ENGR 25 Delta College Stress Strain Curves Questions

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Engineering

ENGR 25

Delta College

ENGR

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ENGR 25 Fall 2021 Midterm Exam 2 November 10, 2021 This is an open book and open notes examination, however the work submitted must be your own. Students are given 24-hours to complete the exam. There are 3 problems, each with multiple parts, for a total of 30 possible points. There is one extra credit problem on the ninth page of this exam, worth 2 extra points. The exam is worth 15% of your final grade for the class. There are a total of 11 pages including this cover page, the list of equations, and the periodic table. Show all of your work for each question. Name:____________________________________________________(please print) Honor Code: All students agree, individually and collectively, that they will not give or receive unpermitted aid during examinations. Engaging in such unpermitted actions is considered cheating. By my signature, I affirm that I have adhered to the letter and spirit of the Honor Code. Signature:____________________________________________________ 1/11 Problem 1 Stress-Strain Curves (10 points) Problem 3 (8 points) a) This titanium specimen was deformed in tension. The specimen failed at the point marked a) X. This was stress-strain deformed in tension. failed at the point marked with an Is titanium this an specimen engineering curve The or aspecimen true stress-strain curve? How do you with an X. Is this an engineering stress–strain curve or a true stress–strain curve. How do you know? (1 point) know? (1 point) b) Would you say this material is tough? Why or why not? Refer to features of the stress–strain curve in your answer. (1 point) b) Approximate the maximum load in Newton’s that this specimen sustained if the initial crosssectional area was 10 mm2. (1 point) c) Approximate the true stress that corresponds to the ultimate tensile strength. (1 point) c) Approximate the true stress that corresponds to the ultimate tensile strength. (1 point) d) Approximate the maximum load in newtons that this specimen sustained if the initial cross– sectional area was 10 mm2. (1 point) 5/7 2/11 Problem 1 continued d) At the moment of fracture, what was the final length of the specimen if the initial length was 50 mm? (1 point) Problem 3 continued e) At the moment of fracture, what was the final length of the specimen if the initial length was 50 mm? (1 point) e) Using the magnification of the elastic region shown below, approximate the elastic modulus. (1 point) f) Using the magnification of the elastic region shown above, approximate the elastic modulus. (1 point) f) Approximate the 0.2% offset yield stress. Show your work on the plot above for full credit. (1 point) g) Approximate the 0.2% offset yield stress. Show your work on the plot for full credit. (1 point) h) What is the amount of plastic strain at failure sustained by this specimen? (1 point) 3/11 Problem 1 continued g) Calculate the total elongation ∆l of the specimen when subjected to 200 MPa if the initial length was 50 mm. You will need to use the elastic modulus computed in part e). (1 point) h) What is the amount of plastic strain at failure sustained by this specimen? You will need to use the elastic modulus computed in part e). (1 point) i) Compute the percent elongation of the specimen if the initial length was 50 mm. You will need to use the final length of the specimen computed in part d). (1 point) j) Suppose the specimen is loaded to a stress of 500 MPa, then the specimen is completely unloaded, and without interrupting the test, it is loaded again to failure. Draw the path that is followed by the data from start to finish on the graph below. You will need to deviate from the Problem path shown. Draw on top of the given data where appropriate. (1 point) 3 (8 points) a) This titanium specimen was deformed in tension. The specimen failed at the point marked with an X. Is this an engineering stress–strain curve or a true stress–strain curve. How do you 4/11 know? (1 point) Problem 2 Defects (10 points) a) Pure nickel has a higher hardness than