The science and engineering of materials 4th edition solution manual

If the lattice parameter for the FCC alloy is 3. BCC iron foil is used to separate a high hydrogen gas from a low hydro- gen gas at oC. The concentration of N at one surface is 0.

The concentration at the inner surface is 0. Calculate the number of grams of nitrogen that are lost from the container per hour. If the concentration of hydrogen at one surface is 0. Note the negative sign for the flux. Solution: Helium atoms diffuse through the chains of the polymer material due to the small size of the helium atoms and the ease at which they dif- fuse between the loosely-packed chains.

Determine the activation energy that controls the tem- perature dependence of conductivity. Explain the process by which the temperature controls conductivity. Explain the difference. Calculate the carbon content at 0. Explain the dif- ference. Assume that the iron is FCC.

If the surface composition is maintained at 0. Calculate the carbon content required at the surface of the steel. The carbon con- tent at the steel surface is zero. To what depth will the steel be decarburized to less than 0. Only the outermost 0.

What is the maximum time that the steel part can operate? If the nitrogen con- tent at the steel surface is 0. The nitrogen content at the surface is 0. In an effort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to oC. What time will be required to give us a similar carburizing treatment? By heating the alloy to oC for 3 hours, diffusion of zinc helps to make the composi- tion more uniform.

What temperature would be required if we wished to perform this homogenization treatment in 30 minutes? To minimize thermal stresses during the process, we plan to reduce the temperature to oC. Which will limit the rate at which sintering can be done: diffusion of mag- nesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? Solution: Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.

The alloy is then heated to various temperatures, permitting grain growth to occur. The times required for the grains to grow to a diameter of 0. Does this correlate with the diffu- sion of zinc in copper? Hint: Note that rate is the reciprocal of time. At oC, h are required to obtain the same degree of bonding, and at oC, bonding requires years. What is the activation energy for the diffusion bonding process? Does it appear that diffusion of gold or diffusion of silver controls the bonding rate?

Hint - note that rate is the reciprocal of time. Determine a whether the wire will plastically deform and b whether the wire will experience necking. Determine a whether the bar will plastically deform and b whether the bar will experience necking.

Express your answer in pounds and newtons. Calculate the modulus of elasticity, both in GPa and psi. The polymer has a modulus of elasticity of , psi.

What force is required to stretch the bar elastically to If the aluminum has a yield strength of MPa, what is the min- imum width of the plate? To account for the elastic strain, what should be the diameter of the opening? The opening in the die must be smaller than the final diameter. What is the length of the cable during lifting? After fracture, the gage length is 3. Plot the data and calculate a the 0. After fracture, the gage length is 2.

The deflection of the center of the bar is measured as a function of the applied load. The data are shown below.

Determine the flexural strength and the flexural modulus. The flexural strength is the stress at fracture, or 24, psi. The flexural modulus can be calculated from the linear curve; picking the first point as an example: FL3 Determine the length and diameter of the bar when a lb load is applied.

Determine the applied load, using the data in Table 6—3. When a force of lb is applied, the specimen deflects 0. Calculate a the flex- ural strength and b the flexural modulus, assuming that no plastic deformation occurs.

The sam- ple breaks when a deflection of 0. Calculate a the force that caused the fracture and b the flexural strength. The flexural modulus for silicon carbide is GPa. Assume that no plastic deformation occurs. Solution: a The force F required to produce a deflection of 0. The polymer part is 2 cm wide, 0. If the flexural modulus is 6. Will the polymer fracture if its flexural strength is 85 MPa? Solution: The minimum distance L between the supports can be calculated from the flexural modulus.

A bar of alumina 0. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs.

Determine the Brinell hardness number HB of the metal. Estimate the tensile strength of the steel. Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 0oC? Solution: Test temperature Impact energy J oC 0. If the part is to be used at 25oC, we would want at least 1. Plot the data and determine a the transition temperature defined by the mean of the absorbed ener- gies in the ductile and brittle regions and b the transition temperature defined as the temperature that provides 10 J absorbed energy.

