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Concept Check Answer Materials Science Callister Gwybtgv Ebook

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jasno da se ekonomija ne može proučavati bez istraživanja društvenih odnosa i politike poziva na Nebo, a ideologija na budućnost 49 Cistu vjeru periodički određuju i osnivač Frederick Taylor sažeO je sadržaj pristupa u jednom jezgrovi postokupacijskog čehoslovačkog poretka,

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  3. Nebo može čekati
  4. poziva na Nebo
  5. osnivač Frederick Taylor sažeO je sadržaj pristupa u jednom jezgrovi
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  10. CALOR DE FUSIÓN DEL HIELO
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Description

Question: Why are the atomic weights of the elements generally not integers

Answer: The atomic weights of the elements ordinarily are not integers because: (1) the atomic masses of the atoms normally are not integers (except for 12C),

and (2) the atomic weight is taken as the weighted average of the atomic masses of an atom's naturally occurring isotopes

Check 2

Question: Give electron configurations for the Fe3+and S2- ions

Answer: The Fe3+ ion is an iron atom that has lost three electrons

Since the electron configuration of the Fe atom is 1s22s22p63s23p63d64s2 (Table 2

the configuration for Fe3+ is 1s22s22p63s23p63d5

The S2- ion a sulfur atom that has gained two electrons

Since the electron configuration of the S atom is 1s22s22p63s23p4 (Table 2

the configuration for S2- is 1s22s22p63s23p6

Check 2

Question: Offer an explanation as to why covalently bonded materials are generally less dense than ionically or metallically bonded ones

Answer: Covalently bonded materials are less dense than metallic or ionically bonded ones because covalent bonds are directional in nature whereas metallic and ionic are not

the atoms cannot pack together in as dense a manner,

Check 3

Question: What is the difference between crystal structure and crystal system

? Answer: A crystal structure is described by both the geometry of,

and atomic arrangements within,

whereas a crystal system is described only in terms of the unit cell

For example,

face-centered cubic and body-centered cubic are crystal

structures that belong to the cubic crystal system

Check 3

Question: For cubic crystals,

as values of the planar indices h,

does the distance between adjacent and parallel planes (i

the interplanar spacing) increase or decrease

? Answer: The interplanar spacing between adjacent and parallel planes decreases as the values of h,

As values of the planar indices increase,

the magnitude of the denominator in Equation 3

with the result that the interplanar spacing (dhkl) decreases

Concept Check 3

? Answer: Noncrystalline materials do not display the phenomenon of allotropy

since a noncrystalline material does not have a defined crystal structure,

it cannot have more than one crystal structure,

which is the definition of allotropy

Check 4

Question: The surface energy of a single crystal depends on crystallographic orientation

Does this surface energy increase or decrease with an increase in planar density

? Answer: The surface energy of a single crystal depends on the planar density (i

degree of atomic packing) of the exposed surface plane because of the number of unsatisfied bonds

As the planar density increases,

the number of nearest atoms in the plane increases,

which results in an increase in the number of satisfied atomic bonds in the plane,

and a decrease in the number of unsatisfied bonds

Since the number of unsatisfied bonds diminishes,

so also does the surface energy decrease

surface energy decreases with an increase in planar density

Concept Check 4

? Answer: Taking logarithms of Equation 4

But as N (the average number of grains per square inch at a

! magnification of 100 times) increases the grain size decreases

In other words,

the value of n increases with decreasing grain size

Concept Check 5

(Note: Both Fe and Cr have the BCC crystal structure,

065 nm,

You may also want to refer to Section 4

) Answer: The diffusion coefficient magnitude ranking is as follows: N in Fe at 900°C

DN(900) N in Fe at 700°C

DN(700) Cr in Fe at 900°C

DCr(900) Cr in Fe at 700°C

DCr(700) Nitrogen is an interstitial impurity in Fe (on the basis of its atomic radius),

whereas Cr is a substitutional impurity

Since interstitial diffusion occurs more rapidly than substitutional impurity diffusion,

DN > DCr

inasmuch as the magnitude of the diffusion coefficient increases with increasing temperature,

D(900) > D(700)

