Solution Manual for Manufacturing Processes for Engineering Materials, 6th Edition
Master your textbook problems with Solution Manual for Manufacturing Processes for Engineering Materials, 6th Edition, providing comprehensive solutions and explanations.
Lucas Taylor
Contributor
4.9
55
4 months ago
Preview (16 of 297)
Sign in to access the full document!
Manufacturing Processes for Engineering Materials, 6th ed.
Serope Kalpakjian
Steven R. Schmid
Serope Kalpakjian
Steven R. Schmid
Chapter 2
Fundamentals of the Mechanical Behavior of
Materials
QUESTIONS
2.1 Can you calculate the percent elongation of materials
based only on the information given in Fig. 2.6? Ex
plain.
Recall that the percent elongation is defined by Eq. (2.6)
on p. 35 and depends on the original gage length (lo) of
the specimen. From Fig. 2.6 on p. 39, only the necking
strain (true and engineering) and true fracture strain
can be determined. Thus, we cannot calculate the per
cent elongation of the specimen; also, note that the
elongation is a function of gage length and increases
with gage length.
2.2 Explain if it is possible for stress-strain curves in ten
sion tests to reach 0% elongation as the gage length is
increased further.
The percent elongation of the specimen is a function of
the initial and final gage lengths. When the specimen is
being pulled, regardless of the original gage length, it
will elongate uniformly (and permanently) until neck
ing begins. Therefore, the specimen will always have
a certain finite elongation. However, note that as the
specimen’s gage length is increased, the contribution
of localized elongation (that is, necking) will decrease,
but the total elongation will not approach zero.
2.3 Explain why the difference between engineering strain
and true strain becomes larger as strain increases. Is
this phenomenon true for both tensile and compressive
strains? Explain.
The difference between the engineering and true
strains becomes larger because of the way the strains
are defined, respectively, as can be seen by inspecting
Eqs. (2.1) and (2.9). This is true for both tensile and
compressive strains.
2.4 Using the same scale for stress, the tensile true-stress
true-strain curve is higher than the engineering stress-
strain curve. Explain whether this condition also holds
for a compression test.
During a compression test, the cross-sectional area
of the specimen increases as the specimen height de
creases (because of volume constancy) as the load is in
creased. Since true stress is defined as ratio of the load
to the instantaneous cross-sectional area of the speci
men, the true stress in compression will be lower than
the engineering stress for a given load, assuming that
friction between the platens and the specimen is negli
gible.
2.5 Which of the two tests, tension or compression, re
quires a higher capacity testing machine than the
other? Explain.
The compression test requires a higher capacity ma
chine because the cross-sectional area of the specimen
increases during the test, which is the opposite of a ten
sion test. The increase in area requires a load higher
than that for the tension test to achieve the same stress
level. Furthermore, note that compression-test spec
imens generally have a larger original cross-sectional
area than those for tension tests, thus requiring higher
forces.
2.6 Explain how the modulus of resilience of a material
changes, if at all, as it is strained: (a) for an elastic, per
fectly plastic material, and (b) for an elastic, linearly
strain-hardening material.
Recall that the modulus of resilience is given by
Eq. (2.5) on p. 34 as S2
y /(2E). (a) If the material is
perfectly plastic, then the yield strength does not in
crease with strain - see Fig. 2.7c on p. 42. Therefore, the
modulus of resilience is unchanged as the material is
strained. (b) For a linear strain hardening material, the
yield strength increases with plastic strain. Therefore
the modulus of resilience will increase with strain.
1
Fundamentals of the Mechanical Behavior of
Materials
QUESTIONS
2.1 Can you calculate the percent elongation of materials
based only on the information given in Fig. 2.6? Ex
plain.
Recall that the percent elongation is defined by Eq. (2.6)
on p. 35 and depends on the original gage length (lo) of
the specimen. From Fig. 2.6 on p. 39, only the necking
strain (true and engineering) and true fracture strain
can be determined. Thus, we cannot calculate the per
cent elongation of the specimen; also, note that the
elongation is a function of gage length and increases
with gage length.
2.2 Explain if it is possible for stress-strain curves in ten
sion tests to reach 0% elongation as the gage length is
increased further.
The percent elongation of the specimen is a function of
the initial and final gage lengths. When the specimen is
being pulled, regardless of the original gage length, it
will elongate uniformly (and permanently) until neck
ing begins. Therefore, the specimen will always have
a certain finite elongation. However, note that as the
specimen’s gage length is increased, the contribution
of localized elongation (that is, necking) will decrease,
but the total elongation will not approach zero.
2.3 Explain why the difference between engineering strain
and true strain becomes larger as strain increases. Is
this phenomenon true for both tensile and compressive
strains? Explain.
The difference between the engineering and true
strains becomes larger because of the way the strains
are defined, respectively, as can be seen by inspecting
Eqs. (2.1) and (2.9). This is true for both tensile and
compressive strains.
2.4 Using the same scale for stress, the tensile true-stress
true-strain curve is higher than the engineering stress-
strain curve. Explain whether this condition also holds
for a compression test.
During a compression test, the cross-sectional area
of the specimen increases as the specimen height de
creases (because of volume constancy) as the load is in
creased. Since true stress is defined as ratio of the load
to the instantaneous cross-sectional area of the speci
men, the true stress in compression will be lower than
the engineering stress for a given load, assuming that
friction between the platens and the specimen is negli
gible.
2.5 Which of the two tests, tension or compression, re
quires a higher capacity testing machine than the
other? Explain.
The compression test requires a higher capacity ma
chine because the cross-sectional area of the specimen
increases during the test, which is the opposite of a ten
sion test. The increase in area requires a load higher
than that for the tension test to achieve the same stress
level. Furthermore, note that compression-test spec
imens generally have a larger original cross-sectional
area than those for tension tests, thus requiring higher
forces.
2.6 Explain how the modulus of resilience of a material
changes, if at all, as it is strained: (a) for an elastic, per
fectly plastic material, and (b) for an elastic, linearly
strain-hardening material.
Recall that the modulus of resilience is given by
Eq. (2.5) on p. 34 as S2
y /(2E). (a) If the material is
perfectly plastic, then the yield strength does not in
crease with strain - see Fig. 2.7c on p. 42. Therefore, the
modulus of resilience is unchanged as the material is
strained. (b) For a linear strain hardening material, the
yield strength increases with plastic strain. Therefore
the modulus of resilience will increase with strain.
1
Chapter 2
Fundamentals of the Mechanical Behavior of
Materials
QUESTIONS
2.1 Can you calculate the percent elongation of materials
based only on the information given in Fig. 2.6? Ex
plain.
Recall that the percent elongation is defined by Eq. (2.6)
on p. 35 and depends on the original gage length (lo) of
the specimen. From Fig. 2.6 on p. 39, only the necking
strain (true and engineering) and true fracture strain
can be determined. Thus, we cannot calculate the per
cent elongation of the specimen; also, note that the
elongation is a function of gage length and increases
with gage length.
2.2 Explain if it is possible for stress-strain curves in ten
sion tests to reach 0% elongation as the gage length is
increased further.
The percent elongation of the specimen is a function of
the initial and final gage lengths. When the specimen is
being pulled, regardless of the original gage length, it
will elongate uniformly (and permanently) until neck
ing begins. Therefore, the specimen will always have
a certain finite elongation. However, note that as the
specimen’s gage length is increased, the contribution
of localized elongation (that is, necking) will decrease,
but the total elongation will not approach zero.
2.3 Explain why the difference between engineering strain
and true strain becomes larger as strain increases. Is
this phenomenon true for both tensile and compressive
strains? Explain.
The difference between the engineering and true
strains becomes larger because of the way the strains
are defined, respectively, as can be seen by inspecting
Eqs. (2.1) and (2.9). This is true for both tensile and
compressive strains.
2.4 Using the same scale for stress, the tensile true-stress
true-strain curve is higher than the engineering stress-
strain curve. Explain whether this condition also holds
for a compression test.
During a compression test, the cross-sectional area
of the specimen increases as the specimen height de
creases (because of volume constancy) as the load is in
creased. Since true stress is defined as ratio of the load
to the instantaneous cross-sectional area of the speci
men, the true stress in compression will be lower than
the engineering stress for a given load, assuming that
friction between the platens and the specimen is negli
gible.
2.5 Which of the two tests, tension or compression, re
quires a higher capacity testing machine than the
other? Explain.
The compression test requires a higher capacity ma
chine because the cross-sectional area of the specimen
increases during the test, which is the opposite of a ten
sion test. The increase in area requires a load higher
than that for the tension test to achieve the same stress
level. Furthermore, note that compression-test spec
imens generally have a larger original cross-sectional
area than those for tension tests, thus requiring higher
forces.
2.6 Explain how the modulus of resilience of a material
changes, if at all, as it is strained: (a) for an elastic, per
fectly plastic material, and (b) for an elastic, linearly
strain-hardening material.
Recall that the modulus of resilience is given by
Eq. (2.5) on p. 34 as S2
y /(2E). (a) If the material is
perfectly plastic, then the yield strength does not in
crease with strain - see Fig. 2.7c on p. 42. Therefore, the
modulus of resilience is unchanged as the material is
strained. (b) For a linear strain hardening material, the
yield strength increases with plastic strain. Therefore
the modulus of resilience will increase with strain.
1
Fundamentals of the Mechanical Behavior of
Materials
QUESTIONS
2.1 Can you calculate the percent elongation of materials
based only on the information given in Fig. 2.6? Ex
plain.
Recall that the percent elongation is defined by Eq. (2.6)
on p. 35 and depends on the original gage length (lo) of
the specimen. From Fig. 2.6 on p. 39, only the necking
strain (true and engineering) and true fracture strain
can be determined. Thus, we cannot calculate the per
cent elongation of the specimen; also, note that the
elongation is a function of gage length and increases
with gage length.
2.2 Explain if it is possible for stress-strain curves in ten
sion tests to reach 0% elongation as the gage length is
increased further.
The percent elongation of the specimen is a function of
the initial and final gage lengths. When the specimen is
being pulled, regardless of the original gage length, it
will elongate uniformly (and permanently) until neck
ing begins. Therefore, the specimen will always have
a certain finite elongation. However, note that as the
specimen’s gage length is increased, the contribution
of localized elongation (that is, necking) will decrease,
but the total elongation will not approach zero.
2.3 Explain why the difference between engineering strain
and true strain becomes larger as strain increases. Is
this phenomenon true for both tensile and compressive
strains? Explain.
The difference between the engineering and true
strains becomes larger because of the way the strains
are defined, respectively, as can be seen by inspecting
Eqs. (2.1) and (2.9). This is true for both tensile and
compressive strains.
2.4 Using the same scale for stress, the tensile true-stress
true-strain curve is higher than the engineering stress-
strain curve. Explain whether this condition also holds
for a compression test.
During a compression test, the cross-sectional area
of the specimen increases as the specimen height de
creases (because of volume constancy) as the load is in
creased. Since true stress is defined as ratio of the load
to the instantaneous cross-sectional area of the speci
men, the true stress in compression will be lower than
the engineering stress for a given load, assuming that
friction between the platens and the specimen is negli
gible.
2.5 Which of the two tests, tension or compression, re
quires a higher capacity testing machine than the
other? Explain.
The compression test requires a higher capacity ma
chine because the cross-sectional area of the specimen
increases during the test, which is the opposite of a ten
sion test. The increase in area requires a load higher
than that for the tension test to achieve the same stress
level. Furthermore, note that compression-test spec
imens generally have a larger original cross-sectional
area than those for tension tests, thus requiring higher
forces.
2.6 Explain how the modulus of resilience of a material
changes, if at all, as it is strained: (a) for an elastic, per
fectly plastic material, and (b) for an elastic, linearly
strain-hardening material.
Recall that the modulus of resilience is given by
Eq. (2.5) on p. 34 as S2
y /(2E). (a) If the material is
perfectly plastic, then the yield strength does not in
crease with strain - see Fig. 2.7c on p. 42. Therefore, the
modulus of resilience is unchanged as the material is
strained. (b) For a linear strain hardening material, the
yield strength increases with plastic strain. Therefore
the modulus of resilience will increase with strain.
1
2.7 If you pull and break a tensile-test specimen rapidly,
where would the temperature be the highest? Explain
why.
Since temperature rise is due to the work input, the
temperature will be highest in the necked region be
cause that is where the strain, hence the energy dissi
pated per unit volume in plastic deformation, is high
est.
2.8 Comment on the temperature distribution if the speci
men in Question 2.7 is pulled very slowly.
If the specimen is pulled very slowly, the temperature
generated will be dissipated throughout the specimen
and to the environment. Thus, there will be no ap
preciable temperature rise anywhere, particularly with
materials with high thermal conductivity.
2.9 In a tension test, the area under the true-stress-true
strain curve is the work done per unit volume (the spe
cific work). Also, the area under the load-elongation
curve represents the work done on the specimen. If
you divide this latter work by the volume of the spec
imen between the gage marks, you will determine the
work done per unit volume (assuming that all defor
mation is confined between the gage marks). Will this
specific work be the same as the area under the true-
stress-true-strain curve? Explain. Will your answer be
the same for any value of strain? Explain.
If we divide the work done by the total volume of the
specimen between the gage lengths, we obtain the av
erage specific work throughout the specimen. How
ever, the area under the true stress-true strain curve
represents the specific work done at the necked (and
fractured) region in the specimen where the strain is a
maximum. Thus, the answers will be different. How
ever, up to the onset of necking (instability), the specific
work calculated will be the same. This is because the
strain is uniform throughout the specimen until neck
ing begins.
2.10 The note at the bottom of Table 2.4 states that as tem
perature increases, C decreases and m increases. Ex
plain why.
The value of C in Table 2.4 on p. 46 decreases with tem
perature because it is a measure of the strength of the
material. The value of m increases with temperature
because the material becomes more strain-rate sensi
tive, due to the fact that the higher the strain rate, the
less time the material has to recover and recrystallize,
hence its strength increases.
2.11 You are given the K and n values of two different mate
rials. Is this information sufficient to determine which
material is tougher? If not, what additional informa
tion do you need, and why?
