S2020-HW5-Solutions-

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Oregon State University, Corvallis *

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Mechanical Engineering

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Feb 20, 2024

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Spring 2020 1/6 MATS 322 Homework #5 Due May 13 th (Wed) at 7 pm (Submit PDF through Canvas-gradescope ) Name: ___________________________________ Student ID #: _____________________________________ Please use this file as a template for your HW assignment and write your answers in the blank space after each question. From Lab For tensile testing, nine 1018 steel tensile bars were homogenized at 760°C and then each was subjected to one of the following heat treatments: x water quenching (WQ) x water quenching followed by a 1 hour anneal at 200°C (WQ + 200) x water quenching followed by a 1 hour anneal at 300°C (WQ + 300) x water quenching followed by a 1 hour anneal at 400°C (WQ + 400) x water quenching followed by a 1 hour anneal at 500°C (WQ + 500) x water quenching followed by a 1 hour anneal at 600°C (WQ + 600) x Oil quenching, (OQ) x Air cooling, (AC) x Furnace cooling (FC) 1. Below (next page) is TTT diagram. This is useful for determining expected microstructures when cooling from above the eutectoid. Neatly and clearly draw the process path for the steel for each cooling process (WQ, OQ, AC, FC). You’ll need to estimate as best you can the time to quench to room temperature based on the lab videos.

Spring 2020 2/6 2. What is the expected microstructure of the WQ, OQ, AC, and FC samples? Your description of each should include the expected phases and an estimate of the phase fraction of each. Solution HT Microstructure Phase Fraction WQ: 100% Martensite 100 % Martensite OQ: 50 % Martensite + 50% Pearlite 50% Martensite, 48.8% Ferrite, 1.2% Fe3C AC: 100% fine Pearlite 98% alpha 2%Fe3C FC: 100% coarse Pearlite 98% alpha 2%Fe3C ࠵? ࠵? = ࠵? ࠵? − ࠵? ࠵? ࠵?࠵?3࠵? ࠵? ࠵? ࠵? − ࠵? ࠵? ࠵?࠵?3࠵? = 0.18% − 6.7% 0.05% − 6.7% = 98.05% Commented [BPM1]: Okay if it’s 100% Martensit e Commented [BPM2]: Range 96%-99% alpha

Spring 2020 3/6 3. What phases are expected in the WQ specimens annealed at 200, 300, 400, 500, and 600°C? Solution WQ + 200: 100% Tempered Martensite WQ + 300: 100% Tempered Martensite WQ + 400: 100% Tempered Martensite WQ + 500: 100% Tempered Martensite WQ + 600: 100% Tempered Martensite 4. How do you expect the microstructure the annealed 1018 steel bars to vary with anneal temperature? Solution Sphere-like precipitates (cementite) for the tempered Martensite is going to be coarse with increasing the annealing temperature. 5. Martensite is on the TTT diagram, but isn’t on the phase diagram of steel. How do you explain the formation of martensite in steel? Solution By the rapid cooling (quenching) of the austenite form of iron, carbon atoms do not have time to diffuse out of the crystal structure in large enough quantities to form cementite (Fe 3 C).

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Spring 2020 4/6 For the following questions use the raw tensile test data posted on Canvas. The data is for a sample that was furnace annealed . See the example excel file on Canvas for how to analyze the data and the important notices at the beginning of the example excel file. Here is some information about the dimensions of the sample, measured before and after testing: Average initial diameter in the gauge length, d o : 5.64 mm Average final diameter near the fracture, d f : 3.44 mm Initial gauge length, L o : 68.5 mm 6. Plot the engineering stress-strain curve. For full credit be to include a descriptive caption and label the plot axes, the 0.2% yield offset, the ultimate tensile strength (UTS), and final elongation.

Spring 2020 5/6 7. Plot the true stress-strain curve. For full credit be to include a descriptive caption and label the plot axes, the 0.2% yield offset, the ultimate tensile strength (UTS), and final elongation. 8. What are the main differences between the engineering stress-strain plot and the true-stress strain plot? Why do they occur? Solution x Engineering stress-strain curve does not give the true nature of the deformation of a metal because it is based on the original dimensions of the specimen x However, the true stress continues to rise because the cross-sectional area decreases and the material work-hardens in the neck region. (= An assessment of the true stress-true strain curve provides a realistic characteristic of the material) x The true-stress-true-strain curves are obtained by converting the tensile stress and its corresponding strain into true values and extending the curve.

Spring 2020 6/6 9. Plot the extensometer data (Strain 1 value in the file) include equation from linear trend line. Solution For full credit on questions 11 – 16 be sure to give your answers using the indicated units! 10. Calculate the modulus of elasticity, E: _ 180 - 220 ___ GPa This comes from the slope of the extensometer data . Don’t forget to convert units to GPa! 11. Calculate the 0.2% Offset Yield Strength, σ y : ____ 275 - 300 _______ MPa This is the intercept between engineering stress-strain plot and 0.2% yield offset plots. 12. Find the ultimate tensile strength, UTS: ____ 450 ______ MPa This is the maximum stress on the engineering stress-strain plot. Commented [BPM3]: Student answers can have a range of numbers

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Spring 2020 7/6 13. Calculate the percent elongation: ____ 18 - 22 _______ %EL This is the last point on the corrected engineering stress-strain plot. 14. Calculate the percent % reduction of area: _____ 62.8 ____ %RA Calculate A o and A f from the diameter measurements and use those to find %RA. 15. Calculate the toughness (total fracture energy): 715.7 (not sure of range) MJ/m 3 This is found from the area under the true stress/strain plot, make sure the final point of the plot is calculated first! For simplicity, you may assume the plot from the UTS to final point of true stress/strain plot is linear. northern . rooqmi.am got a :* :O one ¥ 8 pie are t# tEE# go 61 #