I hope to help any other students that may come across this post. It is a common mistake to get the values backwards since it appears to be backwards from what it should be. I urge you to look up the formula for yourself to verify as well as check out a derivation of this formula.
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Add a comment. Determine the percent vol- ume change that occurs during solidification. Does the cast iron expand or contract during solidification? Solution: 0. If all of this hydrogen precipitated as gas bubbles during solidification and remained trapped in the casting, calculate the volume percent gas in the solid aluminum. Locate the triple point where solid, liquid, and vapor coexist and give the temperature and the type of solid present.
Each of these might be expected to display complete solid solubility. In addition, the Mg—Cd alloys all solidify like isomorphous alloys; however a number of solid state phase transforma- tions complicate the diagram. Which one would be expected to give the higher strength alloy? Is any of the alloying elements expected to have unlimited solid solubility in copper? The Cu—Sr alloy would be expected to be strongest largest size difference. Which one would be expected to give the least reduction in electrical conductivity?
Is any of the alloy elements expected to have unlimited solid solubility in aluminum? None are expected to have unlimited solid solubility, due either to difference in valence, atomic radius or crystal structure. See Figure 9— What is the ratio of the number of nickel atoms to copper atoms in this alloy? Determine a the composition of each phase; and b the original composition of the alloy. How many pounds of tungsten can be added to the bath before any solid forms?
How many pounds of tungsten must be added to cause the entire bath to be solid? The total amount of tungsten that must be in the final alloy is: x 0. The total amount of tungsten required in the final alloy is: x 0. What happens to the fibers? Since the W and Nb are completely soluble in one another, and the temperature is high enough for rapid diffusion, a single solid solution will eventually be produced.
Describe what happens to the system as it is held at this temperature for several hours. Determine a the composi- tion of the first solid to form and b the composition of the last liquid to solidify under equilibrium conditions. Determine a the composition of the first solid to form and b the composition of the last liquid to solidify under equilibrium conditions.
Determine a the liquidus temperature, b the solidus temperature, c the freezing range, d the pouring temperature, e the superheat, f the local solidification time, g the total solidification time, and h the composition of the ceramic.
Determine a the liquidus temperature, b the solidus temperature, c the freezing range, d the pouring temperature, e the superheat, f the local solidification time, g the total solidification time, and h the composition of the alloy. Based on these curves, construct the Mo—V phase diagram. If so, identify them and determine whether they are stoichio- metric or nonstoichiometric. Is either material A or B allotropic? If so, identify them and determine whether they are stoichiometric or nonstoichiometric.
Determine the formula for each compound. Determine the formula for the compound. Solution: a 2. See Figure 10— Determine the composition of the alloy.
Is the alloy hypoeutectic or hypereutectic? Solution: L What fraction of the total a in the alloy is contained in the eutectic microconstituent? Determine a the pouring temperature, b the superheat, c the liquidus temperature, d the eutectic tempera- ture, e the freezing range, f the local solidification time, g the total solidifica- tion time, and h the composition of the alloy.
Use this data to produce the Cu—Ag phase diagram. The maximum solubility of Ag in Cu is 7. The solubilities at room temperature are near zero. Solution: a Yes. Some liquid will form. Determine the liquidus temperature, the first solid to form, and the phases present at room tem- perature for the following compositions.
From the graph, we find that the slope 0. Then 0. Include appropriate temperatures. Solution: a For the Cu—1. Which of the requirements for age hardening is likely not satisfied? Determine whether each of the following alloys might be good candidates for age hardening and explain your answer. For those alloys that might be good candidates, describe the heat treatment required, including recommended temperatures. However, eutectic is also present and the strengthening effect will not be as dramatic as in a.
The alloy is expected to be very brittle. Determine the constants c and n in Equation for this reaction. By comparing this figure with the TTT diagram, Figure 11—21, estimate the temperature at which this transformation occurred.