Although the K and n values may give a good esti
mate of toughness, the true fracture stress and the true
strain at fracture are required for accurate calculation
of toughness. The modulus of elasticity and yield stress
would provide information about the area under the
elastic region; however, this region is very small and is
thus usually negligible with respect to the rest of the
stress-strain curve.
2.12 Modify the curves in Fig. 2.7 to indicate the effects of
temperature. Explain your changes.
These modifications can be made by lowering the slope
of the elastic region and lowering the general height of
the curves. See, for example, Fig. 2.9 on p. 43.
2.13 Using a specific example, show why the deformation
rate, say in m/s, and the true strain rate are not the
same.
The deformation rate is the quantity v in Eqs. (2.16)
and (2.17). Thus, when v is held constant during de
formation (hence a constant deformation rate), the true
strain rate will vary (l increases), whereas the engineer
ing strain rate will remain constant. Hence, the two
quantities are not the same.
2.14 It has been stated that the higher the value of m, the
more diffuse the neck is, and likewise, the lower the
value of m, the more localized the neck is. Explain the
reason for this behavior.
As discussed in Section 2.2.7, with high m values, the
material stretches to a greater length before it fails; this
behavior is an indication that necking is delayed with
increasing m. When necking is about to begin, the
necking region’s strength with respect to the rest of the
specimen increases, due to strain hardening. However,
the strain rate in the necking region is also higher than
in the rest of the specimen, because the material is elon
gating faster there. Since the material in the necked re
gion becomes stronger as it is strained at a higher rate,
the region exhibits a greater resistance to necking. The
increase in resistance to necking thus depends on the
magnitude of m. As the tension test progresses, neck
ing becomes more diffuse, and the specimen becomes
longer before fracture; hence, total elongation increases
with increasing values of m. As expected, the elonga
tion after necking (postuniform elongation) also increases
with increasing m. It has been observed that the value
of m decreases with metals of increasing strength.
2.15 Explain why materials with high m values, such as hot
glass and silly putty, when stretched slowly, undergo
large elongations before failure. Consider events tak
ing place in the necked region of the specimen.
The answer is similar to Answer 2.14 above.
2.16 Assume that you are running four-point bending tests
on a number of identical specimens of the same length
and cross-section, but with increasing distance be
tween the upper points of loading (see Fig. 2.19b).
What changes, if any, would you expect in the test re
sults? Explain.
2
where would the temperature be the highest? Explain
why.
Since temperature rise is due to the work input, the
temperature will be highest in the necked region be
cause that is where the strain, hence the energy dissi
pated per unit volume in plastic deformation, is high
est.
2.8 Comment on the temperature distribution if the speci
men in Question 2.7 is pulled very slowly.
If the specimen is pulled very slowly, the temperature
generated will be dissipated throughout the specimen
and to the environment. Thus, there will be no ap
preciable temperature rise anywhere, particularly with
materials with high thermal conductivity.
2.9 In a tension test, the area under the true-stress-true
strain curve is the work done per unit volume (the spe
cific work). Also, the area under the load-elongation
curve represents the work done on the specimen. If
you divide this latter work by the volume of the spec
imen between the gage marks, you will determine the
work done per unit volume (assuming that all defor
mation is confined between the gage marks). Will this
specific work be the same as the area under the true-
stress-true-strain curve? Explain. Will your answer be
the same for any value of strain? Explain.
If we divide the work done by the total volume of the
specimen between the gage lengths, we obtain the av
erage specific work throughout the specimen. How
ever, the area under the true stress-true strain curve
represents the specific work done at the necked (and
fractured) region in the specimen where the strain is a
maximum. Thus, the answers will be different. How
ever, up to the onset of necking (instability), the specific
work calculated will be the same. This is because the
strain is uniform throughout the specimen until neck
ing begins.
2.10 The note at the bottom of Table 2.4 states that as tem
perature increases, C decreases and m increases. Ex
plain why.
The value of C in Table 2.4 on p. 46 decreases with tem
perature because it is a measure of the strength of the
material. The value of m increases with temperature
because the material becomes more strain-rate sensi
tive, due to the fact that the higher the strain rate, the
less time the material has to recover and recrystallize,
hence its strength increases.
2.11 You are given the K and n values of two different mate
rials. Is this information sufficient to determine which
material is tougher? If not, what additional informa
tion do you need, and why?
Although the K and n values may give a good esti
mate of toughness, the true fracture stress and the true
strain at fracture are required for accurate calculation
of toughness. The modulus of elasticity and yield stress
would provide information about the area under the
elastic region; however, this region is very small and is
thus usually negligible with respect to the rest of the
stress-strain curve.
2.12 Modify the curves in Fig. 2.7 to indicate the effects of
temperature. Explain your changes.
These modifications can be made by lowering the slope
of the elastic region and lowering the general height of
the curves. See, for example, Fig. 2.9 on p. 43.
2.13 Using a specific example, show why the deformation
rate, say in m/s, and the true strain rate are not the
same.
The deformation rate is the quantity v in Eqs. (2.16)
and (2.17). Thus, when v is held constant during de
formation (hence a constant deformation rate), the true
strain rate will vary (l increases), whereas the engineer
ing strain rate will remain constant. Hence, the two
quantities are not the same.
2.14 It has been stated that the higher the value of m, the
more diffuse the neck is, and likewise, the lower the
value of m, the more localized the neck is. Explain the
reason for this behavior.
As discussed in Section 2.2.7, with high m values, the
material stretches to a greater length before it fails; this
behavior is an indication that necking is delayed with
increasing m. When necking is about to begin, the
necking region’s strength with respect to the rest of the
specimen increases, due to strain hardening. However,
the strain rate in the necking region is also higher than
in the rest of the specimen, because the material is elon
gating faster there. Since the material in the necked re
gion becomes stronger as it is strained at a higher rate,
the region exhibits a greater resistance to necking. The
increase in resistance to necking thus depends on the
magnitude of m. As the tension test progresses, neck
ing becomes more diffuse, and the specimen becomes
longer before fracture; hence, total elongation increases
with increasing values of m. As expected, the elonga
tion after necking (postuniform elongation) also increases
with increasing m. It has been observed that the value
of m decreases with metals of increasing strength.
2.15 Explain why materials with high m values, such as hot
glass and silly putty, when stretched slowly, undergo
large elongations before failure. Consider events tak
ing place in the necked region of the specimen.
The answer is similar to Answer 2.14 above.
2.16 Assume that you are running four-point bending tests
on a number of identical specimens of the same length
and cross-section, but with increasing distance be
tween the upper points of loading (see Fig. 2.19b).
What changes, if any, would you expect in the test re
sults? Explain.
2
As the distance between the upper points of loading
in Fig. 2.19b increases, the magnitude of the bending
moment decreases. However, the volume of material
subjected to the maximum bending moment (hence to
maximum stress) increases. Thus, the probability of
failure in the four-point test increases as this distance
increases.
2.17 Would Eq. (2.10) hold true in the elastic range? Explain.
Note that this equation is based on volume constancy,
i.e., Aolo = Al. We know, however, that because the
Poisson’s ratio ν is less than 0.5 in the elastic range, the
volume is not constant in a tension test; see Eq. (2.47)
on p. 71. Therefore, the expression is not valid in the
elastic range.
2.18 Why have different types of hardness tests been devel
oped? How would you measure the hardness of a very
large object?
There are several basic reasons:
1. The overall hardness range of the materials
2. The hardness of their constituents; see Chapter 3;
3. The thickness of the specimen, such as bulk ver
sus foil
4. The size of the specimen with respect to that of the
indenter
5. The surface finish of the part being tested.
2.19 Which hardness tests and scales would you use for
very thin strips of material, such as aluminum foil?
Why?
Because aluminum foil is very thin, the indentations
on the surface must be very small so as not to affect
test results. Suitable tests would be a microhardness
test such as Knoop or Vickers under very light loads
(see Fig. 2.20 on p. 54). The accuracy of the test can
be validated by observing any changes in the surface
appearance opposite to the indented side.
2.20 List and explain the factors that you would consider in
selecting an appropriate hardness test and scale for a
particular application.
Hardness tests mainly have three differences:
1. type of indenter,
2. applied load, and
3. method of indentation measurement (depth or
surface area of indentation, or rebound of inden
ter).
2.21 In a Brinell hardness test, the resulting impression is
found to be an ellipse. Give possible explanations for
this result.
There are several possible reasons for this phe
nomenon, but the two most likely are anisotropy in the
material and the presence of surface residual stresses
in the material.
2.22 Referring to Fig. 2.20, the material for testers are either
steel, tungsten carbide, or diamond. Why isn’t dia
mond used for all of the tests?
While diamond is the hardest material known, it
would not, for example, be practical to make and use
a 10-mm diamond indenter because the costs would
be prohibitive. Consequently, a hard material such as
those listed are sufficient for most hardness tests.
2.23 What role does friction play in a hardness test? Can
high friction between a material and indenter affect a
hardness test? Explain.
The effect of friction has been found to be minimal. In
a hardness test, most of the indentation occurs through
plastic deformation, and there is very little sliding at
the indenter-workpiece interface; see Fig. 2.23 on p. 58.
2.24 Describe the difference between creep and stress relax
ation, giving two examples for each as they relate to
engineering applications.
Creep is the permanent deformation of a part that is
under a load over a period of time, usually occurring at
elevated temperatures. Stress relaxation is the decrease
in the stress level in a part under a constant strain. Ex
amples of creep include:
1. turbine blades operating at high temperatures,
and
2. high-temperature steam linesand furnace compo
nents.
Stress relaxation is observed when, for example, a rub
ber band or a thermoplastic is pulled to a specific
length and held at that length for a period of time. This
phenomenon is commonly observed in rivets, bolts,
and guy wires, as well as thermoplastic components.
2.25 Referring to the two impact tests shown in Fig. 2.26,
explain how different the results would be if the speci
mens were impacted from the opposite directions.
Note that impacting the specimens shown in Fig. 2.26
on p. 61 from the opposite directions would subject
the roots of the notches to compressive stresses, and
thus they would not act as stress raisers. Thus, cracks
would not propagate as they would when under ten
sile stresses. Consequently, the specimens would basi
cally behave as if they were not notched.
2.26 If you remove the layer ad from the part shown in
Fig. 2.27d, such as by machining or grinding, which
way will the specimen curve? (Hint: Assume that
the part in diagram (d) is composed of four horizontal
springs held at the ends. Thus, from the top down, we
have compression, tension, compression, and tension
springs.)
Since the internal forces will have to achieve a state of
static equilibrium, the new part has to bow downward
(i.e., it will hold water). Such residual-stress patterns
3
in Fig. 2.19b increases, the magnitude of the bending
moment decreases. However, the volume of material
subjected to the maximum bending moment (hence to
maximum stress) increases. Thus, the probability of
failure in the four-point test increases as this distance
increases.
2.17 Would Eq. (2.10) hold true in the elastic range? Explain.
Note that this equation is based on volume constancy,
i.e., Aolo = Al. We know, however, that because the
Poisson’s ratio ν is less than 0.5 in the elastic range, the
volume is not constant in a tension test; see Eq. (2.47)
on p. 71. Therefore, the expression is not valid in the
elastic range.
2.18 Why have different types of hardness tests been devel
oped? How would you measure the hardness of a very
large object?
There are several basic reasons:
1. The overall hardness range of the materials
2. The hardness of their constituents; see Chapter 3;
3. The thickness of the specimen, such as bulk ver
sus foil
4. The size of the specimen with respect to that of the
indenter
5. The surface finish of the part being tested.
2.19 Which hardness tests and scales would you use for
very thin strips of material, such as aluminum foil?
Why?
Because aluminum foil is very thin, the indentations
on the surface must be very small so as not to affect
test results. Suitable tests would be a microhardness
test such as Knoop or Vickers under very light loads
(see Fig. 2.20 on p. 54). The accuracy of the test can
be validated by observing any changes in the surface
appearance opposite to the indented side.
2.20 List and explain the factors that you would consider in
selecting an appropriate hardness test and scale for a
particular application.
Hardness tests mainly have three differences:
1. type of indenter,
2. applied load, and
3. method of indentation measurement (depth or
surface area of indentation, or rebound of inden
ter).
2.21 In a Brinell hardness test, the resulting impression is
found to be an ellipse. Give possible explanations for
this result.
There are several possible reasons for this phe
nomenon, but the two most likely are anisotropy in the
material and the presence of surface residual stresses
in the material.
2.22 Referring to Fig. 2.20, the material for testers are either
steel, tungsten carbide, or diamond. Why isn’t dia
mond used for all of the tests?
While diamond is the hardest material known, it
would not, for example, be practical to make and use
a 10-mm diamond indenter because the costs would
be prohibitive. Consequently, a hard material such as
those listed are sufficient for most hardness tests.
2.23 What role does friction play in a hardness test? Can
high friction between a material and indenter affect a
hardness test? Explain.
The effect of friction has been found to be minimal. In
a hardness test, most of the indentation occurs through
plastic deformation, and there is very little sliding at
the indenter-workpiece interface; see Fig. 2.23 on p. 58.
2.24 Describe the difference between creep and stress relax
ation, giving two examples for each as they relate to
engineering applications.
Creep is the permanent deformation of a part that is
under a load over a period of time, usually occurring at
elevated temperatures. Stress relaxation is the decrease
in the stress level in a part under a constant strain. Ex
amples of creep include:
1. turbine blades operating at high temperatures,
and
2. high-temperature steam linesand furnace compo
nents.
Stress relaxation is observed when, for example, a rub
ber band or a thermoplastic is pulled to a specific
length and held at that length for a period of time. This
phenomenon is commonly observed in rivets, bolts,
and guy wires, as well as thermoplastic components.
2.25 Referring to the two impact tests shown in Fig. 2.26,
explain how different the results would be if the speci
mens were impacted from the opposite directions.
Note that impacting the specimens shown in Fig. 2.26
on p. 61 from the opposite directions would subject
the roots of the notches to compressive stresses, and
thus they would not act as stress raisers. Thus, cracks
would not propagate as they would when under ten
sile stresses. Consequently, the specimens would basi
cally behave as if they were not notched.
