Devin Gatherwright
10/12/2012
IET 307: Materials Science
HW 5 (based on chapter 8), Due by 11.55 PM, Sunday, October 21’st, 2012
100 points
1. What factors would you consider in selecting materials for coins based on the theory of fatigue?
Suggest materials for this application. (10 points)
Answer:
The factors I would consider for selecting materials for coins based on what I have learned about the theory
of fatigue would be as follows: The ability of the material to handle a substantial amount of stress; the
ability of the material to be resistant to physical wear; the ability of the material to be resistant to thermal
and corrosion fatigue. I believe that the ability of the material to handle stress is important due to the fact
that coins created by a stamp or press that prints the design on the material. Also, the fact that coins are
handled as currency means they should be able to weather at least some amount of stress without fracturing
or falling apart. I also believe that the ability of a material to be resistant to physical wear such as cracks,
scuffs, and scratches is important due to the simple fact that as currency, the material will be inevitably
roughed up due to circulation. I think it is important for a material to be able to withstand such physical
wear without fracture. Lastly, I believe that a material should be resistant to both thermal and corrosion
fatigue due to the fact that coins are exposed to varying climates and temperatures through circulation. I
believe it is imperative for coins to be able to effectively combat these fatigue fractures successfully. Some
of the materials I would use to create coins would be brass, silver, copper, aluminum, bronze, nickel, and
gold.
2. Determine the critical crack length for a through crack in a thick plate of 7150-T651 aluminum alloy
that is in uniaxial tension. For this alloy KIC = 25.5 Mpa√𝑚 and σ = 400 MPa. Assume 𝑌 = √𝜋 (10
points)
Answer:
𝑎𝑐 =
𝑎𝑐 =
𝑎𝑐 =
1 25.5 𝑀𝑝𝑎√𝑚 2
(
)
𝜋 708981540.4
𝑎𝑐 =
𝜋
1
1 𝐾𝐼𝑐 2
( )
𝜋 𝜎𝑌
1 25.5 𝑀𝑝𝑎√𝑚 2
(
)
𝜋 400 𝑀𝑝𝑎 √𝜋
(. 0012936313√𝑚)
𝑎𝑐 = 4.12 × 10−4 √𝑚
3. Please use 𝐾𝐼𝐶 = 𝑓𝜎√𝑎𝜋 A steel plate satisfies a plane strain condition. It has KIC=85,000 psi in1/2
and f=1.12. X-ray finds that there exist cracks of up to 0.1 inch inside the steel. What is the maximum
tensile stress that the steel can withstand? (10 points)
Answer:
𝜎𝑐 =
𝜎𝑐 =
𝐾𝐼𝑐
𝑌√𝜋𝑎
85,000 𝑝𝑠𝑖 𝑖𝑛1/2
1.12√𝜋(0.1 𝑖𝑛.)
85,000 𝑝𝑠𝑖 𝑖𝑛1/2
𝜎𝑐 =
. 6277590163 𝑖𝑛.
𝜎𝑐 = 135,402.28 𝑝𝑠𝑖
4. Explain in detail brittle and ductile fracture. What is the difference in microscopic morphology
between ductile and brittle fractures? (10 points).
Answer: According to the materials science textbook, brittle fracture can be defined as a “fracture that
occurs by a rapid crack propagation and has no appreciable microscopic deformation.” Furthermore, the
crack that develops as a result of brittle fracture has a direction that is almost perpendicular to the applied
tensile stress and yields a flat fracture surface. Also, the fracture surfaces of a material that fails under
brittle fracture will have very distinctive patterns, such as lines, ridges, and “chevron” markings depending
on the type of material. Also, it should be noted that any signs indicating gross plastic deformation will not
be present. Furthermore, according to the book, most brittle crystalline materials, the crack propagation
relate to the consecutive and repeated breaking of atomic bonds along the specific crystallographic planes,
which is defined as cleavage. According to the book, cleavage is transgranular in nature, which is due to
the fact that the fracture crack passes through the grains of the material, and can be observed
macroscopically. Also, in some alloys that undergo brittle fracture, crack propagation is along the grain
boundaries of the material, which is termed intergranular. These intergranular fractures can be viewed
using a scanning electron microscope that allows the three-dimensional nature of the grains to be seen.