pure copper. The hardnesses of pure nickel and copper are indicated in the plot below. Draw a schematic curve showing how you might expect the hardness of a nickel–copper alloy to vary from pure nickel to pure copper as a function of composition. Explain the shape of your curve with 1–2 sentences. (2 points) Ultimate Tensile Strength (MPa) b) The plot below shows the ultimate tensile strength of a metal that was cold-worked and then annealed at increasingly higher temperatures. Identify the stages of annealing that you see in the data, give the approximate temperature ranges for these stages, and explain the features of the plot that led you to these answers. (3 points) 450 400 350 300 250 200 150 100 500 550 600 650 700 Temperature 5/11 750 (oC) 800 850 Problem 2 continued Consider the data in the table below. Samples 1 and 2 represent the same material that has been processed slightly differently to produce the different properties. Consider each part below as a separate question that has no relationship to previous parts of the question. A one or two sentence answer should be sufficient. Sample Ultimate Tensile Strength (MPa) 1 2 620 400 c) Samples 1 and 2 are identical except that they have different grain sizes. Which sample has the smaller grain size? How do you know? (1 point) d) Samples 1 and 2 are identical except that they have different dislocation densities. Which sample has a lower dislocation density? How do you know? (1 point) e) Samples 1 and 2 are identical except that one has a small alloying addition. Which sample has been alloyed? How do you know? (1 point) f) Which sample is weaker? How do you know? (1 point) g) Samples 1 and 2 are identical except that one has been work hardened. Which sample has been work-hardened? (1 point) 6/11 Problem 1 (11 points) Problem 3 Phase Diagrams (10 points) The binary phase diagram for the gallium and indium system is shown below. The composition The binary phase diagram for the gallium and indium system is shown below. The composition of 100% indium is given by the right–hand axis. of 100% indium is given by the right-hand axis. a) Label the phases present in each region of the phase diagram. Denote α as the Ga–rich solid and β as the In–rich solid phase. to denote the liquid phase. (2 the points) a) Labelphase the phases present in each region ofUse the Lphase diagram. Denote α as Ga-rich solid phase and β as the In-rich solid phase. Use L to denote the liquid phase. (2 points) b) What is the what composition the solid with a melting temperature 350 K? indium? (1 point) (1 point) b) Approximately are the of melting temperatures of pure galliumof and pure c) Draw a schematic diagram of the microstructure of the alloy with an overall composition of 80 atomic percent In when cooled to 288 K. Label your diagram with the phases. (2 points) c) What is the composition of the first solid to form at a temperature of 350 K? (1 point) 2/7 7/11 Problem 3 continued d) Draw a schematic diagram of the microstructure of the alloy with an overall composition of 80 atomic percent In when cooled to 288 K. Label your diagram with the phases. (2 points) e) What are the compositions and amounts of the phases present for an overall composition of 80 atomic percent In at 288 K? Make any necessary approximations from the phase diagram. (2 points) f) What is the overall alloy composition required to have a material consisting of 75% liquid and 25% β at a temperature of 300 K? (2 points) 8/11 Extra Credit (+2 points; optional) Consider the interstitial diffusion of nitrogen in FCC iron. If D0 = 0.0034 cm2/s and the activation energy is 34,600 cal/mol, calculate the diffusion coefficient at 950°C. 9/11 ENGR 25 Equations and Constants N A = 6.022 × 10 23 1 eV = 1.602 × 10–-. J 1 N = 1 Pa × m2 Chapter 5 Arrhenius-type equations: 𝑄 Rate = 𝑐4 exp 7− < 𝑅𝑇 𝑄 𝑛> = 𝑛 exp 7− < 𝑅𝑇 𝑄 𝐷 = 𝐷4 exp 7− < 𝑅𝑇 Fick’s first law: 𝑑𝑐 𝐽 = −𝐷 𝑑𝑥 Fick’s second law: 𝑐C − 𝑐D 𝑥 = erf 7 < 𝑐C − 𝑐4 2√𝐷𝑡 Chapter 6 Engineering Stress (S): 𝐹 𝑆= 𝐴4 Engineering Strain (e): 𝛥𝑙 𝑒= 𝑙4 True Stress (𝜎): 𝜎 = 𝑆(1 + 𝑒) True Strain (𝜀): 𝜀 = ln(1 + 𝑒) Hooke’s Law: 𝑆 = 𝐸𝑒 𝜎 = 𝐸𝜀 10/11 11/11
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