Chapter 2
Fundamentals of the Mechanical
Behavior of Materials
Questions
2.1 Can you calculate the percent elongation of ma- increases. Is this phenomenon true for both ten-
terials based only on the information given in sile and compressive strains? Explain.
Fig. 2.6? Explain.
The difference between the engineering and true
Recall that the percent elongation is defined by strains becomes larger because of the way the
Eq. (2.6) on p. 33 and depends on the original strains are defined, respectively, as can be seen
gage length (lo ) of the specimen. From Fig. 2.6 by inspecting Eqs. (2.1) on p. 30 and (2.9) on
on p. 37 only the necking strain (true and engi- p. 35. This is true for both tensile and com-
neering) and true fracture strain can be deter- pressive strains.
mined. Thus, we cannot calculate the percent
2.4 Using the same scale for stress, we note that the
elongation of the specimen; also, note that the
tensile true-stress-true-strain curve is higher
elongation is a function of gage length and in-
than the engineering stress-strain curve. Ex-
creases with gage length.
plain whether this condition also holds for a
2.2 Explain if it is possible for the curves in Fig. 2.4 compression test.
to reach 0% elongation as the gage length is in- During a compression test, the cross-sectional
creased further. area of the specimen increases as the specimen
height decreases (because of volume constancy)
The percent elongation of the specimen is a
as the load is increased. Since true stress is de-
function of the initial and final gage lengths.
fined as ratio of the load to the instantaneous
When the specimen is being pulled, regardless
cross-sectional area of the specimen, the true
of the original gage length, it will elongate uni-
stress in compression will be lower than the en-
formly (and permanently) until necking begins.
gineering stress for a given load, assuming that
Therefore, the specimen will always have a cer-
friction between the platens and the specimen
tain finite elongation. However, note that as the
is negligible.
specimen’s gage length is increased, the contri-
bution of localized elongation (that is, necking) 2.5 Which of the two tests, tension or compression,
will decrease, but the total elongation will not requires a higher capacity testing machine than
approach zero. the other? Explain.
2.3 Explain why the difference between engineering The compression test requires a higher capacity
strain and true strain becomes larger as strain machine because the cross-sectional area of the
1
, specimen increases during the test, which is the stress-true strain curve represents the specific
opposite of a tension test. The increase in area work done at the necked (and fractured) region
requires a load higher than that for the ten- in the specimen where the strain is a maximum.
sion test to achieve the same stress level. Fur- Thus, the answers will be different. However,
thermore, note that compression-test specimens up to the onset of necking (instability), the spe-
generally have a larger original cross-sectional cific work calculated will be the same. This is
area than those for tension tests, thus requiring because the strain is uniform throughout the
higher forces. specimen until necking begins.
2.10 The note at the bottom of Table 2.5 states that
2.6 Explain how the modulus of resilience of a ma-
as temperature increases, C decreases and m
terial changes, if at all, as it is strained: (1) for
increases. Explain why.
an elastic, perfectly plastic material, and (2) for
an elastic, linearly strain-hardening material. The value of C in Table 2.5 on p. 43 decreases
with temperature because it is a measure of the
2.7 If you pull and break a tension-test specimen strength of the material. The value of m in-
rapidly, where would the temperature be the creases with temperature because the material
highest? Explain why. becomes more strain-rate sensitive, due to the
fact that the higher the strain rate, the less time
Since temperature rise is due to the work input, the material has to recover and recrystallize,
the temperature will be highest in the necked hence its strength increases.
region because that is where the strain, hence
the energy dissipated per unit volume in plastic 2.11 You are given the K and n values of two dif-
deformation, is highest. ferent materials. Is this information sufficient
to determine which material is tougher? If not,
2.8 Comment on the temperature distribution if the what additional information do you need, and
specimen in Question 2.7 is pulled very slowly. why?
