Elastic deformation involves a temporary distortion of the lattice structure that is proportional to the applied stress.
Plastic deformation involves a stress of sufficient magnitude to cause a permanent shift in the relative positions of adjacent atoms in the lattice. Plastic deformation generally involves the mechanism of slip - relative movement of atoms on opposite sides of a plane in the lattice. Grain boundaries block the continued movement of dislocations in the metal during straining. As more dislocations become blocked, the metal becomes more difficult to deform; in effect it becomes stronger.
Materials typically possessing a crystalline structure are metals and ceramics other than glass. Some plastics have a partially crystalline structure. Materials typically having a noncrystalline structure include glass fused silica , rubber, and certain plastics specifically, thermosetting plastics.
Crystalline structures undergo an abrupt volumetric change as they transform from liquid to solid state and vice versa. This is accompanied by an amount of energy called the heat of fusion that must be added to the material during melting or released during solidification.
Noncrystalline materials melt and solidify without the abrupt volumetric change and heat of fusion. Multiple Choice Quiz There are a total of 20 correct answers in the following multiple choice questions some questions have multiple answers that are correct.
To attain a perfect score on the quiz, all correct answers must be given, since each correct answer is worth 1 point. For each question, each omitted answer or wrong answer reduces the score by 1 point, and each additional answer beyond the number of answers required reduces the score by 1 point.
Percentage score on the quiz is based on the total number of correct answers. To achieve design function and quality, the material must be strong; for ease of manufacturing, the material should not be strong, in general. Engineering stress divides the load force on the test specimen by the original area; while true stress divides the load by the instantaneous area which decreases as the specimen stretches.
The tensile strength is the maximum load experienced during the tensile test divided by the original area. The yield strength is the stress at which the material begins to plastically deform. It is usually measured as the. Because of necking that occurs in the test specimen. Strain hardening is the increase in strength that occurs in metals when they are strained. When the material does not strain harden. In a compression test, the specimen cross-sectional are increases as the test progresses; while in a tensile test, the cross-sectional area decreases.
Barreling of the test specimen due to friction at the interfaces with the testing machine platens. What is the test commonly used to determine the strength properties of such materials? A three-point bending test is commonly used to test the strength of brittle materials.
The test provides a measure called the transverse rupture strength for these materials. Hardness is defined as the resistance to indentation of a material. It is tested by pressing a hard object sphere, diamond point into the test material and measuring the size depth, area of the indentation. Different hardness tests and scales are required because different materials possess widely differing hardnesses.
A test whose measuring range is suited to very hard materials is not sensitive for testing very soft materials. The recrystallization temperature is the temperature at which a metal recrystallizes forms new grains rather than work hardens when deformed.
Viscosity is the resistance to flow of a fluid material; the thicker the fluid, the greater the viscosity. A Newtonian fluid is one for which viscosity is a constant property at a given temperature. Most liquids water, oils are Newtonian fluids. Viscoelasticity refers to the property most commonly exhibited by polymers that defines the strain of the material as a function of stress and temperature over time.
It is a combination of viscosity and elasticity. Multiple Choice Quiz There are a total of 18 correct answers in the following multiple choice questions some questions have multiple answers that are correct.
The plastic region is characterized by a power function - the flow curve. This answer may require some justification. Viscoelasticity is usually considered to be a property that combines elasticity and viscosity.
However, in deforming over time it involves plastic flow plasticity. Strictly speaking, the shape return feature in viscoelastic behavior violates the definition of plastic flow; however, many materials considered to be viscoelastic do not completely return to their original shape. Problems 7. Strength and Ductility in Tension 3. During the test the specimen yields under a load of 98, N.
This is the 0. Determine: a yield strength Y, b modulus of elasticity E, and c tensile strength TS. During the test the specimen yields under a load of 32, lb. Determine: b yield strength Y, c modulus of elasticity E, d tensile strength TS.
Solution: a Student exercise. Flow Curve 3. Be sure not to use data after the point at which necking occurred. Determine the flow curve parameters n and K. Without knowing any more about the test, can you estimate the flow curve parameters n and K? Determine the strength coefficient and the strain hardening exponent for this metal. Does this help to show what is meant by the term true strain? The summation process is an approximation of the integration over the range from 75 to mm in b.
As the interval size is reduced, the summation becomes closer to the integration value. Determine the engineering strain and true strain for this test. If the metal had been strained in compression, determine the final compressed length of the specimen such that: a the engineering strain is equal to the same value as in tension it will be negative value because of compression , and b the true strain would be equal to the same value as in tension again, it will be negative value because of compression.
Note that the answer to part a is an impossible result. True strain is therefore a better measure of strain during plastic deformation. Based on this information, calculate the engineering tensile strength for the metal.
Solution: Tensile strength occurs at maximum value of load. Necking begins immediately thereafter. This is a true stress. TS is defined as an engineering stress. From Problem 3. Therefore, 0. Determine the true stress and true strain at failure. However, it should be noted that these values are associated with the necked portion of the test specimen. Determine the true stress and true strain at this maximum load. Assuming that the cross- section increases uniformly, determine the load required to compress the specimen to a height of a 50 mm and b Determine the force required to achieve this compression, assuming that the cross-section increases uniformly.