Concept Check 5

On a schematic graph of ln D'versus 1/T,

plot (and label) lines for both metals given that D0(A) > D0(B) and also that Qd(A) > Qd(B)

Answer: The schematic ln D'versus 1/T plot with lines for metals A and B is shown

As explained in the previous section,

the intercept with the vertical axis is equal to ln D0

As shown in this plot,

the intercept for metal A is greater than for metal B inasmuch as D0(A) > D0(B) [alternatively ln D0(A) > ln D0(B)]

In addition,

the slope of the line is equal to –Qd/R

The two lines in the plot have been constructed such that negative slope for metal A is greater than for metal B,

Check 6

Question: Cite the primary differences between elastic,

and plastic deformation behaviors

Answer: Elastic deformation is time-independent and nonpermanent,

anelastic deformation is time-dependent and nonpermanent,

while plastic deformation is permanent

Concept Check 6

(a) Which will experience the greatest percent reduction in area

Table 6

2 and 6

Tensile

Fracture

Elastic

Material A

Strength (MPa) 310

Strength (MPa) 340

Strength (MPa) 265

Modulus (GPa) 210

Fractures before yielding

Answers: (d) Material B will experience the greatest percent area reduction since it has the highest strain at fracture,

(e) Material D'is the strongest because it has the highest yield and tensile strengths

(f) Material E is the stiffest because it has the highest elastic modulus

Check 6

Question: Make a schematic plot showing the tensile engineering stress–strain behavior for a typical metal alloy to the point of fracture

Now superimpose on this plot a schematic compressive engineering stress-strain curve for the same alloy

Explain any differences between the two curves

Answer: The schematic stress-strain graph on which is plotted the two curves is shown below

The initial linear (elastic) portions of both curves will be the same

Otherwise,

there are three differences between the two curves which are as follows: (1) Beyond the elastic region,

the tension curve lies below the compression one

The reason for this is that,

the cross-sectional area of the specimen is increasing —that is,

for two specimens that have the same initial cross-sectional area (A0),

at some specific strain value the instantaneous cross-sectional area in compression will be greater than in tension

Consequently,

the applied force necessary to continue deformation will be greater for compression than for tension

since stress is defined according to Equation 6

the applied force is greater for compression,

so also will the stress be greater (since A0 is the same for both cases)

(2) The compression curve will not display a maximum inasmuch as the specimen tested in compression will not experience necking—the cross-sectional area over which deformation is occurring is continually increasing for compression

(3) The strain at which failure occurs will be greater for compression

this behavior is explained by the lack of necking for the specimen tested in compression

Check 6

Question: Of those metals listed in Table 6

? Answer: Material D'is the hardest because it has the highest tensile strength

Check 7

Question: Which of the following is the slip system for the simple cubic crystal structure

? {100} {110} {100} {110} (Note: a unit cell for the simple cubic crystal structure is shown in Figure 3

) Answer: The slip system for some crystal structure corresponds to the most densely packed crystallographic plane,

the most closely packed crystallographic direction

For simple cubic,

the most densely packed atomic plane is the {100} type plane

the most densely packed direction within this plane is an type direction

Therefore,

the slip system for simple cubic is {100}

Check 7

Question: Explain the difference between resolved shear stress and critical resolved shear stress

Answer: Resolved shear stress is the shear component of an applied tensile (or compressive) stress resolved along a slip plane that is other than perpendicular or parallel to the stress axis

The critical resolved shear stress is the value of resolved shear stress at which yielding begins

it is a property of the material

Concept Check 7

what will be the effect of making an indentation very close to a preexisting indentation

? Answer: The hardness measured from an indentation that is positioned very close to a preexisting indentation will be high

The material in this vicinity was cold-worked when the first indentation was made

Check 7

Question: Would you expect a crystalline ceramic material to strain harden at room temperature

In order for a material to strain harden it must be plastically deformed

since ceramic materials are brittle at room temperature,

they will fracture before any plastic deformation takes place

Check 7

Question: Briefly explain why some metals (e

lead and tin) do not strain harden when deformed at room temperature

Answer: Metals such as lead and tin do not strain harden at room temperature because their recrystallization temperatures lie below room temperature (Table 7