If the specimen is pulled very slowly, the tem- Although the K and n values may give a good
perature generated will be dissipated through- estimate of toughness, the true fracture stress
out the specimen and to the environment. and the true strain at fracture are required for
Thus, there will be no appreciable temperature accurate calculation of toughness. The modu-
rise anywhere, particularly with materials with lus of elasticity and yield stress would provide
high thermal conductivity. information about the area under the elastic re-
gion; however, this region is very small and is
2.9 In a tension test, the area under the true-stress- thus usually negligible with respect to the rest
true-strain curve is the work done per unit vol- of the stress-strain curve.
ume (the specific work). We also know that
the area under the load-elongation curve rep- 2.12 Modify the curves in Fig. 2.7 to indicate the
resents the work done on the specimen. If you effects of temperature. Explain the reasons for
divide this latter work by the volume of the your changes.
specimen between the gage marks, you will de-
These modifications can be made by lowering
termine the work done per unit volume (assum-
the slope of the elastic region and lowering the
ing that all deformation is confined between
general height of the curves. See, for example,
the gage marks). Will this specific work be
Fig. 2.10 on p. 42.
the same as the area under the true-stress-true-
strain curve? Explain. Will your answer be the 2.13 Using a specific example, show why the defor-
same for any value of strain? Explain. mation rate, say in m/s, and the true strain rate
are not the same.
If we divide the work done by the total volume
of the specimen between the gage lengths, we The deformation rate is the quantity v in
obtain the average specific work throughout the Eqs. (2.14), (2.15), (2.17), and (2.18) on pp. 41-
specimen. However, the area under the true 46. Thus, when v is held constant during de-
2
, formation (hence a constant deformation rate), However, the volume of material subjected to
the true strain rate will vary, whereas the engi- the maximum bending moment (hence to max-
neering strain rate will remain constant. Hence, imum stress) increases. Thus, the probability
the two quantities are not the same. of failure in the four-point test increases as this
distance increases.
2.14 It has been stated that the higher the value of
m, the more diffuse the neck is, and likewise, 2.17 Would Eq. (2.10) hold true in the elastic range?
the lower the value of m, the more localized the Explain.
neck is. Explain the reason for this behavior.
Note that this equation is based on volume con-
As discussed in Section 2.2.7 starting on p. 41, stancy, i.e., Ao lo = Al. We know, however, that
with high m values, the material stretches to because the Poisson’s ratio ν is less than 0.5 in
a greater length before it fails; this behavior the elastic range, the volume is not constant in
is an indication that necking is delayed with a tension test; see Eq. (2.47) on p. 69. There-
increasing m. When necking is about to be- fore, the expression is not valid in the elastic
gin, the necking region’s strength with respect range.
to the rest of the specimen increases, due to
strain hardening. However, the strain rate in 2.18 Why have different types of hardness tests been
the necking region is also higher than in the rest developed? How would you measure the hard-
of the specimen, because the material is elon- ness of a very large object?
gating faster there. Since the material in the
There are several basic reasons: (a) The overall
necked region becomes stronger as it is strained
hardness range of the materials; (b) the hard-
at a higher rate, the region exhibits a greater re-
ness of their constituents; see Chapter 3; (c) the
sistance to necking. The increase in resistance
thickness of the specimen, such as bulk versus
to necking thus depends on the magnitude of
foil; (d) the size of the specimen with respect to
m. As the tension test progresses, necking be-
that of the indenter; and (e) the surface finish
comes more diffuse, and the specimen becomes
of the part being tested.
longer before fracture; hence, total elongation
increases with increasing values of m (Fig. 2.13 2.19 Which hardness tests and scales would you use
on p. 45). As expected, the elongation after for very thin strips of material, such as alu-
necking (postuniform elongation) also increases minum foil? Why?
with increasing m. It has been observed that
the value of m decreases with metals of increas- Because aluminum foil is very thin, the indenta-
ing strength. tions on the surface must be very small so as not
to affect test results. Suitable tests would be a
2.15 Explain why materials with high m values (such microhardness test such as Knoop or Vickers
as hot glass and silly putty) when stretched under very light loads (see Fig. 2.22 on p. 52).
slowly, undergo large elongations before failure. The accuracy of the test can be validated by ob-
Consider events taking place in the necked re- serving any changes in the surface appearance
gion of the specimen. opposite to the indented side.