Estimate the carbon content of the steel. Is the steel hypoeutectoid or hypereutectoid? Estimate the temperature and the overall carbon content of the steel. Solution: In order for g to contain 0. Solution: In order for to contain 1.
At this temperature: 6. For the metallic systems, comment on whether you expect the eutectoid microconstituent to be ductile or brittle. Solution: We can find the interlamellar spacing from Figure 11—20 and then use this spacing to find the strength from Figure 11— Estimate a the transformation temperature and b the interlamellar spacing in the pearlite.
Solution: We can first find the interlamellar spacing from Figure 11—19; then using this interlamellar spacing, we can find the transformation temperature from Figure 11— Using Figure 11—35, determine a the temperature from which the steel was quenched and b the carbon content of the steel. Solution: In order for g and therefore martensite to contain 0. Then: 0. Solution: In order for g and therefore martensite to contain 1. Then: 6.
Estimate the vol- ume change that occurs, assuming that the lattice parameter of the austenite is 3. Does the steel expand or contract during quenching? By describing the changes that occur with decreasing temperature in each reaction, explain why this difference is expected. Solution: In a eutectoid reaction, the original grain boundaries serve as nucleation sites; consequently the primary microconstituent outlines the original grain boundaries and isolates the eutectoid product as a discontinuous constitutent.
In a eutectic reaction, the primary phase nucleates from the liquid and grows. When the liquid composition approaches the eutectic composition, the eutectic constituent forms around the primary constituent, making the eutectic product the continuous constitutent. Solution: a 6. Estimate the total interface area between the ferrite and cementite in a cubic centimeter of each steel. Determine the percent reduction in surface area when the pearlitic steel is spheroidized.
The density of ferrite is 7. Solution: First, we can determine the weight and volume percents of Fe3C in the steel: 0. During quenching, the remaining austenite forms martensite; the final structure is ferrite and martensite. The final structure is ferrite, bainite, and martensite.
The austenite transforms to martensite during quenching. The final structure is tempered martensite. The final structure is all bainite. All of the austenite transforms to martensite during quenching.
This is a martempering heat treatment. The final structure is cementite and marten-site. The final structure is cementite and bainite.
The remaining austenite forms martensite during air cooling. The final structure is cementite, bainite, and martensite. Consequently all of the austenite transforms to marten- site during quenching. Discuss the effect of the carbon content of the steel on the kinetics of nucle- ation and growth during the heat treatment. The longest time is obtained for the , or eutec- toid, steel.
Determine the yield strength and tensile strength that are obtained by this heat treatment. The higher strengths are obtained for the lower tempering temperatures. Estimate the carbon content of the martensite and the austenitizing temperature that was used.
What austenitizing temperature would you recommend? The composition of the ferrite at each of these temperatures is about 0. The carbon content of the martensite that forms is about 0. What might have gone wrong in the heat treatment to cause this low strength? What might have gone wrong in the heat treatment to cause this high hardness?
What micro- structure would be obtained if we had used a steel? What microstructure would be obtained if we had used a steel? If the same cooling rates are used for the other steels, the microstructures are: steel: fine pearlite steel: martensite 12—21 Fine pearlite and a small amount of martensite are found in a quenched steel.
What microstructure would be expected if we had used a low alloy, 0. What microstructure would be expected if we had used a steel? For the same cooling rate, the microstructure in the other steels will be: low alloy, 0. What changes in the microstructure, if any, would be expected if the steel contained an alloying element, such as Mo or Cr?
Solution: The alloying element may shift the eutectoid carbon content to below 0. This in turn means that grain boundary Fe3C will form rather than grain bound- ary ferrite. The grain boundary Fe3C will embrittle the steel.
What range of cooling rates would we have to obtain for the following steels? Are some steels inappropriate? When the part is made from steel, the hardness is only HRC Determine the hardness if the part were made under identical condi- tions, but with the following steels.
