility and decreases strength.
Problem 11: (5%) A bar having a square cross section of 0.1 x 0.1 in is made of a steel
that is assumed to be elastoplastic with E = 30x106 psi and σy = 60 ksi. Determine the
bending moment at which
a) Yield first occurs,
b) When deformation is fully plastic
I = bh3/12 = 8.33x10-6 in4
σ max =
Mymax
I
M=
σyI
ymax
M = 10 lb in
Zp = bh2/4 = 2.5x10-4 in3
M = Zp σy = 15 lb in
Problem 12: (8%) Derive the material index for a rod in tension where we want to
minimize weight without sacrificing yield strength.
The rod must carry a certain load
without yielding. Consider a rod of
length L that carries a force F and
extends by a distance δ. The area A is
free to vary and the density ρ and yield
strength σy are free to vary.
Minimize mass:
Strength:
m = ALρ
What will decrease the mass? m decreases as A or ρ decreases.
σ = (F/A) < σy
What will decrease stress? σ decreased as A increases.
Α conflicts with our two goals. Let’s eliminate it from the equation.
m
F
≥
Lρ σ y
m≥
FLρ
σy
The force and length are not free to vary. Therefore, we want to decrease (ρ/σy) or
increase (σy /ρ).
Material index for tension or tie: Mt = σy /ρ
Problem 13: (3%) Rebar increases concrete’s tensile strength.
Problem 14: (3%) Concrete emits heat when it cures.
Problem 15: (3%) Shape memory alloys remember their crystal structure.
Problem 16: (3%) The inner liner of a helmet is made of a (metal, polymer, ceramic)
that is good at absorbing shock.
Problem 17: (3%) (True, False) Plastic bikes are not popular because they are bulky
and heavy.
Problem 18: (3%) Ferrofluids are fluids that contain nanoscale particles that usually
contain iron.
Problem 19: (3%) This fiber is small, strong, light weight and has a low thermal
expansion. Carbon fiber
E317: TEST#3
Name: _______________________________________
1) (%) Strength is the resistance to plastic deformation.
2) (%) Toughness is the resistance to crack propagation.
3) (%) What are the two modes of fracture? Give an example of a material that under
normal circumstances undergoes that particular mode of fracture.
Ductile Fracture: Extensive plastic deformation around the crack, high energy
absorption and the crack proceeds relatively slowly. (Stable) Metals.
Brittle Fracture: Cracks spread extremely rapidly, with very little plastic
deformation and low energy absorption. Once the crack is started it will
continue without an increase in the stress magnitude. (Unstable) Ceramics
and glasses.
4) (%) What is the transition temperature when speaking about impact testing?
It is the temperature where the material transitions from ductile to brittle
failure.
5) (%) Fatigue failure cannot occur if the applied stress is below the material’s yield
Stress. (True, False).
6) (%) Increasing the mean stress does not affect fatigue life. (True, False)
7) (%) The SN curve shown was generated using a zero mean stress. Based on this
curved answer the following questions.
a. What is the endurance limit of the tool steel? 60 ksi
b. What is the endurance limit of the aluminum? 29 ksi
c. If each material is cycle at 40 ksi for 106 cycles, will both materials fail, will
only one material fail, or will both materials not fail. Specify which
material. The aluminum will fail but the tool steel will not fail.
8) (%) Explain why friction is independent of the apparent or nominal contact area of
the touching surfaces.
Surfaces are not smooth but bumpy and consist of many asperities. When
two surfaces come into contact they touch at relatively few points. The
friction force is related to this actual or real contact area, not the apparent or
nominal contact area.
9) (%) The adhesion theory of friction states that the friction force is related to the shear
strength of the weld junctions created between the two contacting surfaces. (True,
False)
10) (%) Abrasive wear occurs when a harder material digs into a softer material.
Adhesive wear occurs when two similar materials stick together and shear off.
11) (%) A bi-metallic strip deflects when heated because the two metals have different
thermal expansion coefficients.
12) (%) Use Fourier’s law to calculate the rate of heat transfer through the bottom of a
copper pan used to heat water on an electric stove. The thermal conductivity of
copper is 320 W/moC, the thickness of the pan is 3 mm and the temperature
difference between the heated side and water side of the pan is 9oC.
13) (%) Rank the thermal conductivity of a gas, a solid and a liquid in order from low to
high.
λgas < λliquid < λsolid
14) (%) The thermal conductivity of a material is due to two effects; lattice vibrations (λl)
and migration of free electrons (λe).
a. In metals, which component dominates? (λe)
b. In ceramics, which component dominates? (λl)
15) (%) Label the 3 stages of creep that occur before rupture on the figure shown.
16) (%) Name 2 properties that increase the creep resistance of a material.
a. High melting temperature
b. Alloying to interfere with dislocation motion
c. Increased grain size to decrease diffusion
d. Increased modulus of elasticity
e. Heavy cross-linking
17) (%) Auxetic materials have a negative Poisson’s ratio and expand transversely
when pull axially.
18) (%) The common name for solid CO2. Dry Ice
19) (%) The common name for Polyvinyl Butyral laminated glass. Bullet Proof Glass
20) (%) Adhesives are made of polymers.
21) (%)The goal of insulation is to slow down heat flow by trapping air within a solid.
22) Name 3 materials that were used as early insulators.
Mineral wool, Sod, Straw.
Asbestos, Cork, Snow,
E317: FINAL EXAM
Name: ______________________________
1) Fill in the missing words in the figure caption.