2.26 If you remove the layer ad from the part shown in
Fig. 2.27d, such as by machining or grinding, which
way will the specimen curve? (Hint: Assume that
the part in diagram (d) is composed of four horizontal
springs held at the ends. Thus, from the top down, we
have compression, tension, compression, and tension
springs.)
Since the internal forces will have to achieve a state of
static equilibrium, the new part has to bow downward
(i.e., it will hold water). Such residual-stress patterns
3
can be modeled with a set of horizontal tension and
compression springs. Note that the top layer of the ma
terial ad in Fig. 2.27d, which is under compression, has
the tendency to bend the bar upward. When this stress
is relieved (such as by removing a layer), the bar will
compensate for it by bending downward.
2.27 Is it possible to completely remove residual stresses in a
piece of material by the technique described in Fig. 2.29
if the material is elastic, linearly strain hardening? Ex
plain.
By following the sequence of events depicted in
Fig. 2.29 on p. 64, it can be seen that it is not possi
ble to completely remove the residual stresses. Note
that for an elastic, linearly strain hardening material,
σc
� will never catch up with σt
� .
2.28 Referring to Fig. 2.29, would it be possible to elimi
nate residual stresses by compression? Assume that
the piece of material will not buckle under the uniaxial
compressive force.
Yes, by the same mechanism described in Fig. 2.29 on
p. 64.
2.29 List and explain the desirable mechanical properties
for (a) an elevator cable; (b) a bandage; (c) a shoe sole;
(d) a fish hook; (e) an automotive piston; (f) a boat pro
peller; (g) a gas-turbine blade; and (h) a staple.
The following are some basic considerations:
(a) Elevator cable: The cable should not elongate elas
tically to a large extent or undergo yielding as the
load is increased. These requirements thus call for
a material with a high elastic modulus and yield
stress.
(b) Bandage: The bandage material must be compli
ant, that is, have a low stiffness, but have high
strength in the membrane direction. Its inner sur
face must be permeable and outer surface resis
tant to environmental effects.
(c) Shoe sole: The sole should be compliant for com
fort, with a high resilience. It should be tough so
that it absorbs shock and should have high fric
tion and wear resistance.
(d) Fish hook: A fish hook needs to have high
strength so that it doesn’t deform permanently
under load, and thus maintain its shape. It should
be stiff (for better control during its use) and
should be resistant the environment it is used in
(such as salt water).
(e) Automotive piston: This product must have high
strength at elevated temperatures, high physical
and thermal shock resistance, and low mass.
(f) Boat propeller: The material must be stiff (to
maintain its shape) and resistant to corrosion, and
also have abrasion resistance because the pro
peller encounters sand and other abrasive parti
cles when operated close to shore.
(g) Gas turbine blade: A gas turbine blade operates
at high temperatures (depending on its location in
the turbine); thus it should have high-temperature
strength and resistance to creep, as well as to oxi
dation and corrosion due to combustion products
during its use.
(h) Staple: The properties should be closely parallel
to that of a paper clip. The staple should have
high ductility to allow it to be deformed without
fracture, and also have low yield stress so that it
can be bent (as well as unbent when removing it)
easily without requiring excessive force.
2.30 Make a sketch showing the nature and distribution of
the residual stresses in Figs. 2.28a and b before the parts
were cut. Assume that the split parts are free from any
stresses. (Hint: Force these parts back to the shape they
were in before they were cut.)
As the question states, when we force back the split
portions in the specimen in Fig. 2.28a on p. 63, we in
duce tensile stresses on the outer surfaces and com
pressive on the inner. Thus the original part would,
along its total cross section, have a residual stress dis
tribution of tension-compression-tension. Using the
same technique, we find that the specimen in Fig. 2.28b
would have a similar residual stress distribution prior
to cutting.
2.31 It is possible to calculate the work of plastic deforma
tion by measuring the temperature rise in a workpiece,
assuming that there is no heat loss and that the temper
ature distribution is uniform throughout? If the spe
cific heat of the material decreases with increasing tem
perature, will the work of deformation calculated us
ing the specific heat at room temperature be higher or
lower than the actual work done? Explain.
If we calculate the heat using a constant specific heat
value in Eq. (2.62), the work will be higher than it ac
tually is. This is because, by definition, as the specific
heat decreases, less work is required to raise the work
piece temperature by one degree. Consequently, the
calculated work will be higher than the actual work
done.
2.32 Explain whether or not the volume of a metal specimen
changes when the specimen is subjected to a state of
(a) uniaxial compressive stress and (b) uniaxial tensile
stress, all in the elastic range.
For case (a), the quantity in parentheses in Eq. (2.47)
on p. 71 will be negative, because of the compressive
stress. Since the rest of the terms are positive, the prod
uct of these terms is negative and, hence, there will be
a decrease in volume (This can also be deduced intu
itively.) For case (b), it will be noted that the volume
will increase.
2.33 It is relatively easy to subject a specimen to hydrostatic
compression, such as by using a chamber filled with a
liquid. Devise a means whereby the specimen (say, in
4
compression springs. Note that the top layer of the ma
terial ad in Fig. 2.27d, which is under compression, has
the tendency to bend the bar upward. When this stress
is relieved (such as by removing a layer), the bar will
compensate for it by bending downward.
2.27 Is it possible to completely remove residual stresses in a
piece of material by the technique described in Fig. 2.29
if the material is elastic, linearly strain hardening? Ex
plain.
By following the sequence of events depicted in
Fig. 2.29 on p. 64, it can be seen that it is not possi
ble to completely remove the residual stresses. Note
that for an elastic, linearly strain hardening material,
σc
� will never catch up with σt
� .
2.28 Referring to Fig. 2.29, would it be possible to elimi
nate residual stresses by compression? Assume that
the piece of material will not buckle under the uniaxial
compressive force.
Yes, by the same mechanism described in Fig. 2.29 on
p. 64.
2.29 List and explain the desirable mechanical properties
for (a) an elevator cable; (b) a bandage; (c) a shoe sole;
(d) a fish hook; (e) an automotive piston; (f) a boat pro
peller; (g) a gas-turbine blade; and (h) a staple.
The following are some basic considerations:
(a) Elevator cable: The cable should not elongate elas
tically to a large extent or undergo yielding as the
load is increased. These requirements thus call for
a material with a high elastic modulus and yield
stress.
(b) Bandage: The bandage material must be compli
ant, that is, have a low stiffness, but have high
strength in the membrane direction. Its inner sur
face must be permeable and outer surface resis
tant to environmental effects.
(c) Shoe sole: The sole should be compliant for com
fort, with a high resilience. It should be tough so
that it absorbs shock and should have high fric
tion and wear resistance.
(d) Fish hook: A fish hook needs to have high
strength so that it doesn’t deform permanently
under load, and thus maintain its shape. It should
be stiff (for better control during its use) and
should be resistant the environment it is used in
(such as salt water).
(e) Automotive piston: This product must have high
strength at elevated temperatures, high physical
and thermal shock resistance, and low mass.
(f) Boat propeller: The material must be stiff (to
maintain its shape) and resistant to corrosion, and
also have abrasion resistance because the pro
peller encounters sand and other abrasive parti
cles when operated close to shore.
(g) Gas turbine blade: A gas turbine blade operates
at high temperatures (depending on its location in
the turbine); thus it should have high-temperature
strength and resistance to creep, as well as to oxi
dation and corrosion due to combustion products
during its use.
(h) Staple: The properties should be closely parallel
to that of a paper clip. The staple should have
high ductility to allow it to be deformed without
fracture, and also have low yield stress so that it
can be bent (as well as unbent when removing it)
easily without requiring excessive force.
2.30 Make a sketch showing the nature and distribution of
the residual stresses in Figs. 2.28a and b before the parts
were cut. Assume that the split parts are free from any
stresses. (Hint: Force these parts back to the shape they
were in before they were cut.)
As the question states, when we force back the split
portions in the specimen in Fig. 2.28a on p. 63, we in
duce tensile stresses on the outer surfaces and com
pressive on the inner. Thus the original part would,
along its total cross section, have a residual stress dis
tribution of tension-compression-tension. Using the
same technique, we find that the specimen in Fig. 2.28b
would have a similar residual stress distribution prior
to cutting.
2.31 It is possible to calculate the work of plastic deforma
tion by measuring the temperature rise in a workpiece,
assuming that there is no heat loss and that the temper
ature distribution is uniform throughout? If the spe
cific heat of the material decreases with increasing tem
perature, will the work of deformation calculated us
ing the specific heat at room temperature be higher or
lower than the actual work done? Explain.
If we calculate the heat using a constant specific heat
value in Eq. (2.62), the work will be higher than it ac
tually is. This is because, by definition, as the specific
heat decreases, less work is required to raise the work
piece temperature by one degree. Consequently, the
calculated work will be higher than the actual work
done.
2.32 Explain whether or not the volume of a metal specimen
changes when the specimen is subjected to a state of
(a) uniaxial compressive stress and (b) uniaxial tensile
stress, all in the elastic range.
For case (a), the quantity in parentheses in Eq. (2.47)
on p. 71 will be negative, because of the compressive
stress. Since the rest of the terms are positive, the prod
uct of these terms is negative and, hence, there will be
a decrease in volume (This can also be deduced intu
itively.) For case (b), it will be noted that the volume
will increase.
2.33 It is relatively easy to subject a specimen to hydrostatic
compression, such as by using a chamber filled with a
liquid. Devise a means whereby the specimen (say, in
4
Loading page 6...
the shape of a cube or a round disk) can be subjected
to hydrostatic tension, or one approaching this state of
stress. (Note that a thin-walled, internally pressurized
spherical shell is not a correct answer, because it is sub
jected only to a state of plane stress.)
Two possible answers are the following:
1. A solid cube made of a soft metal has all its six
faces brazed to long square bars (of the same cross
section as the specimen); the bars are made of a
stronger metal. The six arms are then subjected to
equal tension forces, thus subjecting the cube to
equal tensile stresses.
2. A thin, solid round disk (such as a coin) and made
of a soft material is brazed between the ends of
two solid round bars of the same diameter as that
of the disk. When subjected to longitudinal ten
sion, the disk will tend to shrink radially. But be
cause it is thin and its flat surfaces are restrained
by the two rods from moving, the disk will be sub
jected to tensile radial stresses. Thus, a state of tri
axial (though not exactly hydrostatic) tension will
exist within the thin disk.
2.34 Referring to Fig. 2.17, make sketches of the state of
stress for an element in the reduced section of the tube
when it is subjected to (a) torsion only; (b) torsion while
the tube is internally pressurized; and (c) torsion while
the tube is externally pressurized. Assume that the
tube is a closed-end tube.
These states of stress can be represented simply by re
ferring to the contents of this chapter as well as the rel
evant materials covered in texts on mechanics of solids.
2.35 A penny-shaped piece of soft metal is brazed to the
ends of two flat, round steel rods of the same diameter
as the piece. The assembly is then subjected to uniax
ial tension. What is the state of stress to which the soft
metal is subjected? Explain.
The penny-shaped soft metal piece will tend to contract
radially due to Poisson’s ratio; however, the solid rods
to which it attached will prevent this from happening.
Consequently, the state of stress will tend to approach
that of hydrostatic tension.
2.36 A circular disk of soft metal is being compressed be
tween two flat, hardened circular steel punches of hav
ing the same diameter as the disk. Assume that the
disk material is perfectly plastic and that there is no
friction or any temperature effects. Explain the change,
if any, in the magnitude of the punch force as the disk
is being compressed plastically to, say, a fraction of its
original thickness.
Note that as it is compressed plastically, the disk
will expand radially, because of volume constancy.
An approximately donut-shaped material will then be
pushed radially outward, which will then exert ra
dial compressive stresses on the disk volume under
the punches. The volume of material directly between
the punches will now subjected to a triaxial compres
sive state of stress. According to yield criteria (see
Section 2.11), the compressive stress exerted by the
punches will thus increase, even though the material
is not strain hardening. Therefore, the punch force will
increase as deformation increases.
2.37 A perfectly plastic metal is yielding under the stress
state σ1, σ2, σ3, where σ1 > σ2 > σ3. Explain what
happens if σ1 is increased.
Consider Fig. 2.32 on p. 70. Points in the interior of
the yield locus are in an elastic state, whereas those on
the yield locus are in a plastic state. Points outside the
yield locus are not admissible. Therefore, an increase
in σ1 while the other stresses remain unchanged would
require an increase in yield stress. This can also be de
duced by inspecting either Eq. (2.38) or Eq. (2.39).
2.38 What is the dilatation of a material with a Poisson’s ra
tio of 0.5? Is it possible for a material to have a Pois
son’s ratio of 0.7? Give a rationale for your answer.
It can be seen from Eq. (2.47) on p. 71 that the dilatation
of a material with ν = 0.5 is always zero, regardless of
the stress state. To examine the case of ν = 0.7, consider
the situation where the stress state is hydrostatic ten
sion. Equation (2.47) would then predict contraction
under a tensile stress, a situation that cannot occur.
2.39 Can a material have a negative Poisson’s ratio? Give a
rationale for your answer.
Solid material do not have a negative Poisson’s ratio,
with the exception of some composite materials (see
Chapter 10), where there can be a negative Poisson’s
ratio in a given direction.
2.40 As clearly as possible, define plane stress and plane
strain.
Plane stress is the situation where the stresses in one of
the direction on an element are zero; plane strain is the
situation where the strains in one of the direction are
zero.
2.41 What test would you use to evaluate the hardness of a
coating on a metal surface? Would it matter if the coat
ing was harder or softer than the substrate? Explain.
The answer depends on whether the coating is rela
tively thin or thick. For a relatively thick coating, con
ventional hardness tests can be conducted, as long as
the deformed region under the indenter is less than
about one-tenth of the coating thickness. If the coat
ing thickness is less than this threshold, then one must
either rely on nontraditional hardness tests, or else
use fairly complicated indentation models to extract
the material behavior. As an example of the former,
atomic force microscopes using diamond-tipped pyra
mids have been used to measure the hardness of coat
ings less than 100 nanometers thick. As an example of
5
to hydrostatic tension, or one approaching this state of
stress. (Note that a thin-walled, internally pressurized
spherical shell is not a correct answer, because it is sub
jected only to a state of plane stress.)
