Tensile engineering stress-strain curves through failure are depicted below.Which specimen features the highest toughness?Engineering Stress, MPa1800-1600-1400-1200-1000-800-600-400-200-Ol0Aluminum Alloy 2024-T810.040.08Stainless Steel (18-8)1340 Steel, Water-Quenched& Tempered at 370 CMagnesium0.12Structural Steel (Mild Steel)Engineering Strain0.161340 Steel, Water-Quenched & Tempered at 370°CAluminum Alloy 2024-T81Structural Steel (Mild Steel)Stainless Steel (18-8)Magnesium0.200.24

Tensile engineering stress-strain curves through failure are depicted below.Which specimen features…

Answered: Tensile engineering stress-strain…

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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98 Z 161423820785837... -> Choose the correct answer for the following: 1- Cast iron the refined form of ? a- Pig iron b- carbon C- molybdenum 2- Proof stress is used when the material ? a- Don't have clear yield point in stress-strain diagram. b- Have clear yield point. C- Have no elasticity 3- Ductile material ? a- Have elongation before failure. b- Doesn't have elongation before failure. c- When failure the section of failure flat. 4- Compression test is the opposite of tensile test ? a- According to direction of applied load. b- Is the same direction of applied load with tensile test. c- Compression test used with metallic materials to determincition Its properties. 5- Impact test ? a- Used to measuring the stress. b- Used to measuring the energy absorbed in bending or breaking the specimen. c- Used to measuring the plasticity. 6- Vacancy diffusion this tape occurs in all metals or alloys a- With Fcc only types of crystal structures b- With Fcc Bcc and Hcp types of crystal…
Three identical fatigue specimen ( denoted A ,B and C ) are fabricated from none ferrous alloy .Each one is subjected to one of the maximum- minimum stress cycles listed in the following table, the frequency is the same for all the three tests.. Which of the following samples will have the longest life and the shortest life Specimen Stress,max ( MPa) Stress , min (MPa) A +450 -150 B +300 -300 C +500 -200
M 1:-A O photo_2021-02-1.. -> 200 stress 150 (MPa) L00 50 0.1 0-2 strain From the tensile stress-strain behavior fer the brass specimen the foibuding:- O the diameter is 12-8 mm Calcalate 1- The Yield stress. 2- Modulus of elasticity. 3 - The Tensile strength . II
A ul ? KBIS M 1:11 O photo_2021-0... -> 200 stress 150 (MPa) (00 50 0-2 strain From the tensile stress-strain behavior for the brass specimen d the diamcter is 12-8 mm Calculate the folluding:- 1- The Yield stress. 2- Modulus of elasticity. 3 - The Tensile strength . II
Part B The stress-strain diagram for a steel alloy having an original diameter of 1.0 in. and a gage length of 7 in. is shown in the figure below. (Eigure 1) Determine the load on the specimen that causes yielding. Express your answer to three significant figures and include appropriate units. HA Py- Value Units Submit Request Answer Part C Figure t oft> Determine the ultimate load the specimen will support. Express your answer to three significant figures and include appropriate units. NO 70 60 50 P. - Value Units 40 30 20 Submit Bequest Answer 10 (in/in) o G04 0m 12 016 020 024 02 a aa omams Provide Feedback
Problem 2: A catastrophic failure occurred to a 7079-T6 aluminum alloy shaft. The maximum tensile stress was specified as 30% of the yield stress (0₂ = 300 MPa) and the stress ratio R was zero. Post-failure examination found that fatigue crack growth striation started at a depth of 2.0 cm from the surface, and the striation spacing was 1.3 x 10-4 cm. The fatigue-crack growth behavior of the aluminum alloy is given in the right figure. The manufacturer claimed that the part was overloaded by the user, but the user denied the allegation and sued the manufacturer. As a mechanical engineer, what is your verdict based on your calculation? da/dN (μm)/cycle 4 6 8 10 2.5TTTTT 1.0 0.25 Titanium Aluminum 0.05- 10 20 30 ksi Vin 40 60 80 100 • 200 Som 200 T -Steel 80 AK (MN/m2) 400 600 1000 • A533 alloy steel Ni-Mo-V alloy steel A HP 9-4-25 alloy steel 7079-T6 aluminum alloy 5456-H321 aluminum alloy Ti-6 Al-4V titanium alloy L 180 T 600 100 480 460 40 uin./cycle
Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the maximum-minimum stress cycles listed in the following table; the frequency is the same for all three Specimen omax(MPa) | Omin (MPa) tests. Rank the fatigue lifetimes of these three specimens from the shortest to the longest. A + 450 150 - 300 - 200 B + 300 C + 500
Calculate the Vickers hardness number (VHN) for a material where a diagonal length of 0.42 mm was produced using a 10 kg loads. VHN = 1.85F d? Answer:
Compute the minimum critical surface crack (edge crack) length if the fatigue life must be a minimum of 5.0 x 106 cycles. Assume the initial maximum edge surface crack length of 1.80 mm and a maximum tensile stress of 160 MPa and minimum stress of 0 MPa. Assume m = 1.8 and A= 7.5 x 1013 in MPa and meter units, Assume Y= 1.25. O a. 4.26 mm O b. 5.38 mm OC. 2.36 mm O d.0.125 m
1. An element is to be subjected to a spectrum of completely reverse stress. It will be cycled at 460 MPa for 29000 cycles. Then the stress will be changed to 440 MPa for 65000 cycles. Finally, the stress will be changed to 315 MPa. How many cycles of operation can be expected at this stress level? Known that the material has endurance limit of 320 MPa, material constant m = 10,9 and critical number of cycles No = 7x10". Ans. Infinity
Please match different stresses to the stress value corresponding to the different points on the Stree-strain diagram of a low carbon steel. Proportional Limit Ultimate Strength Tensile Yield Strength A. A B. B C. D
aerials +968 9192 1059L oving Fo courses / MIME2204tnaSEOTION-3 - ASSESENGENTSE Y1 2040 lo.1 The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 143 kN. - Length of the specimen is 24 mm. - The yield strength is 75 kN/mm². - The percentage of elongation is 44 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 18 kN, iv) Young's Modulus if the elongation is 1.6 mm at 18 kN and (v) Final diameter if the percentage of reduction in area is 26 %. Solution Initial Cross-sectional Area (in mm?) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm2) Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)

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