MAAE2700 Lab 03 Template

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Dec 6, 2023

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Lab Report Template Lab 3 – Quenching & Tempering Student Name: Student No: Date: Lab Group/Group Members: 1. Summary The purpose of the lab was to observe the behaviour of steel material after quenching and tempering. The behaviour of the materials could be related to its microstructural change that can be predicted from the continuous cooling transformation diagram. The hardness of the materials was recorded using a Rockwell C scale and then the samples were heated and quenched, then tempered and air cooled 3 times at different temperatures. The cooling rate of SAE 1045 and H13 were found to be 1.142 °C /sec. During the lab, it was concluded that SAE H13 steel was able to retain more of its hardness throughout the different tempering steps as compared to the plain carbon steel samples. It was then concluded that the higher impurity content reduces the ferrite grain sizes which in turn reduces the slippage between the grains which accounts for hardness of SAE H13. 2. Results and Observations 2.1 Record the hardness of each sample in Table #1 provided below. Ensure to indicate the proper unit of measure for all data in the table. 1

Table 1: Hardness measurements for various samples in the quenched and tempered condition Hardness scale used for all measured values: Rockwell C Test Number Condition Sample 1 SAE 1020 Sample 2 SAE 1045 Sample 3 SAE 1045 Sample 4 SAE 1080 Sample 5 SAE 3140 Sample 6 SAE H13 1 As-Received N/A N/A N/A N/A N/A N/A 2 Heat treated at 1050°C for 15 min. and water quenched 29.67 C 18 C 17 C 45.2 C 22 C 19.33 C 3 After tempering at 300°C for 30 min and air cooled. 23.67 C 15.33 C -- 35 C 21 C 30.5 C 4 After tempering at 700°C for 30 min and air cooled. 9.33 C 8 C -- 13.5 C 10.9 C 24 C 5 After tempering at 1050°C for 15 min. and air cooled -- 41 C -- -- -- 36 C 2.2 For the last heat treatment (i.e. Test #5 - 1050 °C for 15 min. and air cooled) record the approximate time (in seconds) to air cool Sample #2 and Sample #6 from the heat treatment temperature to room temperature in the space below. Calculate a constant cooling rate for both samples. Sample 2: Cooling rate = Change in temperature/ change in time taken = (22.5 - 1050)/ (15 *60) = -1.142 °C /sec Sample 6: Cooling rate = Change in temperature/ change in time taken = (22.5 - 1050)/ (15 *60) = -1.142 °C /sec 2

2.3 For Sample #2 (SAE 1045), draw the continuous cooling curve on the CCT diagram provided in the lab below for the water quenched condition (Test #2), and the air cooled condition from 1050 °C (Test #5). Assume the water quench cooling rate is 100 °C /sec, and the air cooling rate is as calculated in question 2.2. For each cooling condition, name the microstructural products that should have formed and enter them in Table #2. 2.4 For Sample #6 (SAE H13) draw the continuous cooling curve on the CCT diagram provided in the lab for the water quenched condition (Test #2), and the air cooled condition from 1050 °C (Test #5). Assume the water quench cooling rate is 100 °C /sec, and the air cooling rate is as calculated in question 2.2. For each cooling condition, name the microstructural products that should have formed and enter them in Table #2. Table 2: Microstructure of different cooling conditions from the austenizing temperature Test Condition Alloy Microstructural Products 2 – WQ SAE 1045 Martensite 5 – AC SAE 1045 Ferrite, perlite, Martensite 2 – WQ SAE H13 Martensite saturated W/ Carbon 5 – AC SAE H13 Bainite Saturated W/ Carbon 2.5 Plot the hardness versus tempering temperature for each of the samples (SAE 1020 steel, 1045, 1090, 3140, and H13). Attach your plot in an Appendix at the end of this template. Ensure that the x and y-axes are properly labeled, and a legend is constructed in order to distinguish the curve for each material. 2.6 Plot the hardness versus carbon content for the “as-received” and water quenched conditions (i.e. Test #1, and Test #2). Attach your plot in an Appendix at the end of this template. Ensure that the x and yaxes are properly labeled, and a legend is constructed in order to distinguish the curve for each test condition. 3