Two possible answers are the following:
1. A solid cube made of a soft metal has all its six
faces brazed to long square bars (of the same cross
section as the specimen); the bars are made of a
stronger metal. The six arms are then subjected to
equal tension forces, thus subjecting the cube to
equal tensile stresses.
2. A thin, solid round disk (such as a coin) and made
of a soft material is brazed between the ends of
two solid round bars of the same diameter as that
of the disk. When subjected to longitudinal ten
sion, the disk will tend to shrink radially. But be
cause it is thin and its flat surfaces are restrained
by the two rods from moving, the disk will be sub
jected to tensile radial stresses. Thus, a state of tri
axial (though not exactly hydrostatic) tension will
exist within the thin disk.
2.34 Referring to Fig. 2.17, make sketches of the state of
stress for an element in the reduced section of the tube
when it is subjected to (a) torsion only; (b) torsion while
the tube is internally pressurized; and (c) torsion while
the tube is externally pressurized. Assume that the
tube is a closed-end tube.
These states of stress can be represented simply by re
ferring to the contents of this chapter as well as the rel
evant materials covered in texts on mechanics of solids.
2.35 A penny-shaped piece of soft metal is brazed to the
ends of two flat, round steel rods of the same diameter
as the piece. The assembly is then subjected to uniax
ial tension. What is the state of stress to which the soft
metal is subjected? Explain.
The penny-shaped soft metal piece will tend to contract
radially due to Poisson’s ratio; however, the solid rods
to which it attached will prevent this from happening.
Consequently, the state of stress will tend to approach
that of hydrostatic tension.
2.36 A circular disk of soft metal is being compressed be
tween two flat, hardened circular steel punches of hav
ing the same diameter as the disk. Assume that the
disk material is perfectly plastic and that there is no
friction or any temperature effects. Explain the change,
if any, in the magnitude of the punch force as the disk
is being compressed plastically to, say, a fraction of its
original thickness.
Note that as it is compressed plastically, the disk
will expand radially, because of volume constancy.
An approximately donut-shaped material will then be
pushed radially outward, which will then exert ra
dial compressive stresses on the disk volume under
the punches. The volume of material directly between
the punches will now subjected to a triaxial compres
sive state of stress. According to yield criteria (see
Section 2.11), the compressive stress exerted by the
punches will thus increase, even though the material
is not strain hardening. Therefore, the punch force will
increase as deformation increases.
2.37 A perfectly plastic metal is yielding under the stress
state σ1, σ2, σ3, where σ1 > σ2 > σ3. Explain what
happens if σ1 is increased.
Consider Fig. 2.32 on p. 70. Points in the interior of
the yield locus are in an elastic state, whereas those on
the yield locus are in a plastic state. Points outside the
yield locus are not admissible. Therefore, an increase
in σ1 while the other stresses remain unchanged would
require an increase in yield stress. This can also be de
duced by inspecting either Eq. (2.38) or Eq. (2.39).
2.38 What is the dilatation of a material with a Poisson’s ra
tio of 0.5? Is it possible for a material to have a Pois
son’s ratio of 0.7? Give a rationale for your answer.
It can be seen from Eq. (2.47) on p. 71 that the dilatation
of a material with ν = 0.5 is always zero, regardless of
the stress state. To examine the case of ν = 0.7, consider
the situation where the stress state is hydrostatic ten
sion. Equation (2.47) would then predict contraction
under a tensile stress, a situation that cannot occur.
2.39 Can a material have a negative Poisson’s ratio? Give a
rationale for your answer.
Solid material do not have a negative Poisson’s ratio,
with the exception of some composite materials (see
Chapter 10), where there can be a negative Poisson’s
ratio in a given direction.
2.40 As clearly as possible, define plane stress and plane
strain.
Plane stress is the situation where the stresses in one of
the direction on an element are zero; plane strain is the
situation where the strains in one of the direction are
zero.
2.41 What test would you use to evaluate the hardness of a
coating on a metal surface? Would it matter if the coat
ing was harder or softer than the substrate? Explain.
The answer depends on whether the coating is rela
tively thin or thick. For a relatively thick coating, con
ventional hardness tests can be conducted, as long as
the deformed region under the indenter is less than
about one-tenth of the coating thickness. If the coat
ing thickness is less than this threshold, then one must
either rely on nontraditional hardness tests, or else
use fairly complicated indentation models to extract
the material behavior. As an example of the former,
atomic force microscopes using diamond-tipped pyra
mids have been used to measure the hardness of coat
ings less than 100 nanometers thick. As an example of
5
Loading page 7...
the latter, finite-element models of a coated substrate
being indented by an indenter of a known geometry
can be developed and then correlated to experiments.
2.42 List the advantages and limitations of the stress-strain
relationships given in Fig. 2.7.
Several answers that are acceptable, and the student is
encouraged to develop as many as possible. Two pos
sible answers are:
1. There is a tradeoff between mathematical com
plexity and accuracy in modeling material behav
ior
2. Some materials may be better suited for certain
constitutive laws than others
2.43 Plot the data in the inside front cover on a bar chart,
showing the range of values, and comment on the re
sults.
By the student. An example of a bar chart for the elastic
modulus is shown below.0 100 200 300 400 500
Aluminum
Copper
Lead
Magnesium
Molybdenum
Nickel
Steels
Stainless steels
Titanium
Tungsten
Elastic modulus (GPa)
Metallic materials0 200 400 600 800 1000 1200
Ceramics
Diamond
Glass
Rubbers
Thermoplastics
Thermosets
Boron fibers
Carbon fibers
Glass fibers
Kevlar fibers
Spectra fibers
Elastic modulus (GPa)
Non-metallic materials
Typical comments regarding such a chart are:
1. There is a smaller range for metals than for non
metals;
2. Thermoplastics, thermosets and rubbers are or
ders of magnitude lower than metals and other
non-metals;
3. Diamond and ceramics can be superior to others,
but ceramics have a large range of values.
2.44 A hardness test is conducted on as-received metal as a
quality check. The results show indicate that the hard
ness is too high, indicating that the material may not
have sufficient ductility for the intended application.
The supplier is reluctant to accept the return of the ma
terial, instead claiming that the diamond cone used in
the Rockwell testing was worn and blunt, and hence
the test needed to be recalibrated. Is this explanation
plausible? Explain.
Refer to Fig. 2.20 on p. 54 and note that if an inden
ter is blunt, then the penetration, t, under a given load
will be smaller than that using a sharp indenter. This
then translates into a higher hardness. The explanation
is plausible, but in practice, hardness tests are fairly
reliable and measurements are consistent if the test
ing equipment is properly calibrated and routinely ser
viced.
2.45 Explain why a 0.2% offset is used to obtain the yield
strength in a tension test.
The value of 0.2% is somewhat arbitrary and is used
to set some standard. A yield stress, representing the
transition point from elastic to plastic deformation, is
difficult to measure. This is because the stress-strain
curve is not linearly proportional after the propor
tional limit, which can be as high as one-half the yield
strength in some metals. Therefore, a transition from
elastic to plastic behavior in a stress-strain curve is dif
ficult to discern. The use of a 0.2% offset is a con
venient way of consistently interpreting a yield point
from stress-strain curves.
2.46 Referring to Question 2.45, would the offset method be
necessary for a highly strained hardened material? Ex
plain.
The 0.2% offset is still advisable whenever it can be
used, because it is a standardized approach for deter
mining yield stress, and thus one should not arbitrarily
abandon standards. However, if the material is highly
cold worked, there will be a more noticeable ‘kink’ in
the stress-strain curve, and thus the yield stress is far
more easily discernable than for the same material in
the annealed condition.
2.47 Explain why the hardness of a material is related to a
multiple of the uniaxial compressive stress, since both
involve compression of workpiece material.
The hardness is related to a multiple of the uniaxial
compressive stress, not just the uniaxial compressive
stress, because:
6
being indented by an indenter of a known geometry
can be developed and then correlated to experiments.
2.42 List the advantages and limitations of the stress-strain
relationships given in Fig. 2.7.
Several answers that are acceptable, and the student is
encouraged to develop as many as possible. Two pos
sible answers are:
1. There is a tradeoff between mathematical com
plexity and accuracy in modeling material behav
ior
2. Some materials may be better suited for certain
constitutive laws than others
2.43 Plot the data in the inside front cover on a bar chart,
showing the range of values, and comment on the re
sults.
By the student. An example of a bar chart for the elastic
modulus is shown below.0 100 200 300 400 500
Aluminum
Copper
Lead
Magnesium
Molybdenum
Nickel
Steels
Stainless steels
Titanium
Tungsten
Elastic modulus (GPa)
Metallic materials0 200 400 600 800 1000 1200
Ceramics
Diamond
Glass
Rubbers
Thermoplastics
Thermosets
Boron fibers
Carbon fibers
Glass fibers
Kevlar fibers
Spectra fibers
Elastic modulus (GPa)
Non-metallic materials
Typical comments regarding such a chart are:
1. There is a smaller range for metals than for non
metals;
2. Thermoplastics, thermosets and rubbers are or
ders of magnitude lower than metals and other
non-metals;
3. Diamond and ceramics can be superior to others,
but ceramics have a large range of values.
2.44 A hardness test is conducted on as-received metal as a
quality check. The results show indicate that the hard
ness is too high, indicating that the material may not
have sufficient ductility for the intended application.
The supplier is reluctant to accept the return of the ma
terial, instead claiming that the diamond cone used in
the Rockwell testing was worn and blunt, and hence
the test needed to be recalibrated. Is this explanation
plausible? Explain.
Refer to Fig. 2.20 on p. 54 and note that if an inden
ter is blunt, then the penetration, t, under a given load
will be smaller than that using a sharp indenter. This
then translates into a higher hardness. The explanation
is plausible, but in practice, hardness tests are fairly
reliable and measurements are consistent if the test
ing equipment is properly calibrated and routinely ser
viced.
2.45 Explain why a 0.2% offset is used to obtain the yield
strength in a tension test.
The value of 0.2% is somewhat arbitrary and is used
to set some standard. A yield stress, representing the
transition point from elastic to plastic deformation, is
difficult to measure. This is because the stress-strain
curve is not linearly proportional after the propor
tional limit, which can be as high as one-half the yield
strength in some metals. Therefore, a transition from
elastic to plastic behavior in a stress-strain curve is dif
ficult to discern. The use of a 0.2% offset is a con
venient way of consistently interpreting a yield point
from stress-strain curves.
2.46 Referring to Question 2.45, would the offset method be
necessary for a highly strained hardened material? Ex
plain.
The 0.2% offset is still advisable whenever it can be
used, because it is a standardized approach for deter
mining yield stress, and thus one should not arbitrarily
abandon standards. However, if the material is highly
cold worked, there will be a more noticeable ‘kink’ in
the stress-strain curve, and thus the yield stress is far
more easily discernable than for the same material in
the annealed condition.
2.47 Explain why the hardness of a material is related to a
multiple of the uniaxial compressive stress, since both
involve compression of workpiece material.
The hardness is related to a multiple of the uniaxial
compressive stress, not just the uniaxial compressive
stress, because:
6
Loading page 8...
1. The volume of material that is stressed is different
- in a hardness test, the volume that is under stress
is not just a cylinder beneath the inventor.
2. The stressed volume is constrained by the elas
tic material outside of the indentation area. This
often requires material to deform laterally and
counter to the indentation direction - see Fig. 2.21
2.48 Without using the words “stress” or “strain”, define
elastic modulus.
This is actually quite challenging, but historically sig-
nificant, since Thomas Young did not have the benefit
of the concept of strain when he first defined modu
lus of elasticity. Young’s definition satisfies the project
requirement. In Young’s words:
The modulus of the elasticity of any substance is a column
of the same substance, capable of producing a pressure on its
base which is to the weight causing a certain degree of com
pression as the length of the substance is to the diminution
of the length.
There are many possible other definitions, of course.
PROBLEMS
2.49 A strip of metal is originally 1.0 m long. It is stretched
in three steps: first to a length of 1.5 m, then to 2.5 m,
and finally to 3.0 m. Show that the total true strain is
the sum of the true strains in each step, that is, that
the strains are additive. Show that, using engineering
strains, the strain for each step cannot be added to ob
tain the total strain.
The true strain is given by Eq. (2.9) on p. 36 as
l
e = ln lo
Therefore, the true strains for the three steps are:
1.5
e1 = ln = 0.4055
1.0
2.5
e2 = ln = 0.5108
1.5
3.0
e3 = ln = 0.1823
2.5
The sum of these true strains is e = 0.4055 + 0.5108 +
0.1823 = 1.099. The true strain from step 1 to 3 is
3
e = ln = 1.099
1
Therefore the true strains are additive. Using the same
approach for engineering strain as defined by Eq. (2.1),
we obtain e1 = 0.5, e2 = 0.667, and e3 = 0.20. The sum
of these strains is e1 + e2 + e3 = 1.367. The engineering
strain from step 1 to 3 is
l − lo 3 − 1
e = = = 2
lo 1
Note that this is not equal to the sum of the engineering
strains for the individual steps. The following Matlab
code can be used to demonstrate that this is generally
true, and not just a conclusion for the specific deforma
tions stated in the problem.
l0=1;
l1=1.5;
l2=2.5;
l3=3;
etot=(l1-l0)/l0+(l2-l0)/l0+(l3-l0)/l0;
efin=(l3-l0)/l0;
epstot=log(l1/l0)+log(l2/l1)+log(l3/l2);
epsfin=log(l3/l0);
2.50 A paper clip is made of wire 1.00 mm in diameter. If
the original material from which the wire is made is a
rod 15 mm in diameter, calculate the longitudinal and
diametrical engineering and true strains that the wire
has undergone during processing.