The answer is similar to Answer 2.14 above. 2.20 List and explain the factors that you would con-
sider in selecting an appropriate hardness test
2.16 Assume that you are running four-point bend-
and scale for a particular application.
ing tests on a number of identical specimens of
the same length and cross-section, but with in- Hardness tests mainly have three differences:
creasing distance between the upper points of
loading (see Fig. 2.21b). What changes, if any, (a) type of indenter,
would you expect in the test results? Explain. (b) applied load, and
As the distance between the upper points of (c) method of indentation measurement
loading in Fig. 2.21b on p. 51 increases, the (depth or surface area of indentation, or
magnitude of the bending moment decreases. rebound of indenter).
3
, The hardness test selected would depend on the 2.23 Describe the difference between creep and
estimated hardness of the workpiece, its size stress-relaxation phenomena, giving two exam-
and thickness, and if an average hardness or the ples for each as they relate to engineering ap-
hardness of individual microstructural compo- plications.
nents is desired. For instance, the scleroscope,
which is portable, is capable of measuring the Creep is the permanent deformation of a part
hardness of large pieces which otherwise would that is under a load over a period of time, usu-
be difficult or impossible to measure by other ally occurring at elevated temperatures. Stress
techniques. relaxation is the decrease in the stress level in
a part under a constant strain. Examples of
The Brinell hardness measurement leaves a
creep include:
fairly large indentation which provides a good
measure of average hardness, while the Knoop (a) turbine blades operating at high tempera-
test leaves a small indentation that allows, for tures, and
example, the determination of the hardness of
individual phases in a two-phase alloy, as well as (b) high-temperature steam linesand furnace
inclusions. The small indentation of the Knoop components.
test also allows it to be useful in measuring the
Stress relaxation is observed when, for example,
hardness of very thin layers on parts, such as
a rubber band or a thermoplastic is pulled to
plating or coatings. Recall that the depth of in-
a specific length and held at that length for a
dentation should be small relative to part thick-
period of time. This phenomenon is commonly
ness, and that any change on the bottom sur-
observed in rivets, bolts, and guy wires, as well
face appearance makes the test results invalid.
as thermoplastic components.
2.21 In a Brinell hardness test, the resulting impres-
sion is found to be an ellipse. Give possible 2.24 Referring to the two impact tests shown in
explanations for this phenomenon. Fig. 2.31, explain how different the results
would be if the specimens were impacted from
There are several possible reasons for this the opposite directions.
phenomenon, but the two most likely are
anisotropy in the material and the presence of Note that impacting the specimens shown in
surface residual stresses in the material. Fig. 2.31 on p. 60 from the opposite directions
would subject the roots of the notches to com-
2.21 Referring to Fig. 2.22 on p. 52, note that the pressive stresses, and thus they would not act
material for indenters are either steel, tungsten as stress raisers. Thus, cracks would not propa-
carbide, or diamond. Why isn’t diamond used gate as they would when under tensile stresses.
for all of the tests? Consequently, the specimens would basically
behave as if they were not notched.
While diamond is the hardest material known,
it would not, for example, be practical to make 2.25 If you remove layer ad from the part shown in
and use a 10-mm diamond indenter because the Fig. 2.30d, such as by machining or grinding,
costs would be prohibitive. Consequently, a which way will the specimen curve? (Hint: As-
hard material such as those listed are sufficient sume that the part in diagram (d) can be mod-
for most hardness tests. eled as consisting of four horizontal springs held
at the ends. Thus, from the top down, we have
2.22 What effect, if any, does friction have in a hard- compression, tension, compression, and tension
ness test? Explain. springs.)
The effect of friction has been found to be mini- Since the internal forces will have to achieve a
mal. In a hardness test, most of the indentation state of static equilibrium, the new part has to
occurs through plastic deformation, and there bow downward (i.e., it will hold water). Such
is very little sliding at the indenter-workpiece residual-stress patterns can be modeled with
interface; see Fig. 2.25 on p. 55. a set of horizontal tension and compression
4
Fundamentals of the Mechanical
Behavior of Materials
Questions
2.1 Can you calculate the percent elongation of ma- increases. Is this phenomenon true for both ten-
terials based only on the information given in sile and compressive strains? Explain.
