. Data was collected from the tensile test demonstration for Aluminum 6061-T6. Below are the resulting stress-strain diagrams based on the tensile test we performed. o (MPa) Al 6061-T6 Stress-Strain Curve o (MPa) 350 320 300 250 200 150 100 50 0 Estimate 0 0.05 0.1 0.15 & (mm/mm) 300 270 250 220 200 150 140 100 a) The modulus of Elasticity, E b) The yield stress, a, using the 0.2% offset method c) The strain at failure, & failure d) The modulus of resilience (approximately!) 50 0 0 Al6061-T6 Stress-Strain Curve, Magnified 0.002 0.0032 0.004 0.006 0.008 € (mm/mm) The right diagram shows the magnified elastic region of stress-strain diagram to enhance the details.

. Data was collected from the tensile test demonstration for Aluminum 6061-T6. Below are the resulting stress-strain diagrams based on the tensile test we performed. o (MPa) Al 6061-T6 Stress-Strain Curve o (MPa) 350 320 300 250 200 150 100 50 0 Estimate 0 0.05 0.1 0.15 & (mm/mm) 300 270 250 220 200 150 140 100 a) The modulus of Elasticity, E b) The yield stress, a, using the 0.2% offset method c) The strain at failure, & failure d) The modulus of resilience (approximately!) 50 0 0 Al6061-T6 Stress-Strain Curve, Magnified 0.002 0.0032 0.004 0.006 0.008 € (mm/mm) The right diagram shows the magnified elastic region of stress-strain diagram to enhance the details.

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

Review Data taken from a stress-strain test for a ceramic are No elements selected o(ksi) given in the table. The curve is linear between the origin and the first point. 50 40 30 Figure 1 of 1 20 10 σ (ksi) e (in./in.) ex 10-*(in./in.) 0.5 1.0 1.5 2.0 25 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
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
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
640 480 160 0.002 0,004 0.006 0.008 0.010 Strain (mm/mm) A tensile test specimen of Aluminum alloy having a gauge length of 280 mm was tested to fracture. Stress and strain data obtained during the test are shown in the attached Figure. **Answers are approximate - do your best to read the graph. Determine: (a) the modulus of elasticity in GPa (4sigs). Enter this value into D2L answer box. (b) the proportional limit in MPa (3sigs). (c) the ultimate strength in MPa (3sigs). (d) Ductility of material based on percent elongation. (3sigs) Your Answer: Stress (MPa)
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?
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)
Specimen (Rebar) Original Length Length when it was clamped Diameter (mm) Internal Diameter (mm) External 1 300 mm 163mm 10 15.030 2 300 mm 180 mm 10.020 14.386 3 300 mm 197mm 9.40 15.028 Dimension of Specimen Specimen A0 P max Tensile strength Modulus of elasaticity Percent Elongation 1 474 N 179.381 2.776 MPa 19.76 MPa 13.4% 2 475 N 159.070 2.895 MPa 18.74 MPa 16.67% 3 485 N 179.381 2.706 MPa 20.058 Mpa 13.4% Result of testing Observation What is the possible discussion of these tablesP.S THIS IS NOT GRADED, ITS JUST EXAMPLES
Select one or more: a. 28.6 Ob. 22.8 O c. 3.7 Od. No Oe. 4.9 Of. Yes 0 8 9 10 11 12 13 14 15 g. 18.5 295 293 280 288 263 290 298 275
The stress-strain data from a tensile test on a cast-iron specimen are € (10-3) 0 0.20 0.44 0.80 1.0 1.5 2.0 26 32 40 46 49 54 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 2.8 3.4 4.0 5.0 a (kpsi) 0 5 10 16 19 Determine the tangent modulus, E, at a value of a=0 psi. 106 psl. The tangent modulus, Eis X
Part B 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. Determine the modulus of elasticity. VO AEO vec ? E = ksi Submit Request Answer Figure 1 of 1 Part C σ (ksi) e (in./in.) Determine the modulus of resilience. ΑΣφ ? vec 33.2 0.0006 45.5 0.0010 in-lb Up = in 49.4 0.0014 51.5 0.0018 Submit Request Answer 53.4 0.0022
1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 0.0019 -0.0057 13.092 0.0762 21.819 0.1270 0.0095 30.547 32.729 0.1778 0.7620 34.911 3.0480 30.01 mm O F Rigid Plate Cylindrical Alloy -Steel casement Figure k Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.
Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…