Chapter 6. Fracture, Fatigue and Creep.
After studying this chapter you will be able to:
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Distinguish between ductile and brittle fracture.
Describe a stress concentrator (also called stress raiser) and the danger it
represents.
Distinguish between the resistance to deformation and to fracture, and name their
units.
Explain why hard materials are more brittle than soft metals.
Explain why ceramics are unreliable in tensile stresses.
Describe fatigue, its mechanism, its consequences and what can be done to the
material in order to improve fatigue resistance.
Measure and describe creep and name the important engineering situations where
it plays a major role.
Under certain circumstances, materials break. Fracture is a failure that can cost
human lives, lead to the destruction of costly structures or machines or interfere with
manufacture. If we want to avoid fracture, we must know how and under what
conditions materials break and what can be done to prevent it. In this chapter we will
discuss the ductile and brittle fracture of metals and the fracture of ceramics and glasses,
which is always brittle. We will examine the concept of stress concentration and learn
why strong and hard materials are brittle while softer materials are capable of absorbing
much more energy before they break. Finally, we will examine the phenomena of fatigue
and creep which can lead to failure at stresses well below the yield stress, where one
would expect the piece to be safe.
Figure 6.1. Schematic Stress-strain curve of a metal.
6.1.
Ductile Fracture of Metals.
Let us look again at the stress-strain curve of a metal (Figure 6.1). After a certain
amount of plastic deformation, a neck forms in the tensile test specimen, the force for
further deformation decreases and, finally, the piece breaks. When fracture occurs after
extensive plastic deformation, the test piece fails by ductile fracture.
Ductile metals neck and display the cup-cone fracture morphology shown in
Fig.6.2. A. The process of ductile fracture may be broadly viewed in terms of the
sequential microvoid nucleation and growth processes, schematically indicated in Fig.6.3.
Te earliest stage of fracture spawns isolated microscopic cavities. These nucleate at
inclusions, second phase particles and probably at grain boundary junctions. Microvoids
coalesce then form an elliptical crack that spreads outward toward the periphery of the
neck. Finally, an overloaded outer ring of material is all that is left to connect the
specimen halves, and it fails by shear. Further examination reveals equiaxed or spherical
dimples on the flat crater bottom loaded in tension, and elongated ellipsoidal dimples on
the shear lips oriented at 45o (Fig.6.4). In high purity FCC and BCC metals free of
inclusions, necking to ~ 100 percent area reduction (i.e., to a point) is possible.
A
B
Figure 6.2. Fracture of metallic tensile test specimens. (A) Ductile Fracture. (B) Brittle
Fracture.
Figure 6.3. Model of ductile fracture. From left to right: application of stress, nucleation
of microvoids, coalescence of microvoids, formation of elliptical crack, shearing of collar
and separation.
Figure 6.4. Ductile fracture surface of a metal.
6.2.Brittle fracture of metals and ceramics.
Brittle fracture is the most feared of all. It often occurs under static loading
without advance warning of impending catastrophe. Figure 6.2.B shows a metal bar that
failed by brittle fracture. Higher magnification reveals that transgranular cracking often
occurs during brittle fracture, especially at low temperatures. In such cases the crack
propagates across the grain interiors. (Figure 6.5A) When cracks propagate along grain
boundaries, we speak of intergranular fracture. Such failures (Fig.6.5B) reveal clearly
outlined grains that appear to be deeply etched or stand out in relief like rock-candy.
6.2.1. Cracks and stress concentrations.
We are not capable of predicting the initiation of a crack, but once a crack exists,
fracture mechanics describes its propagation. Cracks at the surface or the interior of
objects under load generate local stresses that are much larger than the stress one would
compute from the force and the geometry of the piece. This fact is illustrated in Fig.6.6
where a body (plate) containing elliptical shaped surface and interior holes or cracks, is
pulled by uniaxial forces, F. The undisturbed force lines in the defect-free portion of the
plate mean that the stress distribution is uniform there at a level σa. The force lines are
concentrated in the vicinity of cracks; stress is magnified there because there are
effectively more force lines per unit area. At the tip of the crack of half-length c and
radius of curvature ρ, the local tensile stress is
σ = σa 2(c/ρ)1/2
(6.1)
Figure 6.5. (A) Scanning electron microscope image of transgranular brittle
fracture in Fe-2.5 wt% Si tested at -195oC. (B) Scanning electron microscope image of
intergranular fracture in a nickel-based superalloy. Courtesy of G.F. Vander Voort,
Carpenter Technology Corporation.