Fig. 2.6? Explain.
The difference between the engineering and true
Recall that the percent elongation is defined by strains becomes larger because of the way the
Eq. (2.6) on p. 33 and depends on the original strains are defined, respectively, as can be seen
gage length (lo ) of the specimen. From Fig. 2.6 by inspecting Eqs. (2.1) on p. 30 and (2.9) on
on p. 37 only the necking strain (true and engi- p. 35. This is true for both tensile and com-
neering) and true fracture strain can be deter- pressive strains.
mined. Thus, we cannot calculate the percent
2.4 Using the same scale for stress, we note that the
elongation of the specimen; also, note that the
tensile true-stress-true-strain curve is higher
elongation is a function of gage length and in-
than the engineering stress-strain curve. Ex-
creases with gage length.
plain whether this condition also holds for a
2.2 Explain if it is possible for the curves in Fig. 2.4 compression test.
to reach 0% elongation as the gage length is in- During a compression test, the cross-sectional
creased further. area of the specimen increases as the specimen
height decreases (because of volume constancy)
The percent elongation of the specimen is a
as the load is increased. Since true stress is de-
function of the initial and final gage lengths.
fined as ratio of the load to the instantaneous
When the specimen is being pulled, regardless
cross-sectional area of the specimen, the true
of the original gage length, it will elongate uni-
stress in compression will be lower than the en-
formly (and permanently) until necking begins.
gineering stress for a given load, assuming that
Therefore, the specimen will always have a cer-
friction between the platens and the specimen
tain finite elongation. However, note that as the
is negligible.
specimen’s gage length is increased, the contri-
bution of localized elongation (that is, necking) 2.5 Which of the two tests, tension or compression,
will decrease, but the total elongation will not requires a higher capacity testing machine than
approach zero. the other? Explain.
2.3 Explain why the difference between engineering The compression test requires a higher capacity
strain and true strain becomes larger as strain machine because the cross-sectional area of the
1
, specimen increases during the test, which is the stress-true strain curve represents the specific
opposite of a tension test. The increase in area work done at the necked (and fractured) region
requires a load higher than that for the ten- in the specimen where the strain is a maximum.
sion test to achieve the same stress level. Fur- Thus, the answers will be different. However,
thermore, note that compression-test specimens up to the onset of necking (instability), the spe-
generally have a larger original cross-sectional cific work calculated will be the same. This is
area than those for tension tests, thus requiring because the strain is uniform throughout the
higher forces. specimen until necking begins.
2.10 The note at the bottom of Table 2.5 states that
2.6 Explain how the modulus of resilience of a ma-
as temperature increases, C decreases and m
terial changes, if at all, as it is strained: (1) for
increases. Explain why.
an elastic, perfectly plastic material, and (2) for
an elastic, linearly strain-hardening material. The value of C in Table 2.5 on p. 43 decreases
with temperature because it is a measure of the
2.7 If you pull and break a tension-test specimen strength of the material. The value of m in-
rapidly, where would the temperature be the creases with temperature because the material
highest? Explain why. becomes more strain-rate sensitive, due to the
fact that the higher the strain rate, the less time
Since temperature rise is due to the work input, the material has to recover and recrystallize,
the temperature will be highest in the necked hence its strength increases.
region because that is where the strain, hence
the energy dissipated per unit volume in plastic 2.11 You are given the K and n values of two dif-
deformation, is highest. ferent materials. Is this information sufficient
to determine which material is tougher? If not,
2.8 Comment on the temperature distribution if the what additional information do you need, and
specimen in Question 2.7 is pulled very slowly. why?