The metal yields 0. At a load of , lb, the height has been reduced to 1. Determine: a yield strength Y, b flow curve parameters K and n. Assume that the cross-sectional area increases uniformly during the test. Determine the shear strain for this situation. Determine: a the shear stress, b shear strain, and c shear modulus, assuming the specimen had not yet yielded.
What is the shear strength of the metal? Calculate the shear strength of the metal. Hardness 3. Determine the BHN for the metal. He claims that all hardness tests are based on the same principle as the Brinell test, which is that hardness is always measured as the 17 applied load divided by the area of the impressions made by an indentor.
Solution: a No, the claim is not correct. Not all hardness tests are based on the applied load divided by area, but many of them are. Based on the BHN determined in that problem, estimate the tensile strength of the steel.
However, from a legal standpoint, it is unlikely that the batch can be Viscosity of Fluids 3. The space between them is occupied by a fluid of unknown viscosity. The motion of the plates is resisted by a shear stress of 10 Pa due to the viscosity of the fluid. Assuming that the velocity gradient of the fluid is constant, determine the coefficient of viscosity of the fluid. The motion is resisted by a shear stress of 0. If the velocity gradient in the space between the surfaces is constant, determine the viscosity of the fluid.
Determine the magnitude of the torque due to viscosity that acts to resist the rotation of the shaft. Density is the weight per unit volume. A pure metal element melts at one temperature the melting point , while an alloy begins melting at a certain temperature called the solidus and finally completes the transformation to the molten state at a higher temperature called the liquidus.
Between the solidus and liquidus, the metal is a mixture of solid and liquid. In the heating of a noncrystalline material such as glass, the material begins to soften as temperature increases, finally converting to a liquid at a temperature defined for these materials as the melting point.
Specific heat is defined as the quantity of heat required to raise the temperature of a unit mass of the material by one degree. Thermal conductivity is the capacity of a material to transfer heat energy through itself by Thermal diffusivity is the thermal conductivity divided by the volumetric specific heat. According to Fick's first law, mass diffusion depends on: diffusion coefficient which rises rapidly with temperature so temperature could be listed as an important variable , concentration gradient, contact area, and time.
Resistivity is the material's capacity to resist the flow of an electric current. Metals are better conductors because of metallic bonding, which permits electrons to move easily within the metal.
Ceramics and polymers have covalent and ionic bonding, in which the electrons are tightly bound to particular molecules.
The dielectric strength is defined as the electrical potential required to break down the insulator per unit thickness. An electrolyte is an ionized solution capable of conducting electric current by movement of the ions. Multiple Choice Quiz There are a total of 12 correct answers in the following multiple choice questions some questions have multiple answers that are correct. In these cases, which of the following temperatures marks the beginning of melting?
This is perhaps a trick question. Choices a and b are included in Eq. Temperature e has a strong influence on the diffusion coefficient. Time f figures into the process because it affects the concentration gradient; as time elapses, the concentration gradient is reduced so that the rate of diffusion is reduced. This shaft is to be inserted into a hole in an expansion fit assembly operation. To be readily inserted, the shaft must be reduced in diameter by cooling.
Refer to Table 4. Revise Eq. Solution: Assume a 1 cm3 cube, 1 cm on each side. From Table 4. Conversion: 1. Use Table 4. A tolerance is defined as the total amount by which a specified dimension is permitted to vary. The reasons why surfaces are important include: aesthetics, safety, friction and wear, effect of surface on mechanical and physical properties, mating of components in assembly, and thermal electrical contacts.
The nominal surface is the ideal part surface represented on an engineering drawing. It is assumed perfectly smooth; perfectly flat if referring to a planar surface; perfectly round if referring to a round surface, etc. Surface texture is the random and repetitive deviations from the nominal surface, including roughness, waviness, lay, and flaws. Surface texture refers only to the surface geometry; surface integrity includes not only surface but the altered layers beneath the surface.
Roughness consists of the finely-spaced deviations from the nominal surface, while waviness refers to the deviations of larger spacing.
Roughness deviations lie within waviness deviations. Surface roughness is defined as the average value of the vertical deviations from the nominal surface over a specified surface length. Surface roughness measurement provides only a single value of surface texture. Among its limitations are: 1 it varies depending on direction; 2 it does not indicate lay; 3 its value depends on the roughness width cutoff L used to measure the average.
The changes and injuries include: cracks, craters, variations in hardness near the surface, metallurgical changes resulting from heat, residual stresses, intergranular attack, etc. Energy input resulting from the manufacturing process used to generate the surface. The energy forms can be any of several types, including mechanical, thermal, chemical, and electrical.
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Por fa lo necisito urgente te agradeceria tanto. Gracias Mi correo es medina. Robert L. 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? 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.
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