According to the material science textbook, ductile fracture can be defined as “a mode of fracture that is
attended by extensive gross plastic deformation.” Furthermore, according to the book, ductile fracture has
distinctive features at both macroscopic and microscopic levels. According to the textbook, in a highly
ductile fracture, the material necks down to a point fracture and shows 100% reduction in the area of the
material. However, in most cases, the fracture is preceded by a small amount of necking in the material.
This fracture occurs in three stages or steps: first, after necking begins to occur, small cavities form in the
interior of the cross section. Second, the small cavities begin to enlarge, come to together, and eventually
coalesce, forming an elliptical crack that is perpendicular to the stress direction. Lastly, fracture occurs by
the rapid propagation of the crack around the perimeter of the neck.
The difference in microscopic morphology between ductile and brittle fractures is that ductile fractures
have distinctive features at both a macroscopic and microscopic level whereas a brittle fracture has no
appreciable deformation at a microscopic level.
5. What are the three stages of fatigue? Explain each one of them in detail. (10 points)
Answer: According to the material science textbook, the three stages of fatigue are as follows: (1) crack
initiation, (2) crack propagation, and (3) final failure. In the first stage, a small crack forms at a point of
high stress concentration. Furthermore, according to the textbook, the crack that initiates (or nucleates) is
almost always on the surface of the material at some level of stress concentration. The crack nucleation site
can include surface scratches, keyways, dents, threads, and sharp fillets for example. Furthermore, some
microscopic surface discontinuities can result due to cyclic loading of the material. These surface
discontinuities can act as crack initiation sites. In the second stage, the crack advances in increments with
each stress cycle. According to the book, the fracture surface region that formed during the crack
propagation phase is characterized by two types of markings known as beachmarks and striations. Both of
these markings indicate the position where the crack tip has occurred and appear as concentric ridges that
develop away from the crack initiation site or sites in a semicircular or circular pattern. Beachmarks may
be observed with the naked eye and are generally found on components that have experienced disturbances
during the crack propagation phase. In contrast, striations can only be viewed by an electron microscope.
According to the textbook, striations represent the advance distance of a crack front that occurs during a
single load cycle. The width of the striation depends on the stress range. Likewise, if the stress range
increases, the width of the striation also increases as well. In the third and final stage, the material fails as a
result of the advancing crack reaching its critical size.
6. Draw a typical creep curve for a metal under constant load and at a relatively high temperature,
and indicate on it all three stages of creep. (10 points)
Answer:
7. Define endurance limit. Please look at the S-N curve to the right and figure out the endurance limit
for this material. Also, if we want the part to work for one million cycles, what maximum tensile stress
can we apply? (10 points)
Answer: According to the materials science textbook, endurance limit or fatigue limit as it is often also
known as can be defined as “the maximum stress amplitude level below which a material is able to endure
a more or less infinite number of stress cycles without failure.
Based on the S-N chart given, I would say that the endurance limit of this particular material is around 150
Mpa respectively because the curve starts to become horizontal at that point from what I can see. If we
need to work the part for one million cycles, the maximum stress we can apply around 220 Mpa to the
material, according to the S-N chart.
8. A polystyrene component must not fail when a tensile stress of 1.25 MPa is applied. Determine the
maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/m 2. Assume a
modulus of elasticity of 3.0 GPa. (10 points)
Answer:
𝜎𝑐 = (
𝑎=
2𝐸𝛾𝑠 1/2
)
𝜋𝑎
2𝐸𝑦𝑠
𝜋𝜎 2
𝑁
𝑎=
𝑁
(2)(3.0 ×109 2 )(0.50 )
𝑚
𝑚
𝑁
𝜋(1.25 ×106 2 )2
𝑚
𝑎=
3 × 109
4.91 × 1012
𝑎 = 6.11 × 10−4 𝑚 = 611 𝜇𝑚
9. An aluminum alloy that has a plain strain fracture toughness of 25,000 psi (in.) 1/2 fails when a
stress of 42,000 psi is applied. Observation of the fracture surface indicates that fracture began at the
surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f (or Y) = 1.1 (10
points).