If the specimen is pulled very slowly, the tem- Although the K and n values may give a good
perature generated will be dissipated through- estimate of toughness, the true fracture stress
out the specimen and to the environment. and the true strain at fracture are required for
Thus, there will be no appreciable temperature accurate calculation of toughness. The modu-
rise anywhere, particularly with materials with lus of elasticity and yield stress would provide
high thermal conductivity. information about the area under the elastic re-
gion; however, this region is very small and is
2.9 In a tension test, the area under the true-stress- thus usually negligible with respect to the rest
true-strain curve is the work done per unit vol- of the stress-strain curve.
ume (the specific work). We also know that
the area under the load-elongation curve rep- 2.12 Modify the curves in Fig. 2.7 to indicate the
resents the work done on the specimen. If you effects of temperature. Explain the reasons for
divide this latter work by the volume of the your changes.
specimen between the gage marks, you will de-
These modifications can be made by lowering
termine the work done per unit volume (assum-
the slope of the elastic region and lowering the
ing that all deformation is confined between
general height of the curves. See, for example,
the gage marks). Will this specific work be
Fig. 2.10 on p. 42.
the same as the area under the true-stress-true-
strain curve? Explain. Will your answer be the 2.13 Using a specific example, show why the defor-
same for any value of strain? Explain. mation rate, say in m/s, and the true strain rate
are not the same.
If we divide the work done by the total volume
of the specimen between the gage lengths, we The deformation rate is the quantity v in
obtain the average specific work throughout the Eqs. (2.14), (2.15), (2.17), and (2.18) on pp. 41-
specimen. However, the area under the true 46. Thus, when v is held constant during de-
2
, formation (hence a constant deformation rate), However, the volume of material subjected to
the true strain rate will vary, whereas the engi- the maximum bending moment (hence to max-
neering strain rate will remain constant. Hence, imum stress) increases. Thus, the probability
the two quantities are not the same. of failure in the four-point test increases as this
distance increases.
2.14 It has been stated that the higher the value of
m, the more diffuse the neck is, and likewise, 2.17 Would Eq. (2.10) hold true in the elastic range?
the lower the value of m, the more localized the Explain.
neck is. Explain the reason for this behavior.
Note that this equation is based on volume con-
As discussed in Section 2.2.7 starting on p. 41, stancy, i.e., Ao lo = Al. We know, however, that
with high m values, the material stretches to because the Poisson’s ratio ν is less than 0.5 in
a greater length before it fails; this behavior the elastic range, the volume is not constant in
is an indication that necking is delayed with a tension test; see Eq. (2.47) on p. 69. There-
increasing m. When necking is about to be- fore, the expression is not valid in the elastic
gin, the necking region’s strength with respect range.
to the rest of the specimen increases, due to
strain hardening. However, the strain rate in 2.18 Why have different types of hardness tests been
the necking region is also higher than in the rest developed? How would you measure the hard-
of the specimen, because the material is elon- ness of a very large object?
gating faster there. Since the material in the
There are several basic reasons: (a) The overall
necked region becomes stronger as it is strained
hardness range of the materials; (b) the hard-
at a higher rate, the region exhibits a greater re-
ness of their constituents; see Chapter 3; (c) the
sistance to necking. The increase in resistance
thickness of the specimen, such as bulk versus
to necking thus depends on the magnitude of
foil; (d) the size of the specimen with respect to
m. As the tension test progresses, necking be-
that of the indenter; and (e) the surface finish
comes more diffuse, and the specimen becomes
of the part being tested.
longer before fracture; hence, total elongation
increases with increasing values of m (Fig. 2.13 2.19 Which hardness tests and scales would you use
on p. 45). As expected, the elongation after for very thin strips of material, such as alu-
necking (postuniform elongation) also increases minum foil? Why?
with increasing m. It has been observed that
the value of m decreases with metals of increas- Because aluminum foil is very thin, the indenta-
ing strength. tions on the surface must be very small so as not
to affect test results. Suitable tests would be a
2.15 Explain why materials with high m values (such microhardness test such as Knoop or Vickers
as hot glass and silly putty) when stretched under very light loads (see Fig. 2.22 on p. 52).
slowly, undergo large elongations before failure. The accuracy of the test can be validated by ob-
Consider events taking place in the necked re- serving any changes in the surface appearance
gion of the specimen. opposite to the indented side.