Figure 6. 6. The concentration of tensile stresses around internal and surface cracks in a
uniformly stretched solid. Parallel lines denote uniform stress.
A stress concentration factor k σ, can be defined as the ratio of amplified local- to
background stress, i.e., kσ = σ/σa. For flat elliptical surface cracks the maximum value
of kσ is at the crack tip and given by
kσ = 2 (c /ρ)1/2.
(6.2)
Here c is half the length of the major axis and ρ is the radius of curvature at the crack tip.
It is apparent that longer and sharper cracks raise kσ. This has important
consequences for the design of pieces that must sustain a load and for the mechanical
properties of materials.
1. A crack or a notch is a stress concentrator. If the radius of curvature ρ is
small enough, the local stress is many times larger than the average stress; it
can be large enough to rupture bonds and lead to the propagation of the
crack.
2. The propagation of a crack can be arrested by increasing the radius ρ. In a
plate, for instance, a crack can be arrested by drilling a hole of sufficient
diameter.
3. In soft metals (with a low yield stress) plastic deformation under the local
4.
5.
6.3.
stress σ increases the radius ρ (i.e. blunts the crack) and the stress
concentration: the crack is arrested. In hard materials, this plastic
deformation does not take place, the stress concentration is large and the
crack propagates. As a rule, hard materials are brittle and soft materials
are ductile. A look at figures 5.18 and 5.22 shows that strengthening a metal
decreases its ductility.
Ceramic materials are very hard and do not deform plastically under tensile
stress, but break abruptly after elastic deformation. Ceramic materials and
glasses are brittle.
Ceramic materials contain small cracks as a consequence of their processing.
The length of these cracks cannot be controlled and the stress concentration
factor kσ is poorly known, therefore the average (design) stress σa that leads
to fracture in a ceramic cannot be determined with certainty. Ceramics are
unreliable in tension; they should be designed in such a way that they are
subjected to compressive stresses only. This will be immediately apparent
when we compare the shapes of a stone bridge and a steel bridge.
Fracture Toughness.
The resistance to fracture is known as the critical stress intensity factor or
fracture toughness
__
KIC = Yσ√πc.
(6.3)
The fracture toughness represents the combination of stress and crack length at which a
crack propagates. When K < KIC, the crack is stable. But when K> KIC the crack is
unstable and will rapidly open. Y is a numerical factor that depends on the geometry of
the crack. Y = 1 for the through crack in the body of the specimen (Figure 6.6) and Y =
1.1 for a crack at the surface of a plane specimen. The fracture toughness has dimensions
MPa√m or ksi√in. 1 ksi√in = 1.1 MPa√m.
Values of KIC for a number of materials are entered in Table 6.1. The high values
of KIC for metals are due to the plastic zone at the crack tip. Among the important trends
gleaned from this table are the following:
1. Glasses and simple metal oxides have the lowest values of KIC. They are
typically a 10 -100 times smaller than fracture toughness values for ductile metals.
2. KIC for some of the toughened ceramics are much higher than for glasses but
are still well below (by a factor of ~10) the level for metals.
EX.7-7
What is the ratio of the critical crack sizes in aluminum relative to
aluminum oxide when stressed to their maximum respective elastic stresses?
ANS. At the maximum elastic stress, σ = σy (the yield stress), and
c = 1/π (KIC /σy)2. The required ratio is
From Table 7-5, for Al: KIC = 44 MPa-m1/2 and σy= 345 MPa, and
for Al2O3: KIC = 3.7 MPam1/2 and σy = 270 MPa. (This is not a yield stress,
alumina does not deform plastically; it is the fracture stress of alumina which we
will discuss later).