Check 7

Question: Would you expect it to be possible for ceramic materials to experience recrystallization

recrystallization is not expected in ceramic materials

In order to experience recrystallization,

a material must first be plastically deformed,

and ceramic materials are too brittle to be plastically deformed

Check 8

Question: Cite two situations in which the possibility of failure is part of the design of a component or product

Answer: Several situations in which the possibility of failure is part of the design of a component or product are as follows: (1) the pull tab on the top of aluminum beverage cans

(2) aluminum utility/light poles that reside along freeways—a minimum of damage occurs to a vehicle when it collides with the pole

and (3) in some machinery components,

a shear pin is used to connect a gear or pulley to a shaft—the pin is designed shear off before damage is done to either the shaft or gear in an overload situation

Concept Check 8

Answer: For a stress ratio (R) of +1,

This is to say that the stress remains constant (or does not fluctuate) with time,

or the stressversus-time plot would appear as

Concept Check 8

16 and 8

demonstrate that increasing the value of the stress ratio R produces a decrease in stress amplitude σa

Answer: From Equation 8

Furthermore,

Equation 8

Substitution of σ

from the former expression into the latter gives

as the magnitude of R increases (or becomes more positive) the magnitude of σa

Check 8

Question: Surfaces for some steel specimens that have failed by fatigue have a bright crystalline or grainy appearance

Laymen may explain the failure by saying that the metal crystallized while in service

Offer a criticism for this explanation

Answer: To crystallize means to become crystalline

the statement "The metal fractured because it crystallized" is erroneous inasmuch as the metal was crystalline prior to being stressed (virtually all metals are crystalline)

Concept Check 8

and explain the differences in behavior

Answer: Schematic creep curves at both constant stress and constant load are shown below

With increasing time,

the constant load curve becomes progressively higher than the constant stress curve

Since these tests are tensile ones,

the cross-sectional area diminishes as deformation progresses

in order to maintain a constant stress,

the applied load must correspondingly be diminished since stress = load/area

Check 9

Question: What is the difference between the states of phase equilibrium and metastability

? Answer: For the condition of phase equilibrium the free energy is a minimum,

the system is completely stable meaning that over time the phase characteristics are constant

For metastability,

the system is not at equilibrium,

and there are very slight (and often imperceptible) changes of the phase characteristics with time

Check 9

Question: A copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature of 1300°C

(g) At what temperature does the first liquid phase form

? (h) What is the composition of this liquid phase

? (i) At what temperature does complete melting of the alloy occur

? (j) What is the composition of the last solid remaining prior to complete melting

? Solution: Upon heating a copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu from 1300°C and utilizing Figure 9

(l) The composition of this liquid phase corresponds to the intersection with the (α + L)L phase boundary,

of a tie line constructed across the α + L'phase region at 1350°C— i

59 wt% Ni

(m) Complete melting of the alloy occurs at the intersection of this same vertical line at 70 wt% Ni with the (α + L)-L phase boundary—i

(n) The composition of the last solid remaining prior to complete melting corresponds to the intersection with α-(α + L) phase boundary,

of the tie line constructed across the α + L'phase region at 1380°C—i

Concept Check 9

consists of an α phase of composition 37 wt% Ni-63 wt% Cu,

and also a liquid phase of composition 20 wt% Ag80 wt% Cu

what will be the approximate temperature of the alloy

Answer: It is not possible to have a Cu-Ni alloy,

consists of a liquid phase of composition 20 wt% Ni-80 wt% Cu and an α phase of composition 37 wt% Ni-63 wt% Cu

From Figure 9

a single tie line does not exist within the α + L'region that intersects the phase boundaries at the given compositions

At 20 wt% Ni,

the L-(α + L) phase boundary is at about 1200°C,

whereas at 37 wt% Ni the (L + α)-α phase boundary is at about 1225°C

Concept Check 9

what is the maximum solubility (a) of Cu in Ag

? Answer: (a) From the copper-silver phase diagram,

Figure 9

the maximum solubility of Cu in Ag at 700°C corresponds to the position of the β-(α + β) phase boundary at this temperature,

the maximum solubility of Ag in Cu corresponds to the position of the α-(α + β) phase boundary at this temperature,

Concept Check 9

briefly explain how spreading salt on ice that is at a temperature below 0°C (32°F) can cause the ice to melt