The answer is similar to Answer 2.14 above. 2.20 List and explain the factors that you would con-
sider in selecting an appropriate hardness test
2.16 Assume that you are running four-point bend-
and scale for a particular application.
ing tests on a number of identical specimens of
the same length and cross-section, but with in- Hardness tests mainly have three differences:
creasing distance between the upper points of
loading (see Fig. 2.21b). What changes, if any, (a) type of indenter,
would you expect in the test results? Explain. (b) applied load, and
As the distance between the upper points of (c) method of indentation measurement
loading in Fig. 2.21b on p. 51 increases, the (depth or surface area of indentation, or
magnitude of the bending moment decreases. rebound of indenter).
3
, The hardness test selected would depend on the 2.23 Describe the difference between creep and
estimated hardness of the workpiece, its size stress-relaxation phenomena, giving two exam-
and thickness, and if an average hardness or the ples for each as they relate to engineering ap-
hardness of individual microstructural compo- plications.
nents is desired. For instance, the scleroscope,
which is portable, is capable of measuring the Creep is the permanent deformation of a part
hardness of large pieces which otherwise would that is under a load over a period of time, usu-
be difficult or impossible to measure by other ally occurring at elevated temperatures. Stress
techniques. relaxation is the decrease in the stress level in
a part under a constant strain. Examples of
The Brinell hardness measurement leaves a
creep include:
fairly large indentation which provides a good
measure of average hardness, while the Knoop (a) turbine blades operating at high tempera-
test leaves a small indentation that allows, for tures, and
example, the determination of the hardness of
individual phases in a two-phase alloy, as well as (b) high-temperature steam linesand furnace
inclusions. The small indentation of the Knoop components.
test also allows it to be useful in measuring the
Stress relaxation is observed when, for example,
hardness of very thin layers on parts, such as
a rubber band or a thermoplastic is pulled to
plating or coatings. Recall that the depth of in-
a specific length and held at that length for a
dentation should be small relative to part thick-
period of time. This phenomenon is commonly
ness, and that any change on the bottom sur-
observed in rivets, bolts, and guy wires, as well
face appearance makes the test results invalid.
as thermoplastic components.
2.21 In a Brinell hardness test, the resulting impres-
sion is found to be an ellipse. Give possible 2.24 Referring to the two impact tests shown in
explanations for this phenomenon. Fig. 2.31, explain how different the results
would be if the specimens were impacted from
There are several possible reasons for this the opposite directions.
phenomenon, but the two most likely are
anisotropy in the material and the presence of Note that impacting the specimens shown in
surface residual stresses in the material. Fig. 2.31 on p. 60 from the opposite directions
would subject the roots of the notches to com-
2.21 Referring to Fig. 2.22 on p. 52, note that the pressive stresses, and thus they would not act
material for indenters are either steel, tungsten as stress raisers. Thus, cracks would not propa-
carbide, or diamond. Why isn’t diamond used gate as they would when under tensile stresses.
for all of the tests? Consequently, the specimens would basically
behave as if they were not notched.
While diamond is the hardest material known,
it would not, for example, be practical to make 2.25 If you remove layer ad from the part shown in
and use a 10-mm diamond indenter because the Fig. 2.30d, such as by machining or grinding,
costs would be prohibitive. Consequently, a which way will the specimen curve? (Hint: As-
hard material such as those listed are sufficient sume that the part in diagram (d) can be mod-
for most hardness tests. eled as consisting of four horizontal springs held
at the ends. Thus, from the top down, we have
2.22 What effect, if any, does friction have in a hard- compression, tension, compression, and tension
ness test? Explain. springs.)
The effect of friction has been found to be mini- Since the internal forces will have to achieve a
mal. In a hardness test, most of the indentation state of static equilibrium, the new part has to
occurs through plastic deformation, and there bow downward (i.e., it will hold water). Such
is very little sliding at the indenter-workpiece residual-stress patterns can be modeled with
interface; see Fig. 2.25 on p. 55. a set of horizontal tension and compression
4
