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? 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: It is expected that since aluminum is a metal that it would have metallic bonding. Silicon on the other hand is a metalloid between metal and non-metal and has covalent bonding. 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.
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