Assuming volume constancy, we may write
2 2
lf do 15
= = = 225
lo df 1.00
Letting lo be unity, the longitudinal engineering strain
is e1 = (225 − 1)/1 = 224. The diametral engineering
strain is calculated as
1 − 15
ed = = −0.933
15
The longitudinal true strain is given by Eq. (2.9) on
p. 36 as
l
e = ln = ln (224) = 5.412
lo
The diametral true strain is
1
ed = ln = −2.708
15
Note the large difference between the engineering and
true strains, even though both describe the same phe
nomenon. Note also that the sum of the true strains
(recognizing that the radial strain is a0.5
er = ln =15
−2.708) in the three principal directions is zero, indi
cating volume constancy in plastic deformation.
The following Matlab code is useful:
7
- in a hardness test, the volume that is under stress
is not just a cylinder beneath the inventor.
2. The stressed volume is constrained by the elas
tic material outside of the indentation area. This
often requires material to deform laterally and
counter to the indentation direction - see Fig. 2.21
2.48 Without using the words “stress” or “strain”, define
elastic modulus.
This is actually quite challenging, but historically sig-
nificant, since Thomas Young did not have the benefit
of the concept of strain when he first defined modu
lus of elasticity. Young’s definition satisfies the project
requirement. In Young’s words:
The modulus of the elasticity of any substance is a column
of the same substance, capable of producing a pressure on its
base which is to the weight causing a certain degree of com
pression as the length of the substance is to the diminution
of the length.
There are many possible other definitions, of course.
PROBLEMS
2.49 A strip of metal is originally 1.0 m long. It is stretched
in three steps: first to a length of 1.5 m, then to 2.5 m,
and finally to 3.0 m. Show that the total true strain is
the sum of the true strains in each step, that is, that
the strains are additive. Show that, using engineering
strains, the strain for each step cannot be added to ob
tain the total strain.
The true strain is given by Eq. (2.9) on p. 36 as
l
e = ln lo
Therefore, the true strains for the three steps are:
1.5
e1 = ln = 0.4055
1.0
2.5
e2 = ln = 0.5108
1.5
3.0
e3 = ln = 0.1823
2.5
The sum of these true strains is e = 0.4055 + 0.5108 +
0.1823 = 1.099. The true strain from step 1 to 3 is
3
e = ln = 1.099
1
Therefore the true strains are additive. Using the same
approach for engineering strain as defined by Eq. (2.1),
we obtain e1 = 0.5, e2 = 0.667, and e3 = 0.20. The sum
of these strains is e1 + e2 + e3 = 1.367. The engineering
strain from step 1 to 3 is
l − lo 3 − 1
e = = = 2
lo 1
Note that this is not equal to the sum of the engineering
strains for the individual steps. The following Matlab
code can be used to demonstrate that this is generally
true, and not just a conclusion for the specific deforma
tions stated in the problem.
l0=1;
l1=1.5;
l2=2.5;
l3=3;
etot=(l1-l0)/l0+(l2-l0)/l0+(l3-l0)/l0;
efin=(l3-l0)/l0;
epstot=log(l1/l0)+log(l2/l1)+log(l3/l2);
epsfin=log(l3/l0);
2.50 A paper clip is made of wire 1.00 mm in diameter. If
the original material from which the wire is made is a
rod 15 mm in diameter, calculate the longitudinal and
diametrical engineering and true strains that the wire
has undergone during processing.
Assuming volume constancy, we may write
2 2
lf do 15
= = = 225
lo df 1.00
Letting lo be unity, the longitudinal engineering strain
is e1 = (225 − 1)/1 = 224. The diametral engineering
strain is calculated as
1 − 15
ed = = −0.933
15
The longitudinal true strain is given by Eq. (2.9) on
p. 36 as
l
e = ln = ln (224) = 5.412
lo
The diametral true strain is
1
ed = ln = −2.708
15
Note the large difference between the engineering and
true strains, even though both describe the same phe
nomenon. Note also that the sum of the true strains
(recognizing that the radial strain is a0.5
er = ln =15
−2.708) in the three principal directions is zero, indi
cating volume constancy in plastic deformation.
The following Matlab code is useful:
7
Loading page 9...
d=0.001;
d0=0.015;
l0=1;
lf=(d0/d)ˆ2*l0;
e1=(lf-l0)/l0;
ed=(d-d0)/d0;
epsilon=log(lf/l0);
epsilon_d=log(d/d0);
2.51 A material has the following properties: Sut = 350 MPa
and n = 0.20. Calculate its strength coefficient, K.
Note from Eq. (2.11) on p. 36 that Sut,true = Ken = Knn
because at necking e = n. From Fig. 2.3, P
Sut = Ao ,
where P is the load at necking. The true ultimate ten
sile strength would be
Ao
Sut,true = P/A = Sut A .
From Eq. (2.10),
Ao
ln = e = 0.20
A
Therefore, Ao = exp(0.2) = 1.2214
A
Substituting into the expression for true ultimate
strength,
Sut,true = (350 MPa) (1.2214) = 427 MPa.
The strength coefficient, K, can then be found as
427
K = = 589 MPa.
0.20.2
2.52 Based on the information given in Fig. 2.6, calculate the
ultimate tensile strength of 304 stainless steel.
From Fig. 2.6 on p. 39, the true stress for 304 stainless
steel at necking (where the slope changes; see Fig. 2.7e)
is found to be about 900 MPa, while the true strain is
about 0.4. We also know that the ratio of the original to
necked areas of the specimen is given by
Ao
ln = 0.40
Aneck
or Aneck = e−0.40 = 0.670
Ao
Thus,
Sut = (900)(0.670) = 603 MPa.
2.53 Calculate the ultimate tensile strength (engineering) of
a material whose strength coefficient is 300 MPa and
that necks at a true strain of 0.25.
In this problem, K = 300 MPa and n = 0.25. Following
the same procedure as in Example 2.1 on p. 41, the true
ultimate tensile strength is
σ = (300)(0.25)0.25 = 212 MPa
and
Aneck = Aoe−0.25 = 0.779Ao
Consequently,
Sut = (212)(0.779) = 165 MPa.
2.54 A material has a strength coefficient K = 700 MPa. As
suming that a tensile-test specimen made from this ma
terial begins to neck at a true strain of 0.20, show that
the ultimate tensile strength of this material is 415 MPa.
The approach is the same as in Example 2.1 on p. 41.
Since the necking strain corresponds to the maximum
load and the necking strain for this material is given
as e = n = 0.20, we have, as the true ultimate tensile
strength:
Sut,true = (700)(0.20)0.20 = 507 MPa.
The cross-sectional area at the onset of necking is ob
tained from
Ao
ln = n = 0.20.
Aneck
Consequently,
Aneck = Ao exp(−0.20)
and the maximum load, P , is
P = σA = Sut,trueAo exp(−0.20)
= (507)(0.8187)(Ao) = 415 × 106 Ao
Since Sut = P/Ao, we have Sut = 415 MPa. This is
confirmed with the following Matlab code.
K=700e6;
n=0.2;
Sut_true=K*nˆn;
P=Sut_true*exp(-n)
2.55 A cable is made of four parallel strands of different ma
terials, all behaving according to the equation σ = Ken ,
where n = 0.20. The materials, strength coefficients
and cross-sections are as follows:
Material A: K = 450 MPa, Ao = 7 mm2;
Material B: K = 600 MPa, Ao = 2.5 mm2;
Material C: K = 300 MPa, Ao = 3 mm2;
Material D: K = 750 MPa, Ao = 2 mm2;
(a) Calculate the maximum tensile force that this ca
ble can withstand prior to necking.
(b) Explain how you would arrive at an answer if the
n values of the three strands were different from
each other.
8
d0=0.015;
l0=1;
lf=(d0/d)ˆ2*l0;
e1=(lf-l0)/l0;
ed=(d-d0)/d0;
epsilon=log(lf/l0);
epsilon_d=log(d/d0);
2.51 A material has the following properties: Sut = 350 MPa
and n = 0.20. Calculate its strength coefficient, K.
Note from Eq. (2.11) on p. 36 that Sut,true = Ken = Knn
because at necking e = n. From Fig. 2.3, P
Sut = Ao ,
where P is the load at necking. The true ultimate ten
sile strength would be
Ao
Sut,true = P/A = Sut A .
From Eq. (2.10),
Ao
ln = e = 0.20
A
Therefore, Ao = exp(0.2) = 1.2214
A
Substituting into the expression for true ultimate
strength,
Sut,true = (350 MPa) (1.2214) = 427 MPa.
The strength coefficient, K, can then be found as
427
K = = 589 MPa.
0.20.2
2.52 Based on the information given in Fig. 2.6, calculate the
ultimate tensile strength of 304 stainless steel.
From Fig. 2.6 on p. 39, the true stress for 304 stainless
steel at necking (where the slope changes; see Fig. 2.7e)
is found to be about 900 MPa, while the true strain is
about 0.4. We also know that the ratio of the original to
necked areas of the specimen is given by
Ao
ln = 0.40
Aneck
or Aneck = e−0.40 = 0.670
Ao
Thus,
Sut = (900)(0.670) = 603 MPa.
2.53 Calculate the ultimate tensile strength (engineering) of
a material whose strength coefficient is 300 MPa and
that necks at a true strain of 0.25.
In this problem, K = 300 MPa and n = 0.25. Following
the same procedure as in Example 2.1 on p. 41, the true
ultimate tensile strength is
σ = (300)(0.25)0.25 = 212 MPa
and
Aneck = Aoe−0.25 = 0.779Ao
Consequently,
Sut = (212)(0.779) = 165 MPa.
2.54 A material has a strength coefficient K = 700 MPa. As
suming that a tensile-test specimen made from this ma
terial begins to neck at a true strain of 0.20, show that
the ultimate tensile strength of this material is 415 MPa.
The approach is the same as in Example 2.1 on p. 41.
Since the necking strain corresponds to the maximum
load and the necking strain for this material is given
as e = n = 0.20, we have, as the true ultimate tensile
strength:
Sut,true = (700)(0.20)0.20 = 507 MPa.
The cross-sectional area at the onset of necking is ob
tained from
Ao
ln = n = 0.20.
Aneck
Consequently,
Aneck = Ao exp(−0.20)
and the maximum load, P , is
P = σA = Sut,trueAo exp(−0.20)
= (507)(0.8187)(Ao) = 415 × 106 Ao
Since Sut = P/Ao, we have Sut = 415 MPa. This is
confirmed with the following Matlab code.
K=700e6;
n=0.2;
Sut_true=K*nˆn;
P=Sut_true*exp(-n)
2.55 A cable is made of four parallel strands of different ma
terials, all behaving according to the equation σ = Ken ,
where n = 0.20. The materials, strength coefficients
and cross-sections are as follows:
Material A: K = 450 MPa, Ao = 7 mm2;
Material B: K = 600 MPa, Ao = 2.5 mm2;
Material C: K = 300 MPa, Ao = 3 mm2;
Material D: K = 750 MPa, Ao = 2 mm2;
(a) Calculate the maximum tensile force that this ca
ble can withstand prior to necking.
(b) Explain how you would arrive at an answer if the
n values of the three strands were different from
each other.
8
Loading page 10...
(a) Necking will occur when e = n = 0.20. At
this point the true stresses in each cable are, from
Eq. (2.11) on p. 36,
σA = (450)0.20.2 = 326 MPa
σB = (600)0.20.2 = 435 MPa
σC = (300)0.20.2 = 217 MPa
σD = (760)0.20.2 = 543 MPa
The areas at necking are calculated from Aneck =
Aoe−n (see Example 2.1 on p. 41):
2
AA = (7)e−0.2 = 5.73 mm
2
AB = (2.5)e−0.2 = 2.04 mm
2
AC = (3)e−0.2 = 2.46 mm
2
AD = (2)e−0.2 = 1.64 mm
Hence the total load that the cable can support is
P = (326)(5.73) + (435)(2.04)
+(217)(2.46) + (543)(1.64)
= 4180 N.
The following Matlab code is helpful:
K_A=450e6;
K_B=600e6;
K_C=300e6;
K_D=750e6;
A0_A=7/1e6;
A0_B=2.5/1e6;
A0_C=3/1e6;
A0_D=2/1e6;
n=0.20;
s_A=K_A*nˆn;
s_B=K_B*nˆn;
s_C=K_C*nˆn;
s_D=K_D*nˆn;
A_A=A0_A*exp(-1*n);
A_B=A0_B*exp(-1*n);
A_C=A0_C*exp(-1*n);
A_D=A0_D*exp(-1*n);
P=s_A*A_A+s_B*A_B+s_C*A_C+s_D*A_D;
(b) If the n values of the four strands were differ
ent, the procedure would consist of plotting the
load-elongation curves of the four strands on the
same chart, then obtaining graphically the maxi
mum load. Alternately, a computer program can
be written to determine the maximum load.
2.56 Using only Fig. 2.6, calculate the maximum load in ten
sion testing of a 304 stainless-steel specimen with an
original diameter of 6.0 mm.
Observe from Fig. 2.6 on p. 39 that necking begins at
a true strain of about 0.4 for 304 stainless steel, and
that Sut,true is about 900 MPa (this is the location of
the ‘kink’ in the stress-strain curve). The original cross-
sectional area is Ao = π(0.006 m)2/4 = 2.827×10−5 m2 .
Since n = 0.4, a procedure similar to Example 2.1 on
p. 41 demonstrates that
Ao = exp(0.4) = 1.49
Aneck
Thus 900
Sut = = 604 MPa
1.49
Hence the maximum load is
P = (Sut)(Ao) = (604)(2.827 × 10−5)
or P = 17.1 kN. The following Matlab code is helpful
to investigate other parameters.
n=0.4;
Sut_true=900e6;
A_0=pi*d0*d0/4;
Sut=Sut_true/exp(n);
P=Sut*A_0;
2.57 Using the data given in the inside front cover, calculate
the values of the shear modulus G for the metals listed
in the table.