For aluminum, the crack length is 5.3 10-3 m = 5.3 mm.
In aluminum oxide, the crack length is 6 10-5 m = 60 µm, which is 85 times shorter.
The different response to flaw size of metals relative to ceramics can be
qualitatively understood from this example. Large cracks in the more defect - tolerant
metals can often be detected during inspection and possibly repaired by sealing them
shut. But, the much smaller flaws of critical size in ceramics are more effectively hidden
and easier to overlook, a combination that makes ceramics fracture prone in service.
6.3.The measurement of fracture resistance.
6.3.1. Fracture toughness
To determine KlC values of materials, specimens in the form of plates are
fashioned containing machined cracks of known length as shown in Fig. 6.7a. As tensile
loads are applied the specimen halves open, and the resulting crack extension is
continuously monitored (Fig.6.7b). Noting the critical crack size and stress necessary to
induce fracture, use of Eq.6.3 yields the value for KlC. The measured toughness of a
metal depends on its size. Ductile metals require large specimens for reliable fracture
toughness measurements, brittle materials can be measured with smaller specimens.
Figure 6.7. Measurement of fracture toughness.
6.3.2. Indentation Toughness of Ceramics.
For routine measurements of toughness, the fabrication of ceramic samples as
shown in figure 6.7 is expensive. A more convenient but less reliable measurement is
afforded by high-load Vickers indentations. When a material is brittle and the
indentation is made at sufficiently high load (minimum 20 Newton, but generally 100
Newton or 10 kg), radial cracks form at the tip of the indentation. (A figure 6.8 will be
provided soon.) Empirical equations provide the fracture toughness from a measurement
of the crack length and the hardness provided by the indent. Such measurements, while
not as precise as the one described above, have the advantage of being convenient and
allowing to measure local variations of hardness. The method cannot be used with metals
which are too ductile to form radial cracks.
Figure 6.8. Vickers indentation in a ceramic. The size of the indentation determines the
hardness, as described in Chapter 5. The length of the radial cracks is used to measure the
toughness.
6.3.3. Charpy and Izod Measurements of Notch Toughness.
The widely used Charpy impact test is a standard way to assess toughness in
notched specimens. In this test a standard bar specimen, with a square cross section and a
V- shape notch cut into it, is strained very rapidly to fracture by means of a swinging,
pendulum-like hammer (Fig.6.9). The difference between the initial and final potential
energies of the hammer (measured by the initial and final hammer heights), is the impact
energy absorbed by the specimen. Charpy and Izod test methods differ only in the
position of the sample as shown in figure 6.9.
Figure 6.9.Impact toughness tests. (A) Specimen used for Charpy and Izod impact
tests. (b) A schematic of an impact testing machine. The hammer of weight W is released
from height h breaks the sample as it swings and stops at height h’. The difference in
potential energies: W (h-h’) is the impact toughness. Charpy and Izod tests differ in the
placement of the specimen as shown.
Chirpy and Izard impact toughness values measure the energy needed to fracture
the specimen and are expressed in foot-pounds (ft-lb). They are more convenient but
less reliable than the fracture toughness tests of figure 6.7.
6.3.4.
Rupture Strength of Ceramics.
Ceramics are brittle, they do not deform plastically under tensile stress but
fracture. The strength of ceramics is often measured in a bend test shown in figure 6.10.
F
F
F
F
Figure 6.10. Four point bend test.
A thin plate of ceramic is placed between four cylinders to which forces are
applied as shown in the figure. These forces tend to bend the plate elastically and
generate a tensile stress on the lower surface. Once the stress is large enough, the plate
fractures. The stress at which the plate fractures has several names: flexural strength,
modulus of rupture, fracture strength or bend strength. Flexural strengths of several
ceramics are shown in Table 6.2. Note that the flexural strengths, given in MPa or ksi,
are indicated with a wide uncertainty. This is easy to understand with a look at equation
(6.3). The toughness of ceramics is a reasonably well defined quantity. It involves a
stress (which is measured in the bend test) and a crack length. The dispersion in fracture
stresses stems from the random occurrence of small cracks the length of which is not
known.