Figure: Charge carriers in different materials:
a) Valence electrons in the metallic bond move easily.
b) Covalent bonds must be broken in semiconductors and insulators for an electron
to be able to move.
c) Entire ions must diffuse to carry charge in many ionically bonded materials.
2) A current of 5 A is passed through a 0.5 mm diameter wire 10 m long. Calculate the
power loss if the wire is made of Copper (conductivity = 5.98 x 105 Ohm-1cm-1)
(P = IV, V = IR, R = ρeL/A)
P = I2ρeL/A
PCu = (5 A)2 (1/5.98x105)Ωcm (1m/100cm) 10 m / (π (0.0005 m)2/4) = 21.3 W
3) Increasing temperature increases conductivity. (true, false)
4) Alloying a pure metal decreases conductivity. (true, false)
5) If you need to increase the strength of a pure metal and still retain high conductivity
would you cold work (work harden) the metal or use solid solution hardening?
Work harden the metal.
6) Explain the two step oxidation processes.
When any metals (except inert metals) are exposed to the air an ultra-thin oxide
layer forms immediately.
M(metal) + O(Oxygen) → MO(oxide) + energy
The oxide layer is created in two steps:
Step 1) The metal first forms a metal ion by losing electrons.
Oxidation reaction (Anode): M → M n + + ne −
Step 2) The electrons are absorbed by, for example, the oxygen producing an
oxygen ion.
Reduction reaction (Cathode): (Pure Oxygen) O + ne − → O n −
7) Galvanic corrosion occurs when two dissimilar metals come in contact.
8) Piezoelectric materials create a voltage when stressed.
9) This type of material comes in closed cell and open cell configurations.
foams
Metal
10) These devices have been used in clocks, thermostats, toasters, and thermometers.
They are cheap to produce, but not very accurate. Bi-metallic strips
11) (6%) An Aluminum rod (E = 69 GPa, A = 80 mm2) is loaded in tension by a 4 kN
force. Calculate the strain.
σ = F/A = 4000 N/(80x(10-3)2 m2) = 50 MPa
σ = Eε
ε = 50x106 / 69x109 = 7.25x10-4 mm/mm
12) (8%) Explain the mechanism of an ionic, covalent, metallic and Van Der Waals
bonds.
Ionic Bond: Occurs only between metallic and nonmetallic elements. The
metallic elements gives up its valance electrons to the nonmetallic element. In
the process all the atoms acquire stable configurations and become ions and
an electric charge which are attracted to each other through Coulombic
Forces.
Covalent Bond: In covalent bonding, stable electron configurations are
assumed by the sharing of electrons between adjacent (similar or dissimilar)
atoms.
Metallic Bond: These valance electrons of metals are bonded relatively loosely
to the nucleus. Therefore when a number of atoms are put together, the outer
electrons leave individual atoms to enter a common “electron cloud”. The
remaining non-valence electrons and atomic nuclei form what are called “ion
cores”, which possess a net positive charge.
Van Der Waals Bond: Secondary bonding that exists between virtually all
atoms or molecules. It arises from atomic or molecular dipoles and the
attraction between the oppositely charged ends of these dipoles.
13) (8%) Derive the material index for a pined-pined square beam of length L. We want
to minimize weight without sacrificing stiffness.
Consider a beam of length L that carries a force F, deflects by a distance δ and
operates in the elastic range. The area A is free to vary and the material properties (ρ,
E) are free to vary.
Minimize mass:
Stiffness:
m = ALoρ
What will decrease the mass? m decreases as A or ρ decreases.
S = F/δ
δ= L3F/48EI (δ depends on the boundary conditions)
S = 48EI/L3
What will increase stiffness? S increases as I = A2/12 or E increases.
A conflicts with our two goals. Let’s eliminate it from the equation.
1/ 2
m ⎛ 12 SL3 ⎞
=⎜
⎟
Lρ ⎝ 48 E ⎠
1/ 2
Lρ ⎛ SL3 ⎞
m=
⎜
⎟
2 ⎝ E ⎠
=
S 1/ 2 L5/2 ⎛ ρ ⎞
⎜
⎟
2 ⎝ E1/ 2 ⎠
The stiffness and length are not free to vary. Therefore, we want to decrease (ρ/E1/2) or
increase (E1/2/ρ).
Material index for bending beam: Mb = E1/2/ρ
14) (16%) The following stress-strain graph is the result of a tensile test performed on a
0.505 in diameter aluminum alloy test bar.
a) What is the yield strength? 35000 psi
b) What is the tensile strength? 40000 psi
c) What is the modulus of elasticity? E = 35000/0.0035 = 10000 ksi
d) What is the 0.2% offset proof stress? 35500 psi
e) What is the total strain right before fracture? 0.103
f) What is the plastic strain after fracture? εplas = εtotal - εelastic = 0.103 – 0.0035 =
0.0995
Stress ‐ Strain
45000
40000
35000
Stress (psi)
30000
25000
20000
15000
10000
5000
0
0
0.02
0.04
0.06
0.08
0.1
Strain (ε)
15) (3%) To make crystalline materials stronger, we need to make it harder for
dislocations to move.
16) (5%) When a metal is cold worked, the strength increases and the ductility
decreases.
17) (5%) Annealing increases ductility and decreases strength.
18) (%) Strength is the resistance to plastic deformation.
19) (%) Toughness is the resistance to crack propagation.
20) (%) Fatigue failure cannot occur if the applied stress is below the material’s yield
Stress. (True, False).

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