Answer:
𝑎𝑐 =
𝑎𝑐 =
1 𝐾𝐼𝑐 2
( )
𝜋 𝜎𝑌
1 25,000 𝑝𝑠𝑖 𝑖𝑛.1/2 2
(
)
𝜋 (42,000𝑝𝑠𝑖)(1.1)
𝑎𝑐 =
1
(.5411255411 𝑖𝑛.1/2 )2
𝜋
𝑎𝑐 =
1
(.2928168512 𝑖𝑛. )
𝜋
𝑎𝑐 = 0.932064986 𝑖𝑛.
10. Explain the major factors that affect the fatigue strength of a metal. (10 points)
Answer: According to the materials science textbook, there are several major factors that affect the fatigue
strength of a metal and they are as follows: mean stress level; surface effects such as design factors and
surface treatments; and environmental effects such as thermal fatigue and corrosion fatigue. Mean stress
effects the fatigue strength of a metal because mean stress is a constant and increasing stress amplitude that
ultimately decreases the fatigue life and/or fatigue strength of a material. The mean stress level is
represented on a typical S-N plot, according to the textbook. Furthermore, according to the textbook, if you
desire to increase the fatigue strength/fatigue life of a metal, you need to reduce the mean stress level.
Surface effects are cracks that appear on the surface of the material at stress amplification sites, which
eventually lead to the failure of the material that is being loaded. In order to combat these cracks and
ultimately improve fatigue resistance, the various design factors of the material and surface treatments
must be managed. According to the textbook, geometrical discontinuities on the material surface such as
notches, holes, or grooves can act as fatigue crack initiation sites. According to the textbook, if you wish to
reduce the probability of the fatigue failure of the material, you must reduce the aforementioned structure
irregularities by making design modifications that eliminate abrupt contour changes that lead to sharp
corners. Furthermore, according to the textbook, small scratches and grooves produced during machining
operations can also decrease fatigue life/fatigue strength. According to the book, fatigue life/fatigue
strength can be greatly improved by polishing the surface finish of the material. Furthermore, according to
the book, one of the more effective methods of increasing the fatigue life/fatigue strength of a material is
by imposing residual compressive stresses within an outer surface layer of the material. This way, the
external stress will be reduced in magnitude, therefore nullifying the chance of crack formation. According
to the textbook, residual compressive stresses most commonly used in ductile metals by mechanically
localized plastic deformation within the outermost surface region of the metal. According to the textbook, a
common process used to achieve this is a process known as “shot peening.” In shot peening, hard, small
particles with diameters that range from 0.1 mm to 1.0 mm are literally shot at high velocities onto the
desired surface. The resulting deformations induce compressive stresses down to a depth of one-quarter to
one-half of the chosen shot diameter. Furthermore, according to the book, case hardening is another
technique used to enhance both the fatigue life/fatigue strength and surface hardness of a steel alloy.
According to the textbook, this is achieved by a nitriding or carburizing process where a component is
subjected to a nitrogenous or carbonaceous atmosphere at a high temperature. A nitrogen or carbon rich
outer surface layer is initiated by atomic diffusion from the gaseous phase. According to the book, the case
is normally 1 mm deep and is harder than the material’s inner core. According to the book, the hardness of
the case greatly improves the fatigue properties of a material.
Environmental effects on the fatigue strength/fatigue life of a metal are, according to the textbook, either
thermal fatigue or corrosion fatigue. According to the textbook, thermal fatigue normally occurs at high
temperatures by changing thermal stresses. According to the textbook, the cause of these thermal stresses is
the restraint to the dimensional expansion and contraction that would occur in a structural member with
variations in temperature. According to the textbook, you can prevent thermal fatigue by eliminating or
reducing the restraint source. Corrosion fatigue is, according to the textbook, a type of failure that takes
place as a result of the simultaneous action of a cyclic stress and chemical attack. Corrosive environments
greatly decrease the fatigue life/fatigue strength of a material and often cause small pits to form on the
surface of a material, which serve as crack nucleation sites. Furthermore, according to the textbook, the
crack propagation rate is enhanced due to the corrosive environment. According to the textbook, you can
prevent corrosion fatigue by applying protective surface coatings or selected a material with a high
corrosive resistance.