Explain why online Al is wrong please don't use it.a) "If the most conservative fatigue criterion were used for infinite life design, the first cycle yieldcheck is not needed." Trueor Falseb) It is easier to find the asymptotic value of fatigue strength as the endurance limit on the S-N curveof a brass alloy (Cu70Zn30) than on the S-N curve of an AISI 1045 steel.Trueor False

Answered: Image /qna-images/answer/d79f45fa-395e-4e52-84e0-b422d14e6288.jpg

"If the most conservative fatigue criterion were used for infinite life design, the first cycle yield check is not needed." True or False It is easier to find the asymptotic value of fatigue strength as the endurance limit on the S-N curve of a brass alloy (Cu70Zn30) than on the S-N curve of an AISI 1045 steel. True or False

"If the most conservative fatigue criterion were used for infinite life design, the first cycle yield check is not needed." True or False It is easier to find the asymptotic value of fatigue strength as the endurance limit on the S-N curve of a brass alloy (Cu70Zn30) than on the S-N curve of an AISI 1045 steel. True or False

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Project 2:Table 3 shows the fatigue data for a ductile cast iron to be used for an automobile axle that rotatesat an average rotational speed of 750 revolutions per minute:a) Make an S-N plot (stress amplitude versus logarithm cycles to failure).b) What is the fatigue limit for this alloy?c) Determine the fatigue lifetimes at stress amplitudes of 500 MPa and 400 MPa?d) Estimate the maximum lifetime of continuous driving that are allowable for the following 250MPa and 150 MPa stress levels.
Table Q2 Steel (Low Low Carbon Medium Carbon High Carbon Alloys) (<0.25 wt% C) (0.25 – 0.6 wt% C) (0.6 – 1.4 wt% C) Туре With With With Plain Plain Plain Property Additive Additive Additive Hardness Low Medium Medium High High Very high Ductility High Medium Medium Low Low Very low A mechanical properties table with different type of steels is shown in Table Q2. a) Based on Table Q2, describe the relationship of hardness and ductility. b) Suggest two methods to improve the hardness of steels by referring to Table Q2. c) Elaborate the reasons the two methods in Q2(b) can be used to improve the hardness of steels.
An S-N plot from fatigue testing of a steel is provided below. Use the plot to determine the following values for this steel. 1. Fatigue limit: 2. Fatigue lifetime at a stress amplitude of: 415 MPa: 275 MPa: 3. Fatigue strength at: 5x104 cycles: 5x105 cycles 500 400 300 200 100 0 1.E+04 Stress Amplitude (MPa) 1.E+05 1.E+06 Cycles to Failure 1.E+07 1.E+08
Both for Ferritic and Martimenistic sstainless steel belong to the AISI 400 series. Farris stainless steel's are heat treatable grades and Martinisitic stick stainless steel's can only be strain hearted to improve their mechanical properties. True or false
B) A machine part is made of forged steel with Su = 630 MPa, S'e = 0.5 Su. Estimate the life cycles (L) for a fatigue strength of 294 MPa. For 103 cycles, take S, = 0.9 Su. %3D
70% + | 8 0 4. An application requires ultimate tensile strength and yield strength of a steel at 110 ksi and 91 ksi, respectively. A data table is attached in the back of the test. Answer the following 4 questions: 4.1. Can SAE 1040 steel be selected for this application? 4.2. If "no" is the answer in Part I, the following Part II, III, and IV can be ignored. If "yes" is the answer in Part I, which condition of SAE 1040 should be selected? 4.3. Why is that steel with the condition in part II selected? 4.4. Is the selected steel brittle or ductile? and Why? Page 4 of 6
316 SS underwent fatigue testing to produce the following S-N curve. a.If the material is cyclically loaded at 200 MPa, what is its expected lifebased off the graph below? b. (please see attachment for fatigue constants) If the material is cyclically loaded at 200MPa, based on these material constants what is its expected life?
An application requires ultimate tensile strength and yield strength ofa steel at 110 ksi and 91 ksi, respectively. Answer the following 4 questions:4.1. Can SAE 1040 steel be selected for this application?4.2. If “no” is the answer in Part I, the following Part II, III, and IV can beignored. If “yes” is the answer in Part I, which condition of SAE 1040 shouldbe selected?4.3. Why is that steel with the condition in part II selected?4.4. Is the selected steel brittle or ductile? and Why?
c). Interpret the following diagram? The effect of carbon and heat treatment on the properties of plain carbon steels. 160 140 Tensile 120 strength, 100 Impact energy 100 Annealed 80 80 Normalized 60 60 Annealed 40 `Yield strength 40 20-% Elongation Annealed 20 Normalized< 0.2 0.4 0.6 0.8 1.0 Weight percent carbon Figure-4 Yield and tensile strength (ksi) Normalized % Elongation or impact energy (It • Ib)
Fatigue data of steel is showed in the following stress-cycle curve.Write down the values of steel endurance limit and steel fatigue strength at 10^5 cycles in the spaces provided below.Steel Endurance Limit: _____ ksiteel Fatigue Strength at 10^5 cycles: _____ ksi
Table Q2 Steel (Low Low Carbon Medium Carbon High Carbon Alloys) (<0.25 wt% C) (0.25 – 0.6 wt% C) (0.6 – 1.4 wt% C) Туре With With With Plain Plain Plain Property Additive Additive Additive Hardness Low Medium Medium High High Very high Ductility High Medium Medium Low Low Very low d) Based on Table Q2, describe a suitable class of steels that can be used in the following applications. (i) Railway tracks (i1) Cutting tools
V20. An Ni-based superalloy component must have a creep rupture life time of at least 100 days at 500°C, compute the maximum allowable stress level. (refer Fig below ). Larsen - Miller Parameter (LM) = T[log tr + C] C = Constant depends up on Alloy C=20. 10 20- loetr0) ssans