Plot the transition temperature versus silicon content and discuss the effect of silicon on the toughness of the cast iron. What would be the maximum silicon allowed in the cast iron if a part is to be used at 25oC? Solution: Test temperature Impact energy J oC 2. If the part is to be used at 25oC, we would want a maximum of about 2.

Solution: FCC metals do not normally display a transition temperature; instead the impact energies decrease slowly with decreasing temperature and, in at least some cases such as some aluminum alloys , the energies even increase at low temperatures. The FCC metals can obtain large ductili- ties, giving large areas beneath the true stress-strain curve. Which part is expected to have the higher toughness? Solution: Parts produced by powder metallurgy often contain considerable amounts of porosity due to incomplete sintering; the porosity provides sites at which cracks might easily nucleate.

Parts machined from solid steel are less likely to contain flaws that would nucleate cracks, therefore improving toughness. Would you expect these alloys to be notch-sensitive in an impact test?

Would you expect these alloys to have good toughness? Explain your answers. Solution: The sharp-edged plates of the brittle silicon may act as stress-raisers, or notches, thus giving poor toughness to the alloy. Consequently this type of alloy is expected to have poor toughness but is not expected to be notch sensitive. Suppose that fibers of sili- con carbide SiC, another brittle ceramic with low toughness, could be embedded within the alumina.

Would doing this affect the toughness of the ceramic matrix composite? These materials are discussed in later chapters. Solution: The SiC fibers may improve the toughness of the alumina matrix. The fibers may do so by several mechanisms. By introducing an interface between the fibers and the matrix , a crack may be blocked; to continue growing, the crack may have to pass around the fiber, thus increasing the total energy of the crack and thus the energy that can be absorbed by the material.

In addition, the fibers may begin to pull out of the matrix, particularly if bonding is poor; the fiber pull-out requires energy, thus improving toughness. Finally, the fibers may bridge across the crack, helping to hold the material together and requiring more energy to propagate the crack. The plane strain fracture toughness of the composite is 45 MPa m and the tensile strength is MPa.

Will the flaw cause the composite to fail before the tensile strength is reached? Any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws. Observation of the fracture surface indi- cates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture.

Determine the plane strain fracture toughness of the polymer. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0. Our nondestructive test can detect flaws as small as 0. Thus our NDT test is not satisfactory.

Assuming that the maximum tensile and compressive stresses are equal, determine the maximum load that can be applied to the end of the beam. See Figure 6— Solution: The stress must be less than the endurance limit, 60, psi.

What is the maximum permissible load that can be applied? Solution: From the figure, we find that the fatigue strength must be 22 MPa in order for the polymer to survive one million cycles. The bar must survive for at least cycles. What is the mini- mum diameter of the bar? Solution: From the figure, we find that the fatigue strength must be 35, psi in order for the aluminum to survive cycles.

How many hours will the part survive before breaking? What is the fatigue strength, or maximum stress amplitude, required? What are the maximum stress, the minimum stress, and the mean stress on the part during its use?

What effect would the frequency of the stress application have on your answers? Solution: From the figure, the fatigue strength at one million cycles is 22 MPa. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time. Calculate the growth rate of a surface crack when it reaches a length of 0.

It is to survive for cycles before failure occurs. Calculate a the size of a surface crack required for failure to occur and b the largest initial surface crack size that will permit this to happen. The largest surface cracks initially detected by nondestructive testing are 0. If the critical fracture toughness of the polymer is 2 MPa m , calculate the number of cycles required before failure occurs.

Hint: Use the results of Problem 6— A copper spec- imen creeps at 0. If the creep rate of copper is dependent on self-diffusion, determine the creep rate if the temperature is oC.

The initial stress applied to the material is 10, psi. The diameter of the specimen after fracture is 0. Plots describing the effect of applied stress on creep rate and on rupture time are shown below. How many days will the bar survive without rupturing at oC? What is the maximum load that can be applied? Calculate the minimum diameter of the bar. What is the maximum operating temperature? What is the maximum allowable temperature?