6.4.
Ductile to Brittle Transition Temperature.
Steels often display an alarming drop in notch toughness resistance at the socalled transition temperature. This is shown in Fig.6.11 for several types of ship steels
used during the Second World War, when all too many ships suffered brittle fracture of
varying severity. In these ships the steels underwent a ductile to brittle transition too
close to room temperature. This is one reason among others that, even today, structures
like storage tanks, oil rigs and even ships fracture catastrophically more often in winter
than in summer.
Figure 6.10. Charpy impact energies of various World War II ship steels as a function of
test temperature.
Transition temperatures can be lowered through alloying. Reducing the carbon
content of steel has a potent effect in this regard; it also raises the magnitude of the
(absorbed) impact energy. Alloying with nickel is particularly beneficial in toughening
steel and lowering the transition temperature. Alloying elements in steel affect toughness
in complex ways depending on composition, matrix structure and heat treatment. Ductile
face centered cubic metals are tough even at low temperatures and do not exhibit a
ductile-to-brittle transition.
Table 6.2 from Callister.
6.4. Fatigue
Fatigue is the failure (by fracture) of structures that are subjected to repeated or
cyclic loading. By cyclic loading we mean almost any reasonably periodic stress-time
variation, e.g., (a) axial tension-compression, (b) reversed bending and (c) reversed
torsion or twisting. Undetected flaws or incipient cracks grow to macroscopic dimensions
through incremental propagation during each stress cycle and the component quickly
undergoes fracture without warning when a crack of critical size is reached. This
sequence of events has been repeated in components of rotating equipment such as motor
and helicopter shafts, train wheels and tracks, pump impellers, ship screws and
propellers, and gas turbine discs and blades. Even surgical prostheses implanted into the
human body have suffered fatigue failure in service. What is difficult to design against is
the fact that under cyclic loading, failure can occur significantly below the tensile or
yield stress σy of the material.
Fatigue of materials is measured by means of a fatigue testing machine (Fig.6.12).
A bar specimen that narrows in diameter toward the middle is mounted horizontally and
rotated at high speed with a motor. A hanging load tilts the grips so that reversed bending
moments are transmitted to the specimen, stressing its surface alternately in tension and
compression during rotation. This test provides a sinusoidal loading about a zero mean
stress level (Fig.6.12b). For the case of arbitrary sinusoidal loading, maximum (σmax.),
minimum (σmin.), mean (σm) stress, and stress range (σr) values are defined in
Fig.6.12c. The number of rotations to failure is counted at specific specimen stress
amplitude. Then a new specimen is mounted, a different load is chosen yielding a new
Figure 6.12. (A) Fatigue testing machine. (B) Sinusoidal loading. (D) Arbitrary loading
with definition of mean, maximum, minimum stress and stress amplitude.
Figure 6.13 S-N data for various metals and polymers.
peak stress level and the number of cycles to failure is determined once again, etc. In this
tedious manner sufficient data points are accumulated to generate an S-N (or stressnumber of cycles) curve.
Amid a typically large scatter band, the S-N response for assorted materials is
displayed in Fig.6.13. Here S is the stress and N the number of cycles to failure. Most
steels can rotate indefinitely below a stress level known as the endurance or fatigue
limit (the curve horizontal). Some high strength steels and many non ferrous metals
(e.g., Al and Cu), on the other hand, do not exhibit an endurance limit. Polymers display
similar fatigue characteristics. To guard against fatigue, stress levels should not exceed
1/3 σUTS.