Which, if any, of these steels would be better choices than ? The steel might be the best choice, since it will likely be the least expensive no alloying elements present. Determine a the cooling rate at that location and b the micro- structure and hardness that would be obtained if the part were made of a steel.
What is this cooling rate? What Jominy distance, and hardness are expected for this cooling rate? This cooling rate corresponds to a Jominy distance of about 3. From the hardenability curve, the hardness will be HRC Solution: a unagitated oil: the H-factor for the 1. The Jominy distance is about 3. The steel has a hardness of HRC 46 and the microstructure contains both pearlite and martensite. What is the minimum severity of the quench H coefficient?
What type of quenching medium would you recommend to produce the desired hardness with the least chance of quench cracking? In order to produce this Jominy distance in a 2-in. All of the quenching media described in Table 12—2 will provide this Jominy distance except unagi- tated oil.
Therefore the maximum diameter that will permit this Jominy distance or cooling rate is 1. The maximum diameter allowed is 1. Consequently bars with a maximum diameter of much greater than 2. Determine the hardness and microstructure at the center of a 2-in. For a 1-in. Therefore, if a 2-in. The case depth is defined as the distance below the surface that contains at least 0. See Chapter 5 for review. Plot the percent carbon versus the distance from the surface of the steel.
If the steel is slowly cooled after carburizing, deter- mine the amount of each phase and microconstituent at 0. See Chapter 5. After cooling, hardnesses in the heat-affected zone are obtained at various locations from the edge of the fusion zone. Determine the hard- nesses expected at each point if a steel were welded under the same condi- tions.
Predict the microstructure at each location in the as-welded steel. Thus at a distance of 0. The table below shows the results for all four points in the weldment. What microstructure would be produced if the martensite were then tempered until the equilibrium phases formed? Solution: We must select a combination of a carbon content and austenitizing tem- perature that puts us in the all-austenite region of the Fe—Cr—C phase dia- gram.
If the martensite is tempered until equilibrium is reached, the two phases will be ferrite and M23C6. The M23C6 is typically Cr23C6. Based on the Fe—Cr—Ni—C phase diagram [Figure 12—30 b ], what phase would you expect is causing the magnetic behavior? Why might this phase have formed? What could you do to restore the nonmagnetic behavior? Solution: The magnetic behavior is caused by the formation of a BCC iron phase, in this case the high temperature d-ferrite.
The d-ferrite forms during solidi- fication, particularly when solidification does not follow equilibrium; sub- sequent cooling is too rapid for the d-ferrite to transform to austenite, and the ferrite is trapped in the microstructure. If the steel is subsequently annealed at an elevated temperature, the d-ferrite can transform to austen- ite and the steel is no longer magnetic.
Why is the tensile strength greater than that given by the class number? What do you think is the diameter of the test bar? Although the iron has a nominal strength of 40, psi, rapid cooling can produce the fine graphite and pearlite that give the higher 50, psi strength.
The nominal 40, psi strength is expected for a casting with a diameter of about 1. If the carbon content in the iron is 3. Solution: We get neither primary phase when the carbon equivalent CE is 4. Explain why you expect different hardenabilities. Solution: Plain carbon steels contain very little alloying elements and therefore are expected to have a low hardenability. Malleable cast irons contain on the order of 1.
The left hand side of the table shows the results of these conversions, with the metals ranked in order of cost per volume. The right hand side of the table shows the cost per pound. Titanium is more expensive than nickel on a weight basis, but less expensive than nickel on a volume basis. If you consid- ered the actual density, do you think the difference between the specific strengths would increase or become smaller?
Both should increase since both Li and Si the major alloying elements are less dense than Al. The brittle eutectic, which is the continuous microconstituent, will then make the entire alloy brittle. Solution: The T9 treatment will give the higher strength; in this temper cold work- ing and age hardening are combined, while in T6, only age hardening is done.