Thus The following measurements are made in the plastic region: Change in Force lb Gage length in. Diameter in. The following measurements are made. Change in Force N Gage length cm Diameter cm 16, 0. Determine the strain harden- ing exponent for the metal. The bar, which has an initial diameter of 1 cm and an initial gage length of 3 cm, fails at an engi- neering stress of MPa. No necking occurred. Calculate the true stress when the true strain is 0.

Estimate the total dislocation line present in the photograph and determine the percent increase in the length of dislocations pro- duced by the deformation. Solution: If the length of the original dislocation line is 1 mm on the photograph, then we can estimate the circumference of the dislocation loops. Find the final thickness. Find the final diameter. In a second case, the 2-in.

Determine the final properties of the plate. See Figure 7— Determine the final properties of the bar. Calculate the total percent cold work. The deformation from 0. If we added these three deformations, the total would be This would not be correct. Instead, we must always use the original 1 in. The following table summarizes the actual deformation and properties after each step.

What is the minimum diameter of the original bar? See Figure 7—7. What range of final thicknesses must be obtained? What range of original thicknesses must be used? It is then cold worked further to 1. Calculate the total percent cold work and determine the final properties of the plate?

The strap must be able to support a 35, lb load without plastic deformation. Determine the range of orientations from which the strap can be cut from the rolled sheet. What will be the effect of these particles on the grain growth temperature and the size of the grains at any particular annealing temperature? Solution: These particles, by helping pin the grain boundaries, will increase the grain growth temperature and decrease the grain size.

A suitable temperature might be oC. Measure the slope and compare with the expected relationship between these two temperatures. Is our approximation a good one?

Solution: Converting the recrystallization and melting temperatures to Kelvin, we can obtain the graph shown. The original thickness of the plate is 3 in. Describe the cold work- ing and annealing steps required to make this product.

Compare this process with that you would recommend if you could do the initial deformation by hot working. HW CW The original diameter of the rod is 2 in. Describe the cold working and annealing steps required to make this product. Calculate a the critical radius of the nucleus required, and b the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.

Calculate a the critical radius of the nucleus required, and b the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2. How many atoms would have to group together spontaneously for this to occur? The specific heat of iron is 5.

The specific heat of silver is 3. Calculate the temperature at which nucleation occurred. The specific heat of nickel is 4. Calculate the solidification time for a 0. The times required for the solid-liquid inter- face to reach different distances beneath the casting surface were measured and are shown in the following table. Distance from surface Time in. Explain why this time might differ from the time calculated in part b.

Or we could take two of the data points and solve for c and k. This in turn changes the constants in the equation and increases the time required for complete solidification. Estimate a the secondary dendrite arm spacing and b the local solidification time for that area of the casting.

Solution: a The distance between adjacent dendrite arms can be measured. Estimate the secondary dendrite arm spacing. Assuming that the size of the titanium dendrites is related to solidification time by the same relationship as in aluminum, estimate the solidification time of the powder particle.

Solution: The secondary dendrite arm spacing can be estimated from the pho- tomicrograph at several locations.

Estimate the solidification time of the weld. Determine a the pouring temperature, b the solidification temperature, c the superheat, d the cooling rate just before solidification begins, e the total solidification time, f the local solidification time, and g the probable identity of the metal. Determine a the pouring temperature, b the solidification temperature, c the superheat, d the cooling rate just before solidification begins, e the total solidification time, f the local solidification time, g the undercooling, and h the probable identity of the metal.

Determine the local solidification times and the SDAS at each location, then plot the tensile strength versus distance from the cast- ing surface.

Would you recommend that the casting be designed so that a large or small amount of material must be machined from the surface during finishing? Solution: The local solidification times can be found from the cooling curves and can be used to find the expected SDAS values from Figure 8— Compare the solidifica- tion times for each casting section and the riser and determine whether the riser will be effective.

In a like manner, the area of contact between the thick and thin portions of the casting are not included in the calculation of the casting area.

Consequently the riser will be completely solid before the thick section is solidified; no liquid metal will be available to compensate for the solidification shrinkage. Even though the riser has the longest solidifi- cation time, the thin section isolates the thick section from the riser, pre- venting liquid metal from feeding from the riser to the thick section.