Combating fatigue damage is fairly straight forward once it is recognized that the
following factors affect fatigue strength in adverse ways:
1. Stress Concentrators.
Keyways on shafts, holes, abrupt changes in cross sections,
sharp corners etc., all stress raisers, are design features that should be avoided in
components subjected to repetitive loading. Stress concentrators mean operation at
higher stress levels on the S-N curve (Fig.6.13). This in turn reduces the number of stress
cycles to failure.
2. Polished Surfaces.
Crack nucleation is facilitated at micro-crevices and
grooves on rough surfaces. Grinding, honing and polishing of surfaces remove these
sources of potential cracks.
3. Residual Stresses. Residual stresses at the surface add themselves to the applied
stress: residual tensile stresses reduce fatigue life. Residual compressive stresses in
surface layers, introduced by shot peening or diffusion, higher levels of applied tension
can be tolerated. As a result, fatigue strength and life can be increased.
4. High Strength Surfaces. Strengthening surfaces can enhance fatigue resistance;
carburization and deposition of hard coatings are ways to achieve this.
5. Corrosion and Environmental Attack. Localized chemical attack will always
reduce fatigue life. Metal removed from exposed grains and grain boundaries leave a
generally rough surface, pits and corrosion products behind, that serve as incipient
cracks.
6.5
Creep
At temperatures of about half the melting point (0.5TM) and above, materials
undergo time dependent plastic straining when loaded. This phenomenon is known as
creep; it can occur at stress levels less than the yield strength! The extension of an
involved component may eventually produce a troublesome loss of dimensional tolerance
or even ultimately lead to catastrophic rupture. Turbine blade creep due to the
inhospitable temperatures of a jet engine is an often cited example. High pressure boilers
and steam lines, nuclear reactor fuel cladding, ceramic refractory brick in furnaces are
components and systems that are also susceptible to creep effects.
Creep tests are performed on round tensile-like specimens that are stressed by
fixed suspended loads while being heated by furnaces which coaxially surround them.
The typical response obtained in a creep test is shown in Fig.6.14, where the specimen
elongation or strain is recorded as a function of time. There is an initial elastic extension
or strain the instant load is applied. Then a viscous-like plastic straining ensues in which
the creep strain rate (dε /dt) decreases with time. This primary creep period then merges
with the secondary creep stage where the strain rate is fairly constant. Alternately
known as steady state creep, this region of minimum creep normally occupies most of
the test lifetime. Finally the strain rate increases rapidly in the tertiary creep stage
leading to rupture of the specimen.
Figure 6.14. Schematic of creep. Bottom: elongation as a function of time. Top: strain
rate as a function of time.
Engineers perform two types of tests. The first aims to determine the steady state
creep rate over a suitable matrix of stress and temperature. These tests are performed at
the same stress but at different temperatures, as well as at the same temperature but
different stresses, as shown in Fig.6.15. Specimens are usually not brought to failure in
such tests; accurately predicting their extension is of interest here. The second, known
as the creep rupture test, is conducted at higher stress and temperature levels in order to
accelerate failure. Such test information can either be used to estimate short term life
(e.g., turbine blades in military aircraft), or be extrapolated to lower service temperatures
and stresses (e.g., turbine blades in utility power plants) to predict long term life.
Figure 6. 15. Creep strain versus time in a 0.5 wt% o, 0.23 wt% steel. (A) Constant stress,
variable temperature. (B) Constant temperature, variable stress. From A.J. Kennedy,
Processes of Creep and Fatigue in Metals. Wiley, New York (1963)
6.5.1. Steady State Creep.
Much engineering design is conducted on the basis of steady state creep data.
Since the creep strain rate is accelerated by both temperature and stress, a useful
phenomenological equation that succinctly summarizes the data of Fig.6.15 and reflects
the role of these variables is
ε = A σmexp - Ec / kT.
(6.4)
A and m are constants; the exponential term is the Boltzmann factor that reflects the
thermally activated nature of creep. Ec is the creep activation energy; it is close to the
activation energy for bulk diffusion. m, a constant, typically ranges in value from 1-5.
EX. 6.2
Determine Ec and m for the data shown in Fig.6-14.
ANS.