The important equation is Eq. (2.24) which gives the
shear modulus as
E
G = 2(1 + ν)
The following values can be calculated (mid-range val
ues of ν are taken as appropriate):
Material E (GPa) ν G (GPa)
Al & alloys 69-79 0.32 26-30
Cu & alloys 105-150 0.34 39-56
Pb & alloys 14 0.43 4.9
Mg & alloys 41-45 0.32 15.5-17.0
Mo & alloys 330-360 0.32 125-136
Ni & alloys 180-214 0.31 69-82
Steels 190-200 0.30 73-77
Stainless steels 190-200 0.29 74-77
Ti & alloys 80-130 0.32 30-49
W & alloys 350-400 0.27 138-157
Ceramics 70-1000 0.2 29-417
Glass 70-80 0.24 28-32
Rubbers 0.01-0.1 0.5 0.0033-0.033
Thermoplastics 1.4-3.4 0.36 0.51-1.25
Thermosets 3.5-17 0.34 1.3-6.34
2.58 Derive an expression for the toughness of a material
represented by the equation σ = K (e + 0.2)n and
whose fracture strain is denoted as ef .
Recall that toughness is the area under the stress-strain
curve, hence the toughness for this material would be
9
this point the true stresses in each cable are, from
Eq. (2.11) on p. 36,
σA = (450)0.20.2 = 326 MPa
σB = (600)0.20.2 = 435 MPa
σC = (300)0.20.2 = 217 MPa
σD = (760)0.20.2 = 543 MPa
The areas at necking are calculated from Aneck =
Aoe−n (see Example 2.1 on p. 41):
2
AA = (7)e−0.2 = 5.73 mm
2
AB = (2.5)e−0.2 = 2.04 mm
2
AC = (3)e−0.2 = 2.46 mm
2
AD = (2)e−0.2 = 1.64 mm
Hence the total load that the cable can support is
P = (326)(5.73) + (435)(2.04)
+(217)(2.46) + (543)(1.64)
= 4180 N.
The following Matlab code is helpful:
K_A=450e6;
K_B=600e6;
K_C=300e6;
K_D=750e6;
A0_A=7/1e6;
A0_B=2.5/1e6;
A0_C=3/1e6;
A0_D=2/1e6;
n=0.20;
s_A=K_A*nˆn;
s_B=K_B*nˆn;
s_C=K_C*nˆn;
s_D=K_D*nˆn;
A_A=A0_A*exp(-1*n);
A_B=A0_B*exp(-1*n);
A_C=A0_C*exp(-1*n);
A_D=A0_D*exp(-1*n);
P=s_A*A_A+s_B*A_B+s_C*A_C+s_D*A_D;
(b) If the n values of the four strands were differ
ent, the procedure would consist of plotting the
load-elongation curves of the four strands on the
same chart, then obtaining graphically the maxi
mum load. Alternately, a computer program can
be written to determine the maximum load.
2.56 Using only Fig. 2.6, calculate the maximum load in ten
sion testing of a 304 stainless-steel specimen with an
original diameter of 6.0 mm.
Observe from Fig. 2.6 on p. 39 that necking begins at
a true strain of about 0.4 for 304 stainless steel, and
that Sut,true is about 900 MPa (this is the location of
the ‘kink’ in the stress-strain curve). The original cross-
sectional area is Ao = π(0.006 m)2/4 = 2.827×10−5 m2 .
Since n = 0.4, a procedure similar to Example 2.1 on
p. 41 demonstrates that
Ao = exp(0.4) = 1.49
Aneck
Thus 900
Sut = = 604 MPa
1.49
Hence the maximum load is
P = (Sut)(Ao) = (604)(2.827 × 10−5)
or P = 17.1 kN. The following Matlab code is helpful
to investigate other parameters.
n=0.4;
Sut_true=900e6;
A_0=pi*d0*d0/4;
Sut=Sut_true/exp(n);
P=Sut*A_0;
2.57 Using the data given in the inside front cover, calculate
the values of the shear modulus G for the metals listed
in the table.
The important equation is Eq. (2.24) which gives the
shear modulus as
E
G = 2(1 + ν)
The following values can be calculated (mid-range val
ues of ν are taken as appropriate):
Material E (GPa) ν G (GPa)
Al & alloys 69-79 0.32 26-30
Cu & alloys 105-150 0.34 39-56
Pb & alloys 14 0.43 4.9
Mg & alloys 41-45 0.32 15.5-17.0
Mo & alloys 330-360 0.32 125-136
Ni & alloys 180-214 0.31 69-82
Steels 190-200 0.30 73-77
Stainless steels 190-200 0.29 74-77
Ti & alloys 80-130 0.32 30-49
W & alloys 350-400 0.27 138-157
Ceramics 70-1000 0.2 29-417
Glass 70-80 0.24 28-32
Rubbers 0.01-0.1 0.5 0.0033-0.033
Thermoplastics 1.4-3.4 0.36 0.51-1.25
Thermosets 3.5-17 0.34 1.3-6.34
2.58 Derive an expression for the toughness of a material
represented by the equation σ = K (e + 0.2)n and
whose fracture strain is denoted as ef .
Recall that toughness is the area under the stress-strain
curve, hence the toughness for this material would be
9
Loading page 11...
given by
f
Toughness = σ dE
0
f
= K (E + 0.2)n dE
0
= K (Ef + 0.2)n+1 − 0.2n+1
n + 1
2.59 A cylindrical specimen made of a brittle material 50
mm high and with a diameter of 25 mm is subjected to
a compressive force along its axis. It is found that frac
ture takes place at an angle of 45◦ under a load of 130
kN. Calculate the shear stress and the normal stress,
respectively, acting on the fracture surface.
Assuming that compression takes place without fric
tion, note that two of the principal stresses will be zero.
The third principal stress acting on this specimen is
normal to the specimen and its magnitude is
130, 000
σ3 = − = −264 MPa
(π/4)(0.025)2
The Mohr’s circle for this situation is shown below.
2=90°
The fracture plane is oriented at an angle of 45◦, corre
sponding to a rotation of 90◦ on the Mohr’s circle. This
corresponds to a stress state on the fracture plane of
−σ = τ = 264/2 = 132 MPa.
2.60 What is the modulus of resilience of a highly cold-
worked piece of steel with a hardness of 280 HB? Of
a piece of highly cold-worked copper with a hardness
of 175 HB?
Referring to Fig. 2.22 on p. 57, the value of c in Eq. (2.31)
is approximately 3.2 for highly cold-worked steels and
around 3.4 for cold-worked aluminum. Therefore, ap
proximate c = 3.3 for cold-worked copper. From
Eq. (2.31),
H 280
Sy,steel = = = 87.5 kg/mm2 = 858 MPa
3.2 3.2
H 175
Sy,Cu = = = 53.0 kg/mm2 = 520 MPa
3.3 3.3
From the inside front cover, Esteel = 200 GPa and
ECu = 124 GPa. The modulus of resilience is calcu
lated from Eq. (2.5) on p. 34. For steel:
2
S2
y 858 × 106
Modulus of Resilience = =
2E 2(200 × 109)
or a modulus of resilience for steel of 1.81 MN-m/m3 .
For copper,
2
S2 520 × 106
Modulus of Resilience = y =
2E 2(124 × 109)
or a modulus of resilience for copper of 1.09 MN
m/m3 .
Note that these values are slightly different than the
values given in the text. This is due to the fact that
(a) highly cold-worked metals such as these have a
much higher yield stress than the annealed materials
described in the text; and (b) arbitrary property values
are given in the statement of the problem.
2.61 Calculate the work done in frictionless compression of
a solid cylinder 40 mm high and 15 mm in diameter
to a reduction in height of 50% for the following mate
rials: (a) 1100-O aluminum; (b) annealed copper; (c)
annealed 304 stainless steel; and (d) 70-30 brass, an
nealed.
The work done is calculated from Eq. (2.59) where the
specific energy, u, is obtained from Eq. (2.57) on p. 73.
Since the reduction in height is 50%, the final height is
20 mm and the absolute value of the true strain is
40
E = ln = 0.6931
20
K and n are obtained from Table 2.3 as follows:
Material K (MPa) n
1100-O Al 180 0.20
Cu, annealed 315 0.54
304 Stainless, annealed 1300 0.30
70-30 brass, annealed 895 0.49
u is then calculated from Eq. (2.57). For example, for
1100-O aluminum, where K is 180 MPa and n is 0.20, u
is calculated as
KEn+1 (180)(0.6931)1.2
u = = = 96.6 MN/m3
n + 1 1.2
The volume is calculated as
3
V = πr2l = π(0.0075)2(0.04) = 7.069 × 10−6 m
The work done is the product of the specific work, u,
and the volume, V . Therefore, the results can be tabu
lated as follows.
Material
u
(MN/m3)
Work
(Nm)
1100-O Al 96.6 682
Cu, annealed 168 1186
304 Stainless, annealed 620 4389
70-30 brass, annealed 348 2460
The following Matlab code can be used to confirm re
sults:
10
f
Toughness = σ dE
0
f
= K (E + 0.2)n dE
0
= K (Ef + 0.2)n+1 − 0.2n+1
n + 1
2.59 A cylindrical specimen made of a brittle material 50
mm high and with a diameter of 25 mm is subjected to
a compressive force along its axis. It is found that frac
ture takes place at an angle of 45◦ under a load of 130
kN. Calculate the shear stress and the normal stress,
respectively, acting on the fracture surface.
Assuming that compression takes place without fric
tion, note that two of the principal stresses will be zero.
The third principal stress acting on this specimen is
normal to the specimen and its magnitude is
130, 000
σ3 = − = −264 MPa
(π/4)(0.025)2
The Mohr’s circle for this situation is shown below.
2=90°
The fracture plane is oriented at an angle of 45◦, corre
sponding to a rotation of 90◦ on the Mohr’s circle. This
corresponds to a stress state on the fracture plane of
−σ = τ = 264/2 = 132 MPa.
2.60 What is the modulus of resilience of a highly cold-
worked piece of steel with a hardness of 280 HB? Of
a piece of highly cold-worked copper with a hardness
of 175 HB?
Referring to Fig. 2.22 on p. 57, the value of c in Eq. (2.31)
is approximately 3.2 for highly cold-worked steels and
around 3.4 for cold-worked aluminum. Therefore, ap
proximate c = 3.3 for cold-worked copper. From
Eq. (2.31),
H 280
Sy,steel = = = 87.5 kg/mm2 = 858 MPa
3.2 3.2
H 175
Sy,Cu = = = 53.0 kg/mm2 = 520 MPa
3.3 3.3
From the inside front cover, Esteel = 200 GPa and
ECu = 124 GPa. The modulus of resilience is calcu
lated from Eq. (2.5) on p. 34. For steel:
2
S2
y 858 × 106
Modulus of Resilience = =
2E 2(200 × 109)
or a modulus of resilience for steel of 1.81 MN-m/m3 .
For copper,
2
S2 520 × 106
Modulus of Resilience = y =
2E 2(124 × 109)
or a modulus of resilience for copper of 1.09 MN
m/m3 .
Note that these values are slightly different than the
values given in the text. This is due to the fact that
(a) highly cold-worked metals such as these have a
much higher yield stress than the annealed materials
described in the text; and (b) arbitrary property values
are given in the statement of the problem.
2.61 Calculate the work done in frictionless compression of
a solid cylinder 40 mm high and 15 mm in diameter
to a reduction in height of 50% for the following mate
rials: (a) 1100-O aluminum; (b) annealed copper; (c)
annealed 304 stainless steel; and (d) 70-30 brass, an
nealed.
The work done is calculated from Eq. (2.59) where the
specific energy, u, is obtained from Eq. (2.57) on p. 73.
Since the reduction in height is 50%, the final height is
20 mm and the absolute value of the true strain is
40
E = ln = 0.6931
20
K and n are obtained from Table 2.3 as follows:
Material K (MPa) n
1100-O Al 180 0.20
Cu, annealed 315 0.54
304 Stainless, annealed 1300 0.30
70-30 brass, annealed 895 0.49
u is then calculated from Eq. (2.57). For example, for
1100-O aluminum, where K is 180 MPa and n is 0.20, u
is calculated as
KEn+1 (180)(0.6931)1.2
u = = = 96.6 MN/m3
n + 1 1.2
The volume is calculated as
3
V = πr2l = π(0.0075)2(0.04) = 7.069 × 10−6 m
The work done is the product of the specific work, u,
and the volume, V . Therefore, the results can be tabu
lated as follows.
Material
u
(MN/m3)
Work
(Nm)
1100-O Al 96.6 682
Cu, annealed 168 1186
304 Stainless, annealed 620 4389
70-30 brass, annealed 348 2460
The following Matlab code can be used to confirm re
sults:
10
Loading page 12...
h0=0.040;
d0=0.015;
hf=0.020;
epsilon=log(h0/hf);
K=180e6;
n=0.20;
u=K*epsilonˆ(n+1)/(n+1);
V=pi*d0ˆ2/4*h0;
Work=u*V;
2.62 A tensile-test specimen is made of a material repre
sented by the equation σ = K(E + n)n . (a) Determine
the true strain at which necking will begin. (b) Show
that it is possible for an engineering material to exhibit
this behavior.
(a) In Section 2.2.4 on p. 40, it was noted that insta
bility, hence necking, requires the following con
dition to be fulfilled:
dσ = σ
dE
Consequently, for this material we have
Kn (E + n)n−1 = K (E + n)n
This is solved as n = 0; thus necking begins as
soon as the specimen is subjected to tension.
(b) Yes, this behavior is possible. Consider a tension-
test specimen that has been strained to necking
and then unloaded. Upon loading it again in ten
sion, it will immediately begin to neck.
2.63 Take two solid cylindrical specimens of equal diame
ter, but different heights. Assume that both specimens
are compressed (frictionless) by the same percent re
duction, say 50%. Prove that the final diameters will be
the same.
Identify the shorter cylindrical specimen with the sub
script s and the taller one as t, and their original diam
eter as D. Subscripts f and o indicate final and origi
nal, respectively. Because both specimens undergo the
same percent reduction in height,
htf hsf
=
hto hso
and from volume constancy,
2
htf Dto
=
hto Dtf
and 2
hsf Dso
=
hso Dsf
Because Dto = Dso, note from these relationships that
Dtf = Dsf .
2.64 In a disk test performed on a specimen 50 mm in di
ameter and 2.5 mm thick, the specimen fractures at a
stress of 500 MPa. What was the load on it at fracture?