Shrinkage will occur in the thick section. Compare the volume and diameter of the shrinkage cavity in the copper casting to that obtained when a 4-in.

Solution: Cu: 5. A spherical shrinkage cavity with a diameter of 1. Determine the percent volume change that occurs during solidification. After cooling to room tem- perature, the casting is found to weigh 80 g.

Determine a the volume of the shrink- age cavity at the center of the casting and b the percent shrinkage that must have occurred during solidification. Solution: The density of the magnesium is 1. Determine a the percent shrinkage that must have occurred during solidification and b the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.

Solution: The density of the iron is 7. Determine the length of the casting immediately after solidification is completed. Immediately after solidification, the density of the solid cast iron is found to be 7. 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? Related products Add to wishlist.

Add to wishlist. Products search. Shopping Cart 0 Wishlist 0. No products in the cart. Since covalent bonding has higher binding energy than metallic bonding, we can conclude that the silicon has the higher melting temperature due to the higher strength of the silicon bonds when compared to aluminum.

Titanium is stiffer than aluminum, has a lower thermal expansion coefficient than aluminum, and has a higher melting temperature than aluminum. On the same graph, carefully and schematically draw the potential well curves for both metals.

Be explicit in showing how the physical properties are manifest in these curves. Solution: The well of titanium, represented by A, is deeper higher melting point , has a larger radius of curvature stiffer , and is more symmetric smaller thermal expansion coefficient than the well of aluminum, represented by B.

Solution: It is expected that SiN would have the higher modulus of elasticity due to its bonding nature covalent compared to iron metallic. Covalent bonds result in higher binding energies thus having a direct result for a higher modulus of elasticity. Beryllium and magnesium, both in the 2A column of the periodic table, are lightweight metals.

Which would you expect to have the higher modulus of elasticity? Explain, considering binding energy and atomic radii and using appropriate sketches of force versus interatomic spacing. Solution: MgO has ionic bonds. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity.

Aluminum and silicon are side-by-side in the periodic table. Solution: Silicon has covalent bonds; aluminum has metallic bonds. Therefore, Si should have a higher modulus of elasticity. Steel is coated with a thin layer of ceramic to help protect against corrosion. What do you expect to happen to the coating when the temperature of the steel is increased significantly? When the structure heats, steel expands more than the coating, which may crack and expose the underlying steel to corrosion.

Name at least four allotropes of carbon. Why is graphite electrically conductive while diamond is not if both are pure forms of carbon? Solution: The four allotropes of carbon are diamond, graphite, nanotubes and buckminsterfullerene. In diamond, the carbon atoms are covalently bonded to four other carbon atoms thus leaving no free valence electrons available to conduct electricity.

In graphite, the carbon is arranged in layers where the carbon atoms form 3 strong bonds with other carbon atoms, but have a fourth bond between layers which is a weak van der Waals bond. This results in the fourth electron for each of the carbon atom to be available to conduct electricity.

Bond hybridization in carbon leads to numerous crystalline forms. With only six electrons, how is this possible? Solution: Carbon in graphite form has the electron configuration of 1s2 2s2 2p2 which only allows for 3 bonds. Additions of pressure and heat, hybridization occurs to the point that the electron moves from the 2s to the 2p orbital 1s2 2s1 2p3 allowing access for 4 bonds to occur.

In the search results, click on the text link for the section The text below provides the equation for calculating the weight of plating metal and the density of zinc:. Go to the Periodic Table Tools menu and find zinc atomic no. The molar mass of zinc in 0. The number of moles is calculated as: 2. The deposition methods for Zn are hot-dip galvanizing, spraying, plating, sherardising, and painting with zinc-rich paints.

Reviewing the above table, we find that the plating is preferred for the steel substrate of given geometry and required coating thickness. Millions discover their favorite reads on issuu every month. Give your content the digital home it deserves. Get it to any device in seconds. Solutions manual for science and engineering of materials 7th edition by askeland ibsn