The steady state strain rate is the slope of the strain-time response.
Limiting the calculation to only the maximum and minimum slopes for the
superimposed lines, the following results are tabulated:
Test Temperature (C) Stress (psi)
1.
670 (= 943K)
9000
Strain Rate (h -1)
(2.5-0.28) X10-3/ (370-0) = 6X10-6
2.
3.
4.
615 (= 888K)
645 (= 918K)
645 (= 918K)
9000
11200
6700
(0.56-0.2) X10-3/ (400-0) = 9X10-7
(1.5-0.6) X10-3/ (200-0) = 4.5X10-6
(0.5-0.25) X10-3/ (400-0) =6.25X10-7
To determine Ec, data from tests 1 and 2 are utilized. From Eq.7-29,
ε (1) / ε (3) = (exp -Ec / kT1) / (exp -Ec / kT3) or
Ec = k T1T2 / (T1 - T2) ln [ε(1) / ε(2)]. Substituting,
Ec = 8.31 J/mol-K (943X888 / 55) Xln [6X10-6 / 9X10 -7] = 240 kJ / mol.
Similarly, using test data 3 and 4, ε(3) / ε(4) = (σ3/σ4)m
 / ε(4))

and m = ln (ε(3)
/ ln (σ3/σ4). After substitution
m = ln (4.5X10-6/6.25X10-7) / ln (11200 / 6700) = 1.01.
Knowing Ec and m enables ε to be evaluated at any other temperature and
stress level.
6.5.2. Creep Rupture
Creep or stress rupture test data are normally plotted as the stress versus rupture
time (tR) on a log-log plot as shown in Fig.6.16. Each data point represents one test at a
specific temperature and stress level. Given sufficient testing at different temperatures, a
complete profile of the material response is obtained.
Figure 6.16. Creep - rupture data of Incoloy 800, an iron-based alloy containing 30 wt%
Ni and 19 wt% Cr. Source: Huntington Alloys Inc.
Several mechanisms have been identified for creep, depending on the stress level
and temperature. At stresses below the yield stress σy and at high temperature, diffusion
of atoms or vacancies dominate. Observations of creep failure often reveal the dominant
role grain boundaries play in the phenomenon. By diffusion of atoms in the grain
boundaries, the grains slide past each other. For this reason, modern jet engine turbine
blades are grown as single crystals by a method of directional solidification. Stainless
steels and superalloys containing nickel and cobalt are notoriously creep resistant.
Recapitulation.
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
Ductile fracture occurs after extensive plastic deformation. It occurs by the
formation and coalescence of voids and shearing.
Brittle fracture occurs by the propagation of cracks that usually nucleate at the
surface.
Toughness is the resistance to fracture. The units of fracture toughness are
MPa√m or ksi√in. 1 ksi√in = 1.1 MPa√m.
Cracks, indentations and sharp interior corners constitute stress concentrators
that lead to crack propagation.
The toughness of ceramics is well defined, the fracture stress is not because of
the presence of cracks of unknown length.
Fatigue is the failure of materials that is due to repeated, cyclic application of
a stress. Fatigue failure can occur at stresses below the yield stress.
Steels and some materials exhibit a fatigue limit. This is a stress below which
fatigue does not occur.
Fatigue resistance can be increased by avoiding stress concentrators in design,
polishing the surfaces and introducing compressive residual stresses.
Creep is the slow plastic deformation due to diffusion of atoms at high
temperatures (T > ½ melting temperature). Creep occurs at stresses below the
yield stress. Note the difference: plastic deformation is not dependent on
time; modern manufacturing exploits the fact that metals can be shaped very
rapidly. Creep is a viscous-like deformation and the stress governs the rate of
deformation.
In gas turbines of jet engines, creep is avoided by fabricating the turbine
blades as single crystals.
Problems and questions.
6.1 What is the difference between
a. stress intensity and critical stress intensity
b. strength and toughness
c. cracks in metals and cracks in ceramics
d. transverse - rupture strength and tensile strength?