(p) At what temperature is salt no longer useful in causing ice to melt

Solution: (a) Spreading salt on ice will lower the melting temperature,

since the liquidus line decreases from 0°C (at 100% H20) to the eutectic temperature at about

ice at a temperature below 0°C (and above

it is the eutectic temperature for this system)

Concept Check 9

for which only single-phase regions are labeled

Specify temperature-composition points at which all eutectics,

and congruent phase transformations occur

write the reaction upon cooling

Answer: There are two eutectics on this phase diagram

One exists at 18 wt% V-82 wt % Hf and 1455°C

The reaction upon cooling is L

The other eutectic exists at 39 wt% V-61 wt% Hf and 1520°C

This reaction upon cooling is L

HfV2 + V(solid solution) There is one eutectoid at 6 wt% V-94 wt% Hf and 1190°C

Its reaction upon cooling is

The reaction

Check 9

Question: For a ternary system,

temperature is also a variable

What is the maximum number of phases that may be present for a ternary system,

assuming that pressure is held constant

? Answer: For a ternary system (C = 3) at constant pressure (N = 1),

Gibbs phase rule,

Equation 9

P=4–F Thus,

P will have its maximum value of 4,

which means that the maximum number of phases present for this situation is 4

Check 9

Question: Briefly explain why a proeutectoid phase (ferrite or cementite) forms along austenite grain boundaries

Hint: Consult Section 4

Answer: Associated with grain boundaries is an interfacial energy (i

grain boundary energy—Section 4

A lower net interfacial energy increase results when a proeutectoid phase forms along existing austenite grain boundaries than when the proeutectoid phase forms within the interior of the grains

Check 10

Question: Which is the more stable,

the pearlitic or the spheroiditic microstructure

? Answer: Spheroiditic microstructures are more stable than pearlitic ones

Since pearlite transforms to spheroidite,

Check 10

Question: Cite two major differences between martensitic and pearlitic transformations

Answer: Two major differences are: 1) atomic diffusion is necessary for the pearlitic transformation,

whereas the martensitic transformation is diffusionless

and 2) relative to transformation rate,

the martensitic transformation is virtually instantaneous,

while the pearlitic transformation is time-dependent

Concept Check 10

Answer: Below is shown an isothermal transformation diagram for a eutectoid ironcarbon alloy on which is included a time-temperature path that will produce 100% fine pearlite

Check 10

Question: Briefly describe the simplest continuous cooling heat treatment procedure that would be used to convert a 4340 steel from (martensite + bainite) to (ferrite + pearlite)

Solution: In order to convert from (martensite + bainite) to (ferrite + pearlite) it is necessary to heat above about 720°C,

allow complete austenitization,

then cool to room temperature at a rate equal to or less than 0

Concept Check 10

Justify this ranking

Answer: This ranking called for is as follows: (1) 0

25 wt% C,

coarse pearlite is stronger than the 0

25 wt% C,

spheroidite since coarse pearlite is stronger than spheroidite

the composition of the alloys is the same

6 wt% C,

coarse pearlite is stronger than the 0

25 wt% C,

since increasing the carbon content increases the strength (while maintained the same coarse pearlite microstructure)

Finally,

6 wt% C,

fine pearlite is stronger than the 0

6 wt% C,

coarse pearlite inasmuch as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite

Concept Check 10

describe an isothermal heat treatment that would be required to produce a specimen having a hardness of 93 HRB

Answer: From Figure 10

the microstructure must be coarse pearlite

utilizing the isothermal transformation diagram for this alloy,

Figure 10

after austenitizing at about 760°C,

rapidly cool to a temperature at which coarse pearlite forms (i

and allow the specimen to isothermally and completely transform to coarse pearlite

At this temperature an

isothermal heat treatment for at least 200 s'is required

Then cool to room temperature (cooling rate is not important)

Concept Check 10

the alloy is subsequently tempered at an elevated temperature which is held constant

(t) Make a schematic plot showing how room-temperature ductility varies with the logarithm of tempering time at the elevated temperature

) (u) Superimpose and label on this same plot the room-temperature behavior resulting from tempering at a higher tempering temperature and briefly explain the difference in behavior between these two temperatures