Equation (2.22) is used to solve this problem. Noting
that σ = 500 MPa, d = 50 mm = 0.05 m, and t = 2.5
mm = 0.0025 m, we can write
2P σπdt
σ = → P =
πdt 2
Therefore
(500 × 106)π(0.05)(0.0025)
P = = 98 kN.
2
The following Matlab code allows for variation in pa
rameters for this problem.
d=0.050;
t=0.0025;
sigma=500e6;
P=sigma*pi*d*t/2;
2.65 In Fig. 2.29a, let the tensile and compressive residual
stresses both be 70 MPa, and the modulus of elasticity
of the material be 200 GPa with a modulus of resilience
of 225 kN-m/m3 . If the original length in diagram (a)
is 500 mm, what should be the stretched length in dia
gram (b) so that, when unloaded, the strip will be free
of residual stresses?
Note that the yield stress can be obtained from Eq. (2.5)
on p. 34 as
S2
Mod. of Resilience = MR = y
2E
Thus,
Sy = 2(MR)E = 2 (225 × 103) (200 × 109)
or Sy = 300 MPa. The strain required to relieve the
residual stress is:
σc Sy 70 × 106 300 × 106
E = + = + = 0.00185
E E 200 × 109 200 × 109
Therefore,
lf lf
E = ln = ln = 0.00185
lo 0.500 m
Therefore, lf = 0.50093 m. The following Matlab code
is helpful
sigma_c=70e6;
E=200e9;
MR=225e3;
l0=0.5;
Sy=(2*MR*E)ˆ(0.5);
epsilon=sigma_c/E+Sy/E;
lf=l0*exp(epsilon);
11
d0=0.015;
hf=0.020;
epsilon=log(h0/hf);
K=180e6;
n=0.20;
u=K*epsilonˆ(n+1)/(n+1);
V=pi*d0ˆ2/4*h0;
Work=u*V;
2.62 A tensile-test specimen is made of a material repre
sented by the equation σ = K(E + n)n . (a) Determine
the true strain at which necking will begin. (b) Show
that it is possible for an engineering material to exhibit
this behavior.
(a) In Section 2.2.4 on p. 40, it was noted that insta
bility, hence necking, requires the following con
dition to be fulfilled:
dσ = σ
dE
Consequently, for this material we have
Kn (E + n)n−1 = K (E + n)n
This is solved as n = 0; thus necking begins as
soon as the specimen is subjected to tension.
(b) Yes, this behavior is possible. Consider a tension-
test specimen that has been strained to necking
and then unloaded. Upon loading it again in ten
sion, it will immediately begin to neck.
2.63 Take two solid cylindrical specimens of equal diame
ter, but different heights. Assume that both specimens
are compressed (frictionless) by the same percent re
duction, say 50%. Prove that the final diameters will be
the same.
Identify the shorter cylindrical specimen with the sub
script s and the taller one as t, and their original diam
eter as D. Subscripts f and o indicate final and origi
nal, respectively. Because both specimens undergo the
same percent reduction in height,
htf hsf
=
hto hso
and from volume constancy,
2
htf Dto
=
hto Dtf
and 2
hsf Dso
=
hso Dsf
Because Dto = Dso, note from these relationships that
Dtf = Dsf .
2.64 In a disk test performed on a specimen 50 mm in di
ameter and 2.5 mm thick, the specimen fractures at a
stress of 500 MPa. What was the load on it at fracture?
Equation (2.22) is used to solve this problem. Noting
that σ = 500 MPa, d = 50 mm = 0.05 m, and t = 2.5
mm = 0.0025 m, we can write
2P σπdt
σ = → P =
πdt 2
Therefore
(500 × 106)π(0.05)(0.0025)
P = = 98 kN.
2
The following Matlab code allows for variation in pa
rameters for this problem.
d=0.050;
t=0.0025;
sigma=500e6;
P=sigma*pi*d*t/2;
2.65 In Fig. 2.29a, let the tensile and compressive residual
stresses both be 70 MPa, and the modulus of elasticity
of the material be 200 GPa with a modulus of resilience
of 225 kN-m/m3 . If the original length in diagram (a)
is 500 mm, what should be the stretched length in dia
gram (b) so that, when unloaded, the strip will be free
of residual stresses?
Note that the yield stress can be obtained from Eq. (2.5)
on p. 34 as
S2
Mod. of Resilience = MR = y
2E
Thus,
Sy = 2(MR)E = 2 (225 × 103) (200 × 109)
or Sy = 300 MPa. The strain required to relieve the
residual stress is:
σc Sy 70 × 106 300 × 106
E = + = + = 0.00185
E E 200 × 109 200 × 109
Therefore,
lf lf
E = ln = ln = 0.00185
lo 0.500 m
Therefore, lf = 0.50093 m. The following Matlab code
is helpful
sigma_c=70e6;
E=200e9;
MR=225e3;
l0=0.5;
Sy=(2*MR*E)ˆ(0.5);
epsilon=sigma_c/E+Sy/E;
lf=l0*exp(epsilon);
11
Loading page 13...
2.66 A horizontal rigid bar c-c is subjecting specimen a
to tension and specimen b to frictionless compression
such that the bar remains horizontal. (See the accompa
nying figure.) The force F is located at a distance ratio
of 2:1. Both specimens have an original cross-sectional
area of 0.0001 m2 and the original lengths are a = 200
mm and b = 115 mm. The material for specimen a has
a true-stress-true-strain curve of σ = (700 MPa)E0.5 .
Plot the true-stress-true-strain curve that the material
for specimen b should have for the bar to remain hori
zontal.F
2 1
a
b
cc
x
From the equilibrium of vertical forces and to keep the
bar horizontal, we note that 2Fa = Fb. Hence, in terms
of true stresses and instantaneous areas, we have
2σaAa = σbAb
From volume constancy we also have, in terms of orig
inal and final dimensions
AoaLoa = AaLa
and
AobLob = AbLb
where Loa = (0.200/0.115)Lob = 1.73Lob. From these
relationships we can show that
0.2 Lb
σb = 2 Kσa
0.115 La
Since σa = KE0.5
a where K = 700 MPa, we can now
write 0.4K Lb √
σb = Ea
0.115 La
Hence, for a deflection of x,
0.4K 0.115 − x 0.2 + x
σb = ln
0.115 0.2 + x 0.2
The true strain in specimen b is given by
0.115 − x
Eb = ln 0.115
By inspecting the figure in the problem statement, we
note that while specimen a gets longer, it will con
tinue exerting some force Fa. However, specimen b will
eventually acquire a cross-sectional area that will be
come infinite as x approaches 115 mm, thus its strength
must approach zero. This observation suggests that
specimen b cannot have a true stress-true strain curve
typical of metals, and that it will have a maximum at
some strain. This is seen in the plot of σb shown below.350
280
210
140
70
0 0 0.5 1.0 1.5 2.0 2.5
True stress (MPa)
Absolute value of true strain
2.67 Inspect the curve that you obtained in Problem 2.66.
Does a typical strain-hardening material behave in that
manner? Explain.
Based on the discussions in Section 2.2.3, it is obvious
that ordinary metals would not normally behave in this
manner. However, under certain conditions, the fol
lowing could explain such behavior:
• When specimen b is heated to higher and higher
temperatures as deformation progresses, with its
strength decreasing as x is increased further after
the maximum value of stress.
• In compression testing of brittle materials, such as
ceramics, when the specimen begins to fracture.
• If the material is susceptible to thermal soften
ing, then it can display such behavior with a suf
ficiently high strain rate.
2.68 Show that you can take a bent bar made of an elastic,
perfectly plastic material and straighten it by stretch
ing it into the plastic range. (Hint: Observe the events
shown in Fig. 2.29.)
The series of events that takes place in straightening
a bent bar by stretching it can be visualized by start
ing with a stress distribution as in Fig. 2.29a on p. 64,
which would represent the unbending of a bent sec
tion. As we apply tension, we algebraically add a uni
form tensile stress to this stress distribution. Note that
the change in the stresses is the same as that depicted
in Fig. 2.29d, namely, the tensile stress increases and
reaches the yield stress, Sy . The compressive stress
is first reduced in magnitude, then becomes tensile.
Eventually, the whole cross section reaches the con
stant yield stress, Sy . Because we now have a uniform
stress distribution throughout its thickness, the bar be
comes straight and remains straight upon unloading.
12
to tension and specimen b to frictionless compression
such that the bar remains horizontal. (See the accompa
nying figure.) The force F is located at a distance ratio
of 2:1. Both specimens have an original cross-sectional
area of 0.0001 m2 and the original lengths are a = 200
mm and b = 115 mm. The material for specimen a has
a true-stress-true-strain curve of σ = (700 MPa)E0.5 .
Plot the true-stress-true-strain curve that the material
for specimen b should have for the bar to remain hori
zontal.F
2 1
a
b
cc
x
From the equilibrium of vertical forces and to keep the
bar horizontal, we note that 2Fa = Fb. Hence, in terms
of true stresses and instantaneous areas, we have
2σaAa = σbAb
From volume constancy we also have, in terms of orig
inal and final dimensions
AoaLoa = AaLa
and
AobLob = AbLb
where Loa = (0.200/0.115)Lob = 1.73Lob. From these
relationships we can show that
0.2 Lb
σb = 2 Kσa
0.115 La
Since σa = KE0.5
a where K = 700 MPa, we can now
write 0.4K Lb √
σb = Ea
0.115 La
Hence, for a deflection of x,
0.4K 0.115 − x 0.2 + x
σb = ln
0.115 0.2 + x 0.2
The true strain in specimen b is given by
0.115 − x
Eb = ln 0.115
By inspecting the figure in the problem statement, we
note that while specimen a gets longer, it will con
tinue exerting some force Fa. However, specimen b will
eventually acquire a cross-sectional area that will be
come infinite as x approaches 115 mm, thus its strength
must approach zero. This observation suggests that
specimen b cannot have a true stress-true strain curve
typical of metals, and that it will have a maximum at
some strain. This is seen in the plot of σb shown below.350
280
210
140
70
0 0 0.5 1.0 1.5 2.0 2.5
True stress (MPa)
Absolute value of true strain
2.67 Inspect the curve that you obtained in Problem 2.66.
Does a typical strain-hardening material behave in that
manner? Explain.
Based on the discussions in Section 2.2.3, it is obvious
that ordinary metals would not normally behave in this
manner. However, under certain conditions, the fol
lowing could explain such behavior:
• When specimen b is heated to higher and higher
temperatures as deformation progresses, with its
strength decreasing as x is increased further after
the maximum value of stress.
• In compression testing of brittle materials, such as
ceramics, when the specimen begins to fracture.
• If the material is susceptible to thermal soften
ing, then it can display such behavior with a suf
ficiently high strain rate.
2.68 Show that you can take a bent bar made of an elastic,
perfectly plastic material and straighten it by stretch
ing it into the plastic range. (Hint: Observe the events
shown in Fig. 2.29.)
The series of events that takes place in straightening
a bent bar by stretching it can be visualized by start
ing with a stress distribution as in Fig. 2.29a on p. 64,
which would represent the unbending of a bent sec
tion. As we apply tension, we algebraically add a uni
form tensile stress to this stress distribution. Note that
the change in the stresses is the same as that depicted
in Fig. 2.29d, namely, the tensile stress increases and
reaches the yield stress, Sy . The compressive stress
is first reduced in magnitude, then becomes tensile.
Eventually, the whole cross section reaches the con
stant yield stress, Sy . Because we now have a uniform
stress distribution throughout its thickness, the bar be
comes straight and remains straight upon unloading.
12
Loading page 14...
2.69 A bar 1 m long is bent and then stress relieved. The
radius of curvature to the neutral axis is 0.50 m. The
bar is 25 mm thick and is made of an elastic, perfectly
plastic material with Sy = 500 MPa and E = 207
GPa. Calculate the length to which this bar should be
stretched so that, after unloading, it will become and
remain straight.
A review of bending theory from a solid mechanics
textbook is necessary for this problem. In particular,
it should be recognized that when the curved bar be
comes straight, the engineering strain it undergoes is
given by the expression
t
e = 2ρ
where t is the thickness and ρ is the radius to the neu
tral axis. Hence in this case,
(0.025)
e = = 0.025
2(0.50)
Since Sy = 500 MPa and E = 207 GPa, we find that the
elastic limit for this material is at an elastic strain of
Sy 500 MPa
e = = = 0.00242
E 207 GPa
which is smaller than 0.025. Therefore, we know that
the bar must be loaded in the plastic range. Follow
ing the description in Answer 2.68 above, the strain re
quired to straighten the bar is twice the elastic limit, or
e = (2)(0.00242) = 0.0048
or lf − lo = 0.0048 → lf = 0.005lo + lo
lo
or lf = 1.0049 m.
The following Matlab code is helpful.
l=1;
rho=0.5;
Sy=500e6;
E=207e9;
t=0.025;
epsilon=t/2/rho;
epsilon_y=Sy/E;
epsilon_b=2*epsilon_y;
lf=l/(1-epsilon_b);
2.70 Assume that a material with a uniaxial yield strength
Sy yields under a stress state of principal stresses σ1,
σ2, σ3, where σ1 > σ2 > σ3. Show that the superpo
sition of a hydrostatic stress p on this system (such as
placing the specimen in a chamber pressurized with a
liquid) does not affect yielding. In other words, the ma
terial will still yield according to yield criteria.
This solution considers the distortion-energy criterion,
although the same derivation could be performed with
the maximum shear stress criterion as well. Equation
(2.39) on p. 67 gives
2 2 2
(σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Sy
2
Now consider a new stress state where the principal
stresses are
σI = σ1 + p1
σI = σ2 + p2
I
σ = σ3 + p3
which represents a new loading with an additional hy
drostatic pressure, p. The distortion-energy criterion
for this stress state is
2 2 2
(σ1
I − σ2
I ) + (σ2
I − σ3
I ) + (σ3
I − σ1
I ) = 2Sy
2
or
2
2S2 = [(σ1 + p) − (σ2 + p)]y
2
+ [(σ2 + p) − (σ3 + p)]
2
+ [(σ3 + p) − (σ1 + p)]
which can be simplified as
2 2 2
(σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2S2
y
which is the original yield criterion. Hence, the yield
criterion is unaffected by the superposition of a hydro
static pressure.
