MCEN–2024, Sp'08
SuperQuiz 2 (SQ2) with Solutions
04/04/08
I. Questions related to Elastic Modulus and Intermolecular Bonding
1. Covalently bonded materials, such as carbon (with diamond structure) have a higher
elastic modulus than metals, like copper, because:
A. Covalent materials usually have a higher coordination number (that is, more nearest
neighbors), than do metals.
B. Covalent materials usually have a different kind of bonds between atoms than metals.
C. The stiffness of the intermolecular bonds between atoms is greater in covalent materials
than in metals.
C: The first answer is not correct because covalent bonds are directional which reduces the number of
nearest neighbors (in metals where the bonding is non–directional, the atoms have the highest possible
number of nearest neighbors, which is 12. The second answer is non–specific. The third answer is
correct because despite smaller number of bonds formed in covalent materials their elastic modulus is
higher.
2. The quantity
in the equation
, in crystalline materials, normally lies in the
following range:
A. 0.1 to 1.0 nm
B. 0.01 to 0.1 nm
C. 1 to 10 nm.
A: since ro is the interatomic spacing, which typlically lies in this range for nearly all elements.
3. Polymers have a lower elastic modulus than metals, because (check the most appropriate
answer):
A. carbon and hydrogen are common constituents in polymers.
B. because hydrogen can form only one bond with the carbon atoms.
C. because polymers have a predominantly random rather than a predominantly crystalline
structure.
B: the elastic modulus is related to the number of bonds per atom, and the stiff of the bonds. Since
hydrogen forms only one bond, the number of bonds per atom in organic polymers is lower than in
inorganic materials, which lowers their elastic modulus.
4. The heat–of–sublimation (the energy required to evaporate a solid into the vapor phase),
is related to the S, the stiffness of the spring constant between neighboring atoms, the
volume occupied by an atom in the solid, , and the number of nearest neighbors, Z, in the
following form:
where
is a dimensionless constant, and NA is the Avogadro's number. What is the
correct value for m (hint: compare units on both sides of the equation)?
A. m=1/3
B. m=2/3
C. m=1.0
B. The answer follows from a consideration of the units. The first three parameters in the equation are
dimensionless. S has dimensions of a spring constant, which is Nm-1. The left hand side has units of Joules
or Nm. Since Ω has units of volume, that is m3, the power "m" must be equal to 2/3.
II. Questions related to Young's Modulus, Shear Modulus, Poisson's Ratio, and the
Melting Temperature
5. A uniaxial specimen, having a gage length of 25mm, and a diameter of 5mm is pulled in
tension in the elastic regime. The sample is stretched in tension by a displacement of
0.25mm. As a result the diameter of the specimen contracts by 0.025mm. What is the
Poisson's ratio for this material?
A. 0.2
B. 0.3
C. 0.5
The Poisson's ratio is equal to the transverse strain divided by the longitudinal strain. The transverse strain is
0.025/5 = 0.005. The longitudinal strain is 0.25/25 = 0.01. Therefore the Poisson's ratio is 0.5.
6. A uniaxial specimen, having a gage length of 25mm, and a diameter of 5mm is pulled in
tension in the elastic regime. The sample is stretched in tension by a displacement of
0.25mm. As a result the diameter of the specimen contracts by 0.025mm (same as in Prob.
5). What is the change in the volume of the specimen?
A. The volume increases.
B. The volume decreases.
C. The volume remains unchanged.
C: since the volume remains constant if the Poisson's Ratio = 0.5
III. Questions related to Materials–Selection based upon Elastic Properties
7. You wish to select a material for a beam with a square cross-section in order to minimize
the weight of the beam. The Figure–of–Merit for this selection is given by:
.
The value for this FOM, in relative units, is 0.31 for GFRP (fiber glass), 0.17 for wood, and
0.41 for polyyrethane foam (taken from Table 7.2 on p. 92 in A&J). Based upon this
criterion which material will you select?
A. GFRP
B. Wood
C. Polyurethane Foam
To achieve the minimum weight it is necessary to minimize the FOM. Therefore wood would be the
best choice.
IV. Questions related to Plastic Deformation (Yield Stress etc.)
8. Which of the following statements is true?
A. The yield strength of metals is independent of temperature.
B. The yield strength of metals generally increases with plastic deformation.
C. The yield strength of metals measured in pure shear deformation is always greater than
the yield stress measured in uniaxial tension.
By elimination we can see that the correct answer is "B". We know from the horse–shoe example
that metals become more deformable at higher temperature, and we know from the lectures that the
yield strength in tension is twice the yield strength in pure shear.
9. Which of the following statements is true?
A. True strain is always greater than nominal strain, and true stress is always greater than
nominal stress.
B. True strain is always smaller than nominal strain, and true stress is always greater than
nominal stress.
C. True strain is always greater than nominal strain, and true stress is always smaller than
nominal stress.
D. True strain is always smaller than nominal strain, and true stress is always smaller than
nominal stress.
B: True stress is always greater while true strain is always smaller: you can see that by comparing the
stress–strain curves plotted in nominal stress and strain, and true stress and strain.
10. The work done in plastic deformation of a metal (as in rolling a plate into a sheet)
depends upon:
A. The true–stress, true–strain curve.
B. The speed of rolling.
C. The nominal–stress, nominal–strain curve, obtained in a tensile test.
A: the work done in plastic deformation always depends on the area under the true-stress/true=strain curve.
11. The springback, or the recovered elastic strain after plastic deformation, when the
applied load is removed, can be predicted from:
A. The stresses applied to achieve the required plastic deformation, and the elastic constants
of the material.
B. The total amount of plastic strain applied to the material.
C. The work done to deform the specimen.
A: the elastic recovery depends on the magnitude of the stress applied to the specimen, which will be equal to
the yield stress at the point of unloading.
12. Glass blowing and horse–shoeing are age old practices, but of a different character,
because (please check only one answer, the one that is the most appropriate):
A. metals lower their yield strength when heated to a high temperature.
B. metals are crystalline, while glasses are in a supercooled liquid state.
C. Crystals change from a solid to a liquid just above their melting temperature.
B: since the principal difference in their mechanical behavior arises from the difference in the
structure of crystalline metals and glasses.
V. Questions related to Materials Selection for Yield Limited Design
13. The figure of merit for yield limited design of springs, and other structures that store
elastic energy, is given by:
. What is the underlying criterion in the derivation of
this equation?
A. Minimize the weight.
B. Maximize the yield strength.
C. Maximize the stored elastic energy.
C: since the elastic energy stored is proportional to the FOM equation.
14. The FOM for material selection in the design of a pressure vessel was derived as
. The derivation of this equation was based upon:
A. Maximizing the pressure within the pressure vessel.
B. Minimizing the weight of a shell of a given radius.
C. Minimizing the weight of a shell of a given radius, which can operate at a given pressure.
C: The mass of the pressure vessel depends on the design pressure, and its radius, and the FOM given in
the question. Therefore the radius and the design pressure must be held constant when using the figure of
merit as defined above.
VI. Questions related to the Nanostructural Mechanisms of
Plastic Deformation in Crystals.
15. The fundamental mechanism of plastic deformation in crystals (as illustrated by the
sliding of atom planes over their neighbors) is related to the elastic properties of the crystal
via:
A. the shear modulus (G).
B. the Youngs Modulus (E).
C. the Poisson's Ratio ( )
A: since plastic deformation is fundamentally a shear deformation process.
16. The "ideal", that is the maximum possible yield stress of crystals is estimated to be:
A.
.
B. E /7 .
C.
.
B: as derived in class.. the ideal tensile yield strength is approximately 10% of the Young's Modulus.