6.2. Draw a stone bridge and a steel (railroad) bridge and explain the difference of their
shapes in terms of their materials properties.
6.3. A plate of steel has a yield stress of 1000 MPa. The plate fractured when the tensile
stress reached 800 MPa and it was therefore hypothesized that a surface crack was
present. If the fracture toughness for this steel is 60 MPa-m1/2, approximately what crack
size is suggested?
6.4. Big Bird Helicopter Company requires a stainless steel that is tough enough for use
in the Tropics (120F) as well as the Arctic (-60F). They peruse this book and find the data
of Fig.6.15. At the designed use stress, the component can tolerate a surface flaw of
0.080in at 120F. Manufacturing inspection equipment can only detect flaws that are
larger than 0.050in. Will a fracture unsafe situation arise at -60F for the same loading?
Figure 6.15. Fracture toughness and yield strength as a function of testing temperature.
From Metals Handbook. American Society of Metals. Metals Park, OH
6.5 A steel cylinder containing CO2 gas pressurized to 1500 psi (10 MPa) is being
transported by truck when someone fires a bullet that pierces it making a 1cm hole.
a. Explain the conditions that might cause the steel to fracture violently with fragments
flying off at high velocity.
b. Similarly, explain the conditions that would cause the cylinder to release the gas
relatively harmlessly.
6.6
Steady state creep testing of electrical solder wire yielded the indicated strain
rates under the test conditions given.
Test
Temperature (C)
1
2
3
22.5
22.5
46
Stress (MPa)
Strain Rate (sec -1)
6.99
9.07
6.99
From these limited data determine the creep activation energy (Ec) and
stress exponent, m.
2.50X10-5
6.92X10-5
2.74X10-4
6.7 Incoloy 800 tubes are selected for use in a pressurized chemical reactor. If they are
designed to withstand wall stresses of 2000 psi at a temperature of 900C predict how long
they will survive.
6.8 Suppose the fatigue behavior of a steel is characterized by a two line response when
plotted on an S- logN plot, namely
1. S(MPa) = 1000 - 100 logN ; between N = 0 to N = 106 stress cycles,
2. S(MPa) = 400 ; for N >106 stress cycles.
a. Sketch the S - logN plot.
b. What is the value of the endurance limit?
c. How many stress cycles will the steel probably sustain prior to failure at a stress of 460
MPa?
d. Elimination of surface scratches and grooves changed the endurance limit to 480 MPa.
(Assume plot 1 is not altered). How many stress cycles will the steel probably sustain
prior to failure at a stress of 470 MPa? How many stress cycles will the steel probably
sustain prior to failure at a stress slightly above 480 MPa?
6.9. Assume the same stress develops in WC and Al2O3 tool bits during identical
machining processes. Since crack formation is a cause of tool bit failure what is the ratio
of the critical flaw dimensions in Al2 O3 that can be tolerated relative to WC?
6.10 The fracture toughness of a given steel is 60 MPa-m1/2 and its yield strength is
given by σy(MPa) = 1400 - 4T where T is the temperature in degrees Kelvin. Surface
cracks measuring 0.001m are detected. Under these conditions determine the temperature
at which there may be a ductile-brittle transition in this steel.
6.11 The quenched and aged Ti-6Al-4V alloy used in the Atlas missile has yield points
of 229 ksi at -320F, 165 ksi at -70 F and 120 ksi at 70F. Suppose small cracks 0.02 inch
long were discovered on cryogenic storage containers made from this alloy whose
fracture toughness is 50 ksi-in1/2. Estimate the ductile to brittle fracture transition
temperature of this alloy. Would it be safe to expose these containers to liquid nitrogen
temperatures (77K)?
Assume the fracture toughness is independent of temperature.
6.12 An aluminum aircraft alloy was tested under cyclic loading with ∆σ = 250 MPa
and fatigue failure occurred at 2X105 cycles. If failure occurred in 107 cycles when
∆σ = 190 MPa, estimate how many stress cycles can be sustained when
∆σ = 155 MPa.