2.71 Give two different and specific examples in which the
maximum-shear-stress and the distortion-energy crite
ria give the same answer.
In order to obtain the same answer for the two yield
criteria, we refer to Fig. 2.32 on p. 70 for plane stress
and note the coordinates at which the two diagrams
meet. Examples are: simple tension, simple compres
sion, equal biaxial tension, and equal biaxial compres
sion. Thus, acceptable answers would include (a) wire
rope, as used on a crane to lift loads; (b) spherical pres
sure vessels, including balloons and gas storage tanks;
and (c) shrink fits.
2.72 A thin-walled spherical shell with a yield strength Sy
is subjected to an internal pressure p. With appropriate
equations, show whether or not the pressure required
to yield this shell depends on the particular yield crite
rion used.
Here we have a state of plane stress with equal biax
ial tension. The answer to Problem 2.71 leads one to
immediately conclude that both the maximum shear
stress and distortion energy criteria will give the same
results. We will now demonstrate this more rigorously.
The principal membrane stresses are given by
pr
σ1 = σ2 = 2t
and
σ3 = 0
13
radius of curvature to the neutral axis is 0.50 m. The
bar is 25 mm thick and is made of an elastic, perfectly
plastic material with Sy = 500 MPa and E = 207
GPa. Calculate the length to which this bar should be
stretched so that, after unloading, it will become and
remain straight.
A review of bending theory from a solid mechanics
textbook is necessary for this problem. In particular,
it should be recognized that when the curved bar be
comes straight, the engineering strain it undergoes is
given by the expression
t
e = 2ρ
where t is the thickness and ρ is the radius to the neu
tral axis. Hence in this case,
(0.025)
e = = 0.025
2(0.50)
Since Sy = 500 MPa and E = 207 GPa, we find that the
elastic limit for this material is at an elastic strain of
Sy 500 MPa
e = = = 0.00242
E 207 GPa
which is smaller than 0.025. Therefore, we know that
the bar must be loaded in the plastic range. Follow
ing the description in Answer 2.68 above, the strain re
quired to straighten the bar is twice the elastic limit, or
e = (2)(0.00242) = 0.0048
or lf − lo = 0.0048 → lf = 0.005lo + lo
lo
or lf = 1.0049 m.
The following Matlab code is helpful.
l=1;
rho=0.5;
Sy=500e6;
E=207e9;
t=0.025;
epsilon=t/2/rho;
epsilon_y=Sy/E;
epsilon_b=2*epsilon_y;
lf=l/(1-epsilon_b);
2.70 Assume that a material with a uniaxial yield strength
Sy yields under a stress state of principal stresses σ1,
σ2, σ3, where σ1 > σ2 > σ3. Show that the superpo
sition of a hydrostatic stress p on this system (such as
placing the specimen in a chamber pressurized with a
liquid) does not affect yielding. In other words, the ma
terial will still yield according to yield criteria.
This solution considers the distortion-energy criterion,
although the same derivation could be performed with
the maximum shear stress criterion as well. Equation
(2.39) on p. 67 gives
2 2 2
(σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Sy
2
Now consider a new stress state where the principal
stresses are
σI = σ1 + p1
σI = σ2 + p2
I
σ = σ3 + p3
which represents a new loading with an additional hy
drostatic pressure, p. The distortion-energy criterion
for this stress state is
2 2 2
(σ1
I − σ2
I ) + (σ2
I − σ3
I ) + (σ3
I − σ1
I ) = 2Sy
2
or
2
2S2 = [(σ1 + p) − (σ2 + p)]y
2
+ [(σ2 + p) − (σ3 + p)]
2
+ [(σ3 + p) − (σ1 + p)]
which can be simplified as
2 2 2
(σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2S2
y
which is the original yield criterion. Hence, the yield
criterion is unaffected by the superposition of a hydro
static pressure.
2.71 Give two different and specific examples in which the
maximum-shear-stress and the distortion-energy crite
ria give the same answer.
In order to obtain the same answer for the two yield
criteria, we refer to Fig. 2.32 on p. 70 for plane stress
and note the coordinates at which the two diagrams
meet. Examples are: simple tension, simple compres
sion, equal biaxial tension, and equal biaxial compres
sion. Thus, acceptable answers would include (a) wire
rope, as used on a crane to lift loads; (b) spherical pres
sure vessels, including balloons and gas storage tanks;
and (c) shrink fits.
2.72 A thin-walled spherical shell with a yield strength Sy
is subjected to an internal pressure p. With appropriate
equations, show whether or not the pressure required
to yield this shell depends on the particular yield crite
rion used.
Here we have a state of plane stress with equal biax
ial tension. The answer to Problem 2.71 leads one to
immediately conclude that both the maximum shear
stress and distortion energy criteria will give the same
results. We will now demonstrate this more rigorously.
The principal membrane stresses are given by
pr
σ1 = σ2 = 2t
and
σ3 = 0
13
Loading page 15...
Using the maximum shear-stress criterion, we find that
σ1 − 0 = Sy .
Hence 2tSy
p = r
Using the distortion-energy criterion, we have
(0 − 0)2 + (σ2 − 0)2 + (0 − σ1)2 = 2S2
y
Since σ1 = σ2, then this gives σ1 = σ2 = Sy , and the
same expression is obtained for pressure.
2.73 Show that, according to the distortion-energy criterion,
the yield strength in plane strain is 1.15Sy , where Sy is
the uniaxial yield strength of the material.
A plane-strain condition is shown in Fig. 2.35d, where
σ1 is the yield stress of the material in plane strain
(S� )y , σ3 is zero, and E2 = 0. From Eq. 2.43b, we find
that σ2 = σ1/2. Substituting these into the distortion-
energy criterion given by Eq. (2.37),
2 2σ1 σ1
σ1 − + − 0 + (0 − σ1)2 = 2S2
y
2 2
and 3σ1
2
= 2Sy
2
2
hence 2
σ1 = √ Sy ≈ 1.15Sy
3
2.74 What would be the answer to Problem 2.73 if the max
imum shear stress criterion were used?
Because σ2 is an intermediate stress and using
Eq. (2.38) on p. 67, the answer would be
σ1 − 0 = Sy .
Hence, the yield stress in plane strain will be equal to
the uniaxial yield stress, Sy .
2.75 A closed-end, thin-walled cylinder of original length
l thickness t, and internal radius r is subjected to an
internal pressure p. Using the generalized Hooke’s
law equations, show the change, if any, that occurs in
the length of this cylinder when it is pressurized. Let
ν = 0.25.
A closed-end, thin-walled cylinder under internal pres
sure is subjected to the following principal stresses:
pr
σ1 = 2t ; pr
σ2 = t ; σ3 = 0
where the subscript 1 is the longitudinal direction, 2
is the hoop direction, and 3 is the thickness direction.
From Hooke’s law given by Eq. (2.34) on p. 66,
1
E1 = [σ1 − ν (σ2 + σ3)]
E
1 pr 1 pr
= − + 0
E 2t 3 t
pr
= 6tE
Since all the quantities are positive (note that in order
to produce a tensile membrane stress, the pressure is
positive as well), the longitudinal strain is finite and
positive. Thus the cylinder becomes longer when pres
surized, as it can also be deduced intuitively.
2.76 A round, thin-walled tube is subjected to tension in the
elastic range. Show that both the thickness and the di
ameter decrease as tension increases.
The stress state in this case is σ1, σ2 = σ3 = 0. From the
generalized Hooke’s law equations given by Eq. (2.34),
and denoting the axial direction as 1, the hoop direc
tion as 2, and the radial direction as 3, we have for the
hoop strain:
1 νσ1
E2 = [σ2 − ν (σ1 + σ3)] = −
E E
Therefore, the diameter is negative for a tensile (posi
tive) value of σ1. For the radial strain, the generalized
Hooke’s law gives
1 νσ1
E3 = [σ3 − ν (σ1 + σ2)] = −
E E
Therefore, the radial strain is also negative and the wall
becomes thinner for a positive value of σ1.
2.77 Take a long cylindrical balloon and, with a thin felt-tip
pen, mark a small square on it. What will be the shape
of this square after you blow up the balloon, (a) a larger
square; (b) a rectangle with its long axis in the circum
ferential direction; (c) a rectangle with its long axis in
the longitudinal direction; or (d) an ellipse? Perform
this experiment, and, based on your observations, ex
plain the results, using appropriate equations. Assume
that the material the balloon is made up of is perfectly
elastic and isotropic and that this situation represents
a thin-walled closed-end cylinder under internal pres
sure.
This is a simple graphic way of illustrating the gen
eralized Hooke’s law equations. A balloon is a read
ily available and economical method of demonstrating
these stress states. It is also encouraged to assign the
students the task of predicting the shape numerically;
an example of a valuable experiment involves partially
inflating the balloon, drawing the square, then expand
ing it further and having the students predict the di
mensions of the square.
Although not as readily available, a rubber tube can be
used to demonstrate the effects of torsion in a similar
manner.
2.78 Take a cubic piece of metal with a side length lo and
deform it plastically to the shape of a rectangular par
allelepiped of dimensions l1, l2, and l3. Assuming that
the material is rigid and perfectly plastic, show that
volume constancy requires that the following expres
sion be satisfied: E1 + E2 + E3 = 0.
14
σ1 − 0 = Sy .
Hence 2tSy
p = r
Using the distortion-energy criterion, we have
(0 − 0)2 + (σ2 − 0)2 + (0 − σ1)2 = 2S2
y
Since σ1 = σ2, then this gives σ1 = σ2 = Sy , and the
same expression is obtained for pressure.
2.73 Show that, according to the distortion-energy criterion,
the yield strength in plane strain is 1.15Sy , where Sy is
the uniaxial yield strength of the material.
A plane-strain condition is shown in Fig. 2.35d, where
σ1 is the yield stress of the material in plane strain
(S� )y , σ3 is zero, and E2 = 0. From Eq. 2.43b, we find
that σ2 = σ1/2. Substituting these into the distortion-
energy criterion given by Eq. (2.37),
2 2σ1 σ1
σ1 − + − 0 + (0 − σ1)2 = 2S2
y
2 2
and 3σ1
2
= 2Sy
2
2
hence 2
σ1 = √ Sy ≈ 1.15Sy
3
2.74 What would be the answer to Problem 2.73 if the max
imum shear stress criterion were used?
Because σ2 is an intermediate stress and using
Eq. (2.38) on p. 67, the answer would be
σ1 − 0 = Sy .
Hence, the yield stress in plane strain will be equal to
the uniaxial yield stress, Sy .
2.75 A closed-end, thin-walled cylinder of original length
l thickness t, and internal radius r is subjected to an
internal pressure p. Using the generalized Hooke’s
law equations, show the change, if any, that occurs in
the length of this cylinder when it is pressurized. Let
ν = 0.25.
A closed-end, thin-walled cylinder under internal pres
sure is subjected to the following principal stresses:
pr
σ1 = 2t ; pr
σ2 = t ; σ3 = 0
where the subscript 1 is the longitudinal direction, 2
is the hoop direction, and 3 is the thickness direction.
From Hooke’s law given by Eq. (2.34) on p. 66,
1
E1 = [σ1 − ν (σ2 + σ3)]
E
1 pr 1 pr
= − + 0
E 2t 3 t
pr
= 6tE
Since all the quantities are positive (note that in order
to produce a tensile membrane stress, the pressure is
positive as well), the longitudinal strain is finite and
positive. Thus the cylinder becomes longer when pres
surized, as it can also be deduced intuitively.
2.76 A round, thin-walled tube is subjected to tension in the
elastic range. Show that both the thickness and the di
ameter decrease as tension increases.
The stress state in this case is σ1, σ2 = σ3 = 0. From the
generalized Hooke’s law equations given by Eq. (2.34),
and denoting the axial direction as 1, the hoop direc
tion as 2, and the radial direction as 3, we have for the
hoop strain:
1 νσ1
E2 = [σ2 − ν (σ1 + σ3)] = −
E E
Therefore, the diameter is negative for a tensile (posi
tive) value of σ1. For the radial strain, the generalized
Hooke’s law gives
1 νσ1
E3 = [σ3 − ν (σ1 + σ2)] = −
E E
Therefore, the radial strain is also negative and the wall
becomes thinner for a positive value of σ1.
2.77 Take a long cylindrical balloon and, with a thin felt-tip
pen, mark a small square on it. What will be the shape
of this square after you blow up the balloon, (a) a larger
square; (b) a rectangle with its long axis in the circum
ferential direction; (c) a rectangle with its long axis in
the longitudinal direction; or (d) an ellipse? Perform
this experiment, and, based on your observations, ex
plain the results, using appropriate equations. Assume
that the material the balloon is made up of is perfectly
elastic and isotropic and that this situation represents
a thin-walled closed-end cylinder under internal pres
sure.
This is a simple graphic way of illustrating the gen
eralized Hooke’s law equations. A balloon is a read
ily available and economical method of demonstrating
these stress states. It is also encouraged to assign the
students the task of predicting the shape numerically;
an example of a valuable experiment involves partially
inflating the balloon, drawing the square, then expand
ing it further and having the students predict the di
mensions of the square.
Although not as readily available, a rubber tube can be
used to demonstrate the effects of torsion in a similar
manner.
2.78 Take a cubic piece of metal with a side length lo and
deform it plastically to the shape of a rectangular par
allelepiped of dimensions l1, l2, and l3. Assuming that
the material is rigid and perfectly plastic, show that
volume constancy requires that the following expres
sion be satisfied: E1 + E2 + E3 = 0.
14
Loading page 16...
13 more pages available. Scroll down to load them.
Preview Mode
Sign in to access the full document!
100%
Study Now!
XY-Copilot AI
Unlimited Access
Secure Payment
Instant Access
24/7 Support
Document Chat
Document Details
Subject
Mechanical Engineering