We do not have data in Table 13—5 for —H However, —H19 has a tensile strength of 61, psi and H18 should be psi less, or 59, psi. Compare the amount of b that will form in this alloy with that formed in a alloy. Solution: The alloy contains 2. See Figure 13—5. Solution: The alloy contains 4. Explain your answer. See Figures 13—3 and phase diagrams from Chapters 10 and Solution: The exact values will differ depending on the alloys we select for com- parison.
The table below provides an example. Calculate the minimum diameter of the bar if it is made of a AZ80A—T5 magnesium alloy and b —T6 aluminum alloy. Calculate the weight of the bar and the approximate cost based on pure Al and Mg in each case. What is the maximum force that can be applied if the rod is made of a aluminum, b magnesium, and c beryllium?
How do we explain this statement in view of the phase diagrams in Figure 13—6? Solution: This is possible due to slow kinetics of transformation at low temperatures. Explain the differences observed. Mg has the HCP structure, a low strain hardening coefficient, and a limited ability to be cold worked. Recommend a heat treatment, including appropriate temperatures. Calculate the amount of each phase after each step of the treatment. What precautions must be taken when a leaded wrought alloy is hot worked or heat treated?
Solution: The lead rich phase may melt during hot working or may form stringers during cold working. We must be sure that the temperature is low enough to avoid melting of the lead phase. Would there be a difference in the resistance of the alloy to crack nucleation compared to crack growth?
Solution: The fracture toughness should be relatively good. The acicular, or Widmanstatten, microstructure forces a crack to follow a very tortuous path, which consumes a large amount of energy.
This microstructure is less resistant to crack nucleation. The acicular struc- ture may concentrate stresses that lead to easier formation of a crack. Solution: The g phase is more numerous and also more uniformly and closely spaced; consequently the g should be more effective than the smaller number of coarse carbides at blocking slip at low temperatures.
Determine the volume percent of the Ni3Al precipitate in the nickel matrix. The weight of the Ni3Al is then: Which precipitate likely formed first? Which precipitate formed at the higher tem- perature? What does our ability to perform this treatment suggest concerning the effect of temperature on the solubility of Al and Ti in nickel?
Solution: The larger precipitate forms first and at the higher temperatures The solubility of Al and Ti in Ni decreases as temperature decreases; at a high temperature, the Al and Ti form the g , but some Al and Ti still remain in solution in the matrix. As the temperature decreases, the solubility decreases as well and more of the g can form. However, when titanium is welded, both the front and back sides of the welded metal must be protected. Why must these extra precautions be taken when joining titanium?
Therefore the titanium must be protected until the metal cools below this critical temperature. Since both sides of the tita- nium plate will be heated by the welding process, special provisions must be made to shield all sides of the titanium until the metal cools sufficiently.
Describe the changes in microstructures during the heat treatment for each alloy, including the amount of each phase. What is the matrix and what is the precipitate in each case? Which is an age-hardening process? Which is a quench and temper process? Calculate the amount of a and u in the eutectoid microconstituent.
Try to explain their order. Solution: Strength Strength-to- psi Density weight ratio Ti , 4. Al 73, 2. Mg 40, 1. Cu , 8. Monel , 8. W , Titanium is both strong and relatively low density. Cu, Ni, W are strong but dense. Al and Mg have modest strength but light weight. Solution: Solubility Structure Valence Atom size difference 1. Solubilities tend to decrease as atom size difference increases. What happens when the pro- tective coating on a tungsten part expands more than the tungsten?
What happens when the protective coating on a tungsten part expands less than the tungsten? Solution: If the protective coating expands more than tungsten, compressive stresses will build up in the coating and the coating will flake. If the protective coating expands less than tungsten, tensile stresses will build up in the coating and the coating will crack and become porous.
A ceramic part is produced by sintering alumina powder. It weighs 80 g when dry, 92 g after it has soaked in water, and 58 g when suspended in water.
Calculate the apparent porosity, the true porosity, and the closed pores. A sintered SiC part is pro- duced, occupying a volume of cm3 and weighing g. After soaking in water, the part weighs g. Calculate the bulk density, the true porosity, and the volume fraction of the total porosity that consists of closed pores. Explain whether this material will provide good glass forming tendencies. Above what temperature must the ceramic be heated to be all-liquid?
