Question 4A large consignment of AA 6061 T6 aluminum alloy bars was received by a purchaser. To verifythat the consignment met his specifications, the purchaser sent twelve (12) of the bars to anindependent testing company for determination of tensile properties. The yield strength results,in MPa, determined for the twelve (12) bar samples were as follows:257252 255 257 256 250 252 260 253 258 257Using this yield strength results:(a) Calculate the mean and standard deviation for the sample.(b) Estimate the standard deviation of the mean (SDOM) for the sample.253(c) Determine the 95% confidence interval on the mean yield strength for the consignment ofbars.

Step 1: Step 2: Step 3: Step 4:

Answered: Question 4 A large consignment of AA…

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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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
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
6. The following engineering stress-strain data were obtained for 0.2% C plain carbon steel. (a) Plot the engineering stress-strain curve (b) Determine the ultimate tensile strength for the alloy (c) Determine the percent elongation at fracture (d) Plot the true stress-strain curve Engineering strain, in./in. Engineering stress, ksi 30 0.001 55 0.002 60 0.005 68 0.01 72 0.02 74 0.04 75 0.06 76 0.08 75 0.1 73 0.12 69 0.14 65 0.16 56 0.18 51 0.19(fracture)
Review Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. No elements selected a(ksi) 50 + 40 - 30 Figure 1 of 1 20 10 + o (ksi) € (in./in.) ex 10(in./in.) 0.5 1.0 1.5 2.0 2.5 33.2 0.0006 45.5 0.0010 Press (ENTER) to select this element. Press ESC) to return to the main menu. Press CTRL+Q) to quit the application. 49.4 0.0014 51.5 0.0018 53.4 0.0022
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Compare full stress-strain curve for steel, brass, copper and aluminium alloy.
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Results: material iron brass aluminum copper NH VINI Discussion: vickers hardnes test Load(kg) 50 50 30 30 Diameter of trace d1 d2 0.58 0.65 0.68 0.74 0.94 0.86 0.72 0.76 d1+d2 2 VHN Kg/mm² 1- Calculate the Vickers hardness number for the previous samples and compare the results and discuss them according to the table. 2- Draw a graphic relationship between the diameter of the trace and the hardness of Vickers. 3- What is the condition for using the Vickers hardness device?
Which material should be chosen when high strength is required?
1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves were generated, both curves generated from the same set of data. Use the graphs to fill in the mechanical properties of the material tested in the box below. Don't forget units! Stress vs Strain Stress, psi Stress, psi 80000 70000 60000 50000 40000 30000 20000 10000 0 0.00 80000 70000 60000 50000 40000 30000 20000 10000 0.02 0 0.000 0.002 0.04 0.004 0.06 0.006 0.08 0.10 Strain Stress vs Strain 0.008 0.12 Elastic Modulus, E: 0.2% Offset Yield Strength, oo: Tensile Strength, ou: Breaking Strength, of: % Elongation: 0.14 0.010 0.012 0.014 Strain 0.16 0.18 0.016 0.018 0.20 0.020
Given the following data for five different materials answer parts A B C D E & F with an explanation

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