93%6/6A-10 Typical properties of carbon and alloy steels.Ultimate strength, s,Yield strength, s,PercentConditionksiMPaksiMPаelongationMaterial SAE no.Annealed5739343.2963610204484833136Hot rolledCold drawn10206575517644412010205175135230AnnealedHot rolled1040759062160414251040Cold drawn97669825651610408769364119WQT 700WQT 900WQT 1100WQT 1300104012711881490621221040107738805522410408760063434321040Annealed8961454372251080130314197212OQT 700OQT 900OQT 1100OQT 130010801891791234129889131080108014510001037101711780770483231080Annealed876005135226114177295655141121931141Cold drawn1721186OQT 700OQT 900OQT 1100OQT 1300114113311461007129889151141116800976692011419464868469281141956556041426AnnealedOQT 700OQT 900OQT 1100OQT 13004140146211932311593212124140187128917315414014710141319031841401188141016962341407244027617AnnealedOQT 700OQT 900OQT 1100OQT 13005160105263181323816419.5160196135117912341251601491027132910175160115793103710235160"Other properties are approximately the same for all carbon and alloy steels:Modulus of elasticity in tension = 30 000 000 psi (207 GPa).Modulus of elasticity in shear = 11 500 000 psi (80 GPa).Density = 0.283 lb_/in.' (7680 kg/m²).OQT means oil quenched and tempered. WQT means water quenched and tempered.AAОстétv A AaW27G Seype URI 70%+ | 8 04.An application requires ultimate tensile strength and yield strength ofa steel at 110 ksi and 91 ksi, respectively. A data table is attached in the back ofthe 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 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?Page 4 of 6

4.1. Can SAE 1040 steel be selected for this application? Ans :Yes, SAE 1040 steel can be selected…

A-10 Typical properties of carbon and alloy steels. Ultimate strength, s, Yield strength, s, Percent Material SAE no. Condition ksi MPa ksi MPа elongation Annealed 57 393 43 296 36 1020 65 448 48 331 36 Hot rolled Cold drawn 1020 75 517 64 441 20 1020 Annealed 75 517 51 352 30 1040 Hot rolled 90 621 60 414 25 1040 Cold drawn 97 669 82 565 16 1040 93 641 19 WQT 700 WQT 900 WQT 1100 1040 127 876 621 552 118 814 90 22 1040 107 738 80 24 1040 600 63 434 32 wOT 1200

A-10 Typical properties of carbon and alloy steels. Ultimate strength, s, Yield strength, s, Percent Material SAE no. Condition ksi MPa ksi MPа elongation Annealed 57 393 43 296 36 1020 65 448 48 331 36 Hot rolled Cold drawn 1020 75 517 64 441 20 1020 Annealed 75 517 51 352 30 1040 Hot rolled 90 621 60 414 25 1040 Cold drawn 97 669 82 565 16 1040 93 641 19 WQT 700 WQT 900 WQT 1100 1040 127 876 621 552 118 814 90 22 1040 107 738 80 24 1040 600 63 434 32 wOT 1200

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

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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?
q1: Describe tensile test in detail considering the steps. . define stress , strain (equation and units) . elasticity modulus, elastic and plastic zones . toughness, resilience of matiral Demonstrate stress strain curve for your answer
5.3) A cold-worked carbon steel was tested in a tensile test and the results are shown in Table 1 below: Table 1. Tensile test results Original diameter 10,00 mm Final diameter 7,60 mm Original gauge length 102,00 mm Final gauge length 114,30 mm Load at upper yield 31,21 kN Maximum force 36,25 kN Load at fracture 28,65 kN Factor of safety 2 For Questions 5.3.1 to 5.3.8, calculate the following tensile test properties based on the information provided in Table 1 above: 5.3.1) Stress at upper yield. ANSWER TO QUESTION 5.3.1 5.3.2) Ultimate tensile stress. ANSWER TO QUESTION 5.3.2
Answer the following questions based on the given information in the table and further calculation (if needed for any missing parameter, e.g. strain at yield point). Assume all metals started with the same shape and size before the tensile tests occurred. Material A B C D E Yield Strength (MPa) 105 600 430 355 The material that experienced the greatest percent reduction in area is material The hardest material is material The hardest material has Tensile Strength (MPa) The material that experienced the greatest percent reduction in area has 135 850 The material that experienced the largest change in shape before plastic deformation occurred is material The stiffest material is material The stiffest material has 535 365 Fracture before yielding The material that experienced the largest change in shape before plastic deformation occurred has Strain at Fracture [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] 0.45 0.15 0.20 0.25 Fracture Strength (MPa)…
Match the items in columns I and II. Column-I Column-II 1. Fluidity P. Charpy Q. Knoop 2. Microhardness R. Spiral test S. Cupping test 3. Formability 4. Toughness 5. Permeability (a) P-4, Q-5, R-3, S-2 (b) P-3, Q-5, R-1, S-4 (c) P-2, Q-4, R-3, S-5 (d) P-4, Q-2, R-1, S-3
An application requires ultimate tensile strength and yield strength ofa steel at 110 ksi and 91 ksi, respectively. Answer the following 4 questions: Can SAE 1040 steel be selected for this application? 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? Why is that steel with the condition in part II selected? Is the selected steel brittle or ductile? and Why?
Q3b. The stress-strain characteristic curves for materials A, B and C are shown below. Which one statement most accurately describes the relative toughness and ductility for materials A, B & C? Material A Material B Material C Strain Select one: a. Material A has low toughness, Material B has high ductility and Material C has low toughness. b. Material A has high toughness, Material B has low ductility and Material C has low toughness. c. Material A has low ductility, Material B has low toughness and Material C has low toughness. d. Material A has high ductility, Material B has low toughness and Material C has high toughness. Stress
a . Sketch stress strain curve if the result shown in table represent the force and extension happened in steel, and show Mechanical properties that we get from tensile test on curve? (10p)Note : Lo=80 mm , Do=10 mm , use excel to plot the curve ExtensionLoad(mm) (N)0 0.900.83 4694.341.67 4831.412.50 4781.083.33 4918.834.17 4926.585.00 5257.075.83 5437.016.66 5575.888.33 5775.189.16 5847.5210.83 5965.4111.67 6010.5312.50 6042.5713.33 6072.2614.16 6092.9315.00 6113.2416.67 6140.3617.50 6146.3718.33 6148.1419.16 6149.1725.00 5940.2125.83 5675.3326.67 4725.52b. What is meant by modulus of rigidity? if it increases what does happen to material? (2p)
As the tempering temperature of a heat treated steel increases the associated values of Brinnell hardness also increase, while the values of tensile strength and yield strength decrease. True or false
Using the tensile test simulation tool, a. generate the stress-strain curve for aluminum b. Indicate the following points in the stress-strain curve for aluminum and give the corresponding values: limit of proportionality elastic limit 0.2% offset yield stress (include the graph illustrating how this was determined) ultimate stress fracture stress c. Calculate modulus of elasticity. d. compare aluminum with nylon (include the related graph) and answer the following: Which has higher tensile strength? Provide the necessary values to support the answer. Which is stiffer? Support your answer with calculations.
The engineering stress-strain curve below was obtained for a precipitation hardened Aluminum alloy. What is the approximate Yield Strength for this alloy in psi? Engineering Stress Based on Original Area (psi) 50,000 45,000 40,000 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 O 0.02 0.04 0.06 Aluminum 6061-T6 0.08 0.1 0.12 Engineering Strain (in/in) 0.14 X 0.16 0.18
Question 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…