Compare this to the case when Li2O is added to SiO2. Solution: We can first calculate the required mole fraction of BaO required to produce an O:Si ratio of 2. The weight percent BaO is therefore: The weight percent Li2O is therefore: Will this material pro- vide good glass-forming tendencies?
Draw a sketch similar to Figure 14—10 showing the effect of each of these oxides on the silicate network.
Which oxide is a modifier and which is an intermediate? Solution: PbO2 provides the same number of metal and oxygen atoms to the network as does silica; the PbO2 does not disrupt the silicate network; therefore the PbO2 is a intermediate.
PbO does not provide enough oxygen to keep the network intact; conse- quently PbO is a modifier. Calculate the O:Si ratio and determine whether the material has good glass-forming tendencies. Estimate the liquidus temperature of the material using Figure 14— To determine the liquidus, we must find the weight percentages of each constituent.
Show schematically how the hydrogen per- oxide will initiate the polymer chains. In order to obtain a degree of polymerization of by the combination mechanism, the number of mols of hydrogen peroxide x required for each chain is: Only 0. Therefore the amount required is: Try to draw a sketch of the reaction and the acetal polymer by both mechanisms.
Calculate a the amount of ethylene glycol required, b the amount of byproduct evolved, and c the amount of polyester produced. One molecule of the alcohol is produced for each monomer that is attached to the growing polymer chain. Would you expect polyethylene to polymerize at a faster or slower rate than a polyester?
Solution: In both cases, we would expect polyethylene to polymerize at a faster rate. The ethylene monomer is smaller than the methyl methacrylate monomer and therefore should diffuse more quickly to the active ends of the grow- ing chains.
In polyester, two different monomers must diffuse to the active end of the chain in order for polymerization to continue; this would also be expected to occur at a slower rate than diffusion of only ethylene monomer. The monomer for terephthalic acid is shown below. One molecule of water is produced for each monomer that is attached to the polymer chain. The amount of water evolved is then: If all of the polymer chains are the same length, calculate a the degree of polymerization and b the number of chains in 1 g of the polymer.
If all of the polymer chains are the same length, calculate a the molecular weight of the chains and b the total number of chains in g of the polymer. If each chain contains repeat units, calculate a the number of polyethylene chains in a ft length of rope and b the total length of chains in the rope, assuming that carbon atoms in each chain are approximately 0. The weight of the 10 ft length of rope is 0.
Calculate the molecular weight of the polymer produced using 1 kg of ethylene and 3 kg of propylene, giving a degree of polymerization of Solution: We will consider that each repeat unit—whether ethylene or propylene— counts towards the degree of polymerization.
Determine a the weight average molecular weight and degree of polymerization and b the number average molecular weight and degree of polymerization.
Solution: We do not want a large number of small chains in the polymer; the small chains, due to less entanglement, will reduce the mechanical properties. What is the approximate rela- tionship between these two critical temperatures? Do the condensation thermoplas- tics and the elastomers also follow the same relationship?
The condensation thermoplastics and elastomers also follow this relationship. CHAPTER 15 Polymers 15—21 List the addition polymers in Table 15—5 that might be good candidates for making the bracket that holds the rear view mirror onto the outside of an automobile, assum- ing that temperatures frequently fall below zero degrees Celsius.
Explain your choices. Solution: Because of the mounting of the rear view mirror, it is often subject to being bumped; we would like the mounting material to have reasonable ductility and impact resistance so that the mirror does not break off the automobile. Both polyethylene and polypropylene have low glass transition temperatures and might be acceptable choices.
In addition, acetal polyoxymethylene has a low glass transition temperature and from Table 15—5 is twice as strong as polyethylene and polypropylene. Finally all of the elastomers listed in Table 15—8 might be appropriate choices. Explain their differences, based on the structure of the monomer.