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2.7 Describe what you observed in hardness changes after tempering at 300°C for 30 min. for all five samples. Explain any observed changes in hardness from the previous heat treatment. After tempering at 300 °C for 30 min and then air cooled, 4 out of 5 samples were observed to have a decrease in their hardness when it is being compared to the previous heat treatment which was 1050 °C for 15 min then water quenched. It is only SAE H13 that had an increase in its hardness. The higher carbon content steel retains higher hardness. The decrease in hardness for 4 of the samples after tempering results as the higher temperature treatment allows for the material to reorganize itself to a lower stress orientation and reduce its irregularities and orientation. The breakdown of martensite into ferrite and cementite and grain growth causes the material to weaken while increasing its ductility. 2.8 Describe what you observed in hardness changes after tempering at 700°C for 30 min. for all five samples. Explain any observed changes in hardness from the previous heat treatment. After tempering at 700 °C for 30 min and then air cooled, all 5 samples were observed to have a further decrease in the hardness as compared to the previous treatment which was 300 °C with sample SAE 1080 with the largest decrease in the hardness and SAE 1045 with the least decrease in hardness. 4

2.9 Describe what you observed in hardness changes after tempering at 1050°C for 15 min. for the SAE 1045 and SAE H13 samples. Explain any observed changes in hardness from the previous heat treatment. After tempering at 1050 °C for 30 min and then air cooled, there has been an increase in the hardness as compared to the previous heat treatment which was 700 °C for 30 min with sample 2 having a higher increase in hardness as compared to sample 6. 3. Provide answers to the questions given by the TA and attach them to the end of this template. 1.What fundamental phenomena cause a phase change during tempering? Describe the microstructure change that occurs and the effect on strength and toughness. The fundamental phenomenon that causes a phase change during tempering is the decomposition of martensite at the ferrite- cementite phase. After quenching, the sample is left with a distorted martensite. During tempering, the martensite decomposes resulting in a more stable microstructure with a mixture of ferrite and cementite. This leads to larger grains of approximately the same dimension, and this is due to the recrystallization process that occurs during tempering. This has a significant effect on the toughness and strength of the material. There is a reduction in hardness and increase in ductility. It means that the material becomes less brittle and more resistant to fracture, thereby improving its toughness. Tempering will lower the strength of the quenched sample but increase the toughness but in the end, both the strength and toughness will increase. 2.Discuss the relationship between quench hardness and carbon content. In general, the higher the carbon content in steel, the hardness of the material after quenching tends to be greater. When steel is quenched, it undergoes a rapid cooling process which hardens the steel by transforming the crystal structure to martensite. Hence there is the formation of a hard and brittle phase(martensite). Martensite is responsible for the increased hardness of the material. 3.If the sample 1080 sample was held at 700 °C for many hours instead of 30mins, what would happen to the strength, ductility, and toughness of the alloy. Explain why? There will be changes in the mechanical properties primarily because of the microstructural transformations. There will be a decrease in the strength of the alloy because of the growth of larger grains, which are less resistant to deformation. There is an increase in the ductility because of grain growth and the development of a more equiaxed and finer grain structure. Finer grains are associated with increased ductility because they can undergo more plastic deformation before failure. The toughness of the material will increase because they can absorb energy better before fracturing, making it tougher and more resistant to brittle fracture. APPENDIX A 5

APPENDIX B 6

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8 0 5 10 15 20 25 30 35 40 45 50 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Hardness(C) carbon content(%wt) Graph of hardness against carbon content -10 0 10 20 30 40 50 0 200 400 600 800 1000 1200 Hardness(C) Tempering temperature( °C) Graph of hardness against tempering temperature SAE 1020 SAE 1045 SAE 1080 SAE 3140 SAE H13