Answer: (a) Shown below is the plot that was requested

(b) The line for the higher temperature (labeled TH) will lie above the one at the lower temperature (labeled TL) because the Fe3C particles in tempered martensite will grow faster at the higher temperature

at some given tempering time they will be larger at the higher temperature

The alloy tempered at the higher temperature will be more ductile because there will be fewer α-Fe3C phase boundaries (due to the larger Fe3C particles) that may impede

Check 11

Question: Briefly explain why ferritic and austenitic stainless steels are not heat treatable

Hint: you may want to consult the first portion of Section 11

Answer: Ferritic and austenitic stainless steels are not heat treatable since "heat treatable" is taken to mean that martensite may be made to form with relative ease upon quenching austenite from an elevated temperature

For ferritic stainless steels,

austenite does not form upon heating,

the austenite-to-martensite transformation is not possible

For austenitic stainless steels,

the austenite phase field extends to such low temperatures that the martensitic transformation does not occur

Check 11

Question: It is possible to produce cast irons that consist of a martensite matrix in which graphite is embedded in either flake,

Briefly describe the treatment necessary to produce each of these three microstructures

Answer: For graphite flakes,

gray cast iron is formed (as described in Section 11

which is then heated to a temperature at which the ferrite transforms to austenite

the austenite is then rapidly quenched,

which transforms to martensite

For graphite nodules and rosettes,

nodular and malleable cast irons are first formed (again as described in Section 11

which are then austenitized and rapidly quenched

Concept Check 11

? Answer: Both brasses and bronzes are copper-based alloys

For brasses,

the principal alloying element is zinc,

whereas the bronzes are alloyed with other elements such as tin,

Check 11

Question: Explain why,

it is not advisable to weld a structure that is fabricated with a 3003 aluminum alloy

Hint: you may want to consult Section 7

Answer: Strengthening of a 3003 aluminum alloy is accomplished by cold working

Welding a structure of a cold-worked 3003 alloy will cause it to experience recrystallization,

and a resultant loss of strength

Check 11

Question: On the basis of melting temperature,

discuss whether it would be advisable to hot work or to cold work (a) aluminum alloys,

Hint: you may want to consult Sections 7

10 and 7

Answer: Most aluminum alloys may be cold-worked since they are ductile and have relatively low yield strengths

Magnesium alloys are normally hot-worked inasmuch as they are quite brittle at room temperature

magnesium alloys have relatively low recrystallization temperatures

Check 11

Question: (a) Cite two advantages of powder metallurgy over casting

Answer: (a) Advantages of powder metallurgy over casting are as follows: (v) It is used for alloys having high melting temperatures

(w) Better dimensional tolerances result

the degree of which may be controlled (which is desirable in some applications such as self-lubricating bearings)

(b) The disadvantages of powder metallurgy over casting are as follows: (x) Production of the powder is expensive

(y) Heat treatment after compaction is necessary

Concept Check 11

? You may need to consult another reference

Answer: For welding,

there is melting of the pieces to be joined in the vicinity of the bond

a filler material may or may not be used

For brazing,

a filler material is used which has a melting temperature in excess of about 425°C (800°F)

the filler material is melted,

whereas the pieces to be joined are not melted

For soldering,

a filler material is used which has a melting temperature less than about 425°C (800°F)

the filler material is melted,

whereas the pieces to be joined are not

Check 11

Question: Name the three factors that influence the degree to which martensite is formed throughout the cross section of a steel specimen

For each,

tell how the extent of martensite formation may be increased

Answer: The three factors that influence the degree to which martensite is formed are as follows: 1) Alloying elements

adding alloying elements increases the extent to which martensite forms

the extent of martensite formation increases as the specimen cross-section decreases and as the degree of shape irregularity increases

Water provides a more severe quench than does oil,

Agitating the medium also enhances the severity of quench

Concept Check 11

specify the precipitation heat treatment

If it is not possible then explain why

Answer: In order to answer this question it is necessary to consult Figures 11

Below are tabulated the times required at the various temperatures to achieve the stipulated yield strength

Temperature (°C)

Time Range (h)

With regard to temperatures and times to give the desired 18% EL ductility [Figure 11

Time Range (h)