ME-371/571 ENGINEERING MATERIALS

Transcription

1 ME-371/571 ENGINEERING MATERIALS Problem Set 2 1. An SAE-AISI 1035 steel alloy is slowly cooled from 950 C to room What is the pro-eutectoid phase, and at what temperature would it first appear? What are the compositions and amounts of the phases present in this alloy just above 727 C, just below 727 C, and at room temperature? Sketch the microstructure of this alloy at room The proeutectoid phase is Ferrite or α-pbase and will appear at about 800ºC. Compositon of this phase is approximately 0.01wt%C. At 727ºC+ΔT the phases present are α (0.022wt%C) and γ (0.76 wt%c). The amounts of the two phases can be calculated using the lever rule ( ) Wα = = 0.56 = 56% ( ) W = 0.44 = 44% γ 2. Determine the SAE-AISI designation for an alloy that contains approximately 95-wt% ferrite and 5-wt% cementite. 1 of 5

2 At room temperature, the composition of ferrite is 0.002% C and that of cementite is 6.7% C. Using the lever rule W α ( 6.7 C0 ) ( ) = 0.95 C0 = ( ) = 0.34 Therefore the SAE-AISI designation for the steel will be Explain the terms pro-eutectoid and eutectoid ferrite. = Pro-eutectoid ferrite: In a hypo-eutectoid steel, pro-eutectoid ferrite forms between the γ solvus line and the eutectoid temperature Eutectoid ferrite: The ferrite that forms during the eutectoid reaction, and is typically one of the components of lamellar pearlite. 4. Explain what makes stainless steels stainless. Why do welded stainless steel components sometimes corrode? Stainless steels contain Chromium. Chromium metal oxidizes easily and forms a thin protective oxide layer. If there is a sufficient amount of Cr in a steel, the oxide layer that forms protects the steel from oxidation, i.e. making it stainless. In some welded stainless steel, under certain conditions, the chromium precipitates as a carbide along grain boundaries, leaving the surrounding area depleted in Cr. This can lead to corrosion in the Cr depleted zones. 5. Why does gray cast iron have good vibrational damping characteristics? Gray cast iron contains graphite flakes that absorb the energy of vibrations 6. What is the purpose of adding small amounts of lead to brass? Free machining brass contains small amounts of lead which allow machining chips to break off 7. What are the basic steps in precipitation hardening? 2 of 5

3 The three steps in precipitation hardening are: heat treatment, which involves heating the alloy to a temperature at which all precipitates dissolve in the matrix resulting in a single phase Quenching to retain the solute atoms in a supersaturated solution in the solvent matrix Aging, which involves allowing the solute atoms to precipitate in the form of a fine dispersion either by heating the alloy for a period of time (artificial aging), or allowing the alloy to remain at room temperature for a longer period of time (natural aging) 8. What is the meaning of the terms normalizing, full annealing, and stress relief annealing?. What will be the resulting microstructure in eutectoid steels subject to the above heat treatments? Normalizing: Heat to 55ºC above A3 temperature to fully austenitize the sample and air cool. Results in fine pearlite Full annealing: Heat to 50ºC above A3 temperature for hypo-eutectoid steel or above eutectoid temperature (hyper-eutectoid steel) and furnace cool. Results in coarse pearlite. Stress relief annealing: Heat to a temperature that allows for internal stresses to be relieved. This temperature is lower than normalizing or full annealing and does not change the microstructure from what it was prior to annealing See pages 388 and 389 in textbook 9. Why does ETP copper with a small amount of oxygen become brittle above 400 C? ETP copper contains a small amount of oxygen as Cu 2 O. When heated to 400ºC in a hydrogen containing atmosphere, hydrogen atoms diffuse into the metal and reduce the Cu 2 O to Cu. The resulting H 2 O (steam) has a much higher volume than Cu 2 O resulting in blisters forming in the copper. This leads to brittleness 10. Three samples of eutectoid steel are austenitized and then given the following heat treatments: (a) Instantaneously quenched to 600 C, held for 2 minutes and then cooled to room Pearlite (b) Instantaneously quenched to 400 C, held for 2 minutes and then cooled to room Bainite (c) Instantaneously quenched to 100 C, held for 2 minutes and then cooled to room Martensite The microstructures listed above can be determined from the isothermal transformation diagram for eutectoid steel (Fig 10.22). In terms of hardness (a)<(b)<(c). Martensite is the hardest and pearlite the softest. 3 of 5

4 11. The surface of forging made from 4340 steel was unexpectedly subjected to a quench rate of 100 C/s (at 700 C). The forging is specified to have a hardness between Rockwell C46 and C48. Is the forging within specifications? Briefly explain your answer. Figure shows that at a cooling rate of 100ºC/s, 4340 steel has a hardness of about Rockwell C57. This is higher than the specified values. The forging is not within specifications 12. The yield strength and density of a steel are 1100 MPa and 7.8 g/cm 3, respectively. The density of aluminum alloys is approximately 2.7 g/cm 3. An aerospace engineer is faced with a choice between the steel and an aluminum alloy for a structural airframe part. (a) Would it be advantageous to choose an aluminum alloy for the application? (b) If so, which aluminum alloy is suitable for this application, and what is the heat treatment procedure required for obtaining a yield strength of at least 415 MPa. For aerospace applications, weight is one of the critical factors. An aluminum alloy with strength to density ratio greater than that of steel can be found, then it would be advantageous to replace the steel with aluminum. The density of aluminum is (2.7/7.8) or about (1/3) that of steel. Therefore, if there is an aluminum alloy which is more than (1/3) as strong as steel, then the (strength/density) ratio of aluminum will be better than that of steel, making aluminum a better choice. Alloys AA7075-T6 and AA2024-T4 both have strengths greater than (1100/3) MPa and are therefore suitable candidates. 13. In your textbook, aluminum, copper, and magnesium alloys have been classified into wrought alloys and casting alloys. What factors determine whether an alloy is suitable for casting or for deforming (wrought) into a final shape? Casting alloys are those that can be used in the as-cast state, i.e., the liquid metal can be solidified in a mold close in shape to the final product shape. For an alloy to be a good casting alloy it should have good fluidity and low shrinkage upon solidification Wrought alloys generally need to be mechanically worked to the final shape. These alloys need to be sufficiently ductile at the forming 14. Aluminum alloys have been further classified into heat-treatable and non-heat-treatable alloys. What is the basis for this classification? How would non-heat-treatable alloys be strengthened? 4 of 5

5 Heat treatable alloys are strengthened by precipitation hardening. Their composition has been optimized to provide the precipitates that will perform this function. Non-heat treatable alloys are strengthened by cold working or strain hardening 15. Suggest possible iron-based alloys that might be used for the following applications: (a) File cabinet A low carbon steel such as 1010 or 1020 because strength is not important and this material is relatively inexpensive. (b) Wood chisel A high-carbon steel such as 1080 because of high strength requirements (c) Metal mixing bowl A ferritic stainless steel corrosion resistance to preserve appearance and relatively inexpensive AISI 409 (d) Fire hydrant A gray cast iron. It is one of the least expensive materials and casting is one of the least expensive methods of making metallic components. 5 of 5