ME3150-01 Spring 2024-Quiz 3 Solution

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California Polytechnic State University, Pomona *

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Mechanical Engineering

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Apr 3, 2024

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ME3150-01 Quiz 3 Solution Name__________________ Score___/25 1) Ductile-to-Brittle Transition Temperature (DBTT) is determined by _ Charpy Impact or Izod Impact ______ test. (2 pts) 2) A Fe-C steel contains 2 wt% Cementite at 1000 o C. the composition of the steel is__ 0.134 wt.% C + 99.866 wt.% Fe ____ (2 pts) Solution to Problem #2: Let x be the concentration of carbon. By Level Rule, x 0.022 6.7 0.022 = 0.02 → x≈ 0.134 wt . % The concentration of Fe is 1-C%=99.866% 3) The measured stress at the center through crack tip is three times as high as the far-field stress. If the length of the defect is 1 mm, what is the radius of curvature at the crack tip? (2 pts) Solution to Problem #3: σ tip = 2 σ ( far field ) a ρ 3 = σ tip σ ( far field ) = 2 0.5 × 1.0 ( mm ) ρ → ρ = 2 9 0.222 ( mm ) 4) Referring to the Cu-Zn phase diagram, if the initial composition of brass contains 30 wt %Cu and cooled down from liquid to 500 o C, what is the total amount of γ in the 100 g sample? How much is the eutectoid γ? What is the total amount of copper can be extracted from the ɛ phase? (9 pts) Solution to Problem #4 We can divide the problem into three parts. Part (a): To find the total amount of γ. Referring to the slide #26 in Chapter 9, at 500 o C, the microstructure contains two phases: ɛ and γ. Using level rule generates:
γ = 77 70 77 67 × 100 ( g ) = 70 ( g ) Part (b): The eutectoid γ comes from the phase transformation of δ →ϵ + γ . The maximum portion of δ can be obtained by Level Rule at 560 o C as δ % 70 69.5 74.5 69.5 = 1 9 11.1% The structure was just cooled down to 560 o C allowed eutectoid structure formation and the relative fraction of eutectoid γ in the microstructure of ϵ + γ is: γ eutoid = 74.5 69.5 77 69.5 2 3 0.667 The eutectoid microstructure will not change much if the Cu-Zn alloy continues to cool down to reach 500 o C. Therefore, the amount of eutectoid γ in 100 grams of alloy is 11.1%×0.667×100 (g)=7.4 (g) Part (c): From the phase diagram, it can be determined that the ɛ phase consists of 23% Cu and 77% of Zn. Approximately. By Level Rule, the per centage of ɛ in ϵ + γ at to 500 o C is: ε % = 70 67 77 67 × 100% = 30% Therefore, the copper amount in the ɛ can be determined as: 23%×30%×100 (g)=6.9 (g). 5) Considering the loading rate effect on the fracture toughness of materials. The higher the loading rate, the higher the fracture toughness of materials. True____ or False__ X ___. (2 pts) Because high speed loading causes high brittleness .
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