Solution: From Table 15—5: polyethylene. As the complexity of the monomers increases, the glass transition temperature increases. How is this condition expected to affect their mechanical properties compared with those of glassy polymers? Solution: Both polyethylene and polypropylene have glass transition temperatures below room temperature and therefore are presumably in the leathery condition. As a consequence, they are expected to have relatively low strengths compared to most other thermoplastic polymers.
Determine the number of propylene repeat units in each unit cell of crystalline polypropylene. Solution: From Table 15 — 6, we find the lattice parameters for orthorhombic polypropylene. Determine the number of vinyl chloride repeat units, hydrogen atoms, chlorine atoms, and carbon atoms in each unit cell of crystalline PVC.
Solution: From Table 15—3, we can find the lattice parameters for orthorhombic polyvinyl chloride. Calculate the percent crystallinity in the sample. Would you expect that the structure of this sample has a large or small amount of branching? Solution: From Example 15—7, we find that the density of completely crystalline polyethylene is 0. The density of completely amorphous poly- ethylene was also given in the example as 0.
Therefore: Hint: find the density of completely crystalline PVC from its lattice parameters, assuming four repeat units per unit cell. The lattice parameters are given in Table 15—3. Solution: a Linear polyethylene is more likely to crystallize than branched polyethyl- ene.
The branching prevents close packing of the polymer chains into the crystalline structure. The propylene monomers have larger side groups than poly- ethylene and, of course, different repeat units are present in the polymer chains. These factors make close packing of the chains more difficult, reducing the ease with which crystallization occurs. In isotactic polypropylene, the side groups are aligned, making the polymer chain less random, and permitting the chains to pack more closely in a crystalline manner.
At a constant strain, the stress drops to psi after h. If the stress on the part must remain above psi in order for the part to function properly, determine the life of the assembly. For a particular application, a part made of the same polymer must maintain a stress of psi after 12 months.
What should be the original stress applied to the polymer for this application? At an applied stress of psi, the figure indicates that the polymer ruptures in 0. Explain each of your choices. The polyethylene chains are symmetrical, with small hydrogen side groups, and consequently will deform rapidly when an impact load is applied. The more loosely packed chains in LD polyethylene can more easily move when an impact load is applied.
Consequently chain sliding will be more quickly accomplished in PTFE. Solution: a Improved impact properties are obtained by increasing the amount of butadiene monomer; the elastomer provides large amounts of elastic strain, which helps to absorb an impact blow. Acrylonitrile has poor ductility when polymerized. From the curve, calcu- late and plot the modulus of elasticity versus strain and explain the results.
Solution: We obtain the modulus of elasticity by finding the slope of the tangent drawn to the stress-strain curve at different values of strain.
Initially the modulus decreases as some of the chains become untwisted. However eventually the modulus increases again as the chains become straight and higher stresses are required to stretch the bonds within the chains. Is this typical? Each time one double bond is broken, two active sites are created and sulfur atoms then join two repeat units.
Therefore, on the average, there is one set of cross-linking sulfur groups per each repeat unit. In other words, the number of moles of isoprene is equal to the number of sulfur atom groups if every cross-linking site is uti- lized.
If each cross-linking strand contains an aver- age of four sulfur atoms, calculate the fraction of the unsaturated bonds that must be broken.
These monomers can be joined into chains by condensation reactions, then cross-linked by breaking unsaturated bonds and inserting a styrene molecule as the cross-linking agent. Show how a linear chain composed of these three monomers can be produced.
Solution: Polymers that are heavily cross-linked to produce a strong three dimen- sional network structure are called thermosetting polymers. Unsaturated bonds are introduced into the linear polymer chain through the maleic acid.
If the maleic acid were not present, cross-linking could not occur. Warranty Registrationwww. All Rights Reserved. For use with machines having Code Numbers, , , ,, Need Help? Call 1. This instruction manual is issued on 1st Jan
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