Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. What tension force is required to cause yielding in a bar of the same material but with a diameter of 20 mm? P, kN AL mm 7.22 0.0839 14.34 0.1636 21.06 26.8 31.7 34.1 35.0 0.241 0.308 0.380 0.484 0.614 36.0 0.924 36.5 1.279 36.9 37.2 1.622 1.994arrow_forwardI need the answer as soon as possiblearrow_forwardResults: 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?arrow_forward
- 2-15 Brinell hardness tests were made on a random sample of 10 steel parts during process- ing. The results were Ha values of 230, 232(2), 234, 235(3), 236(2), and 239. Estimate the mean and standard deviation of the ultimate strength in kpsi. Ans: 50.111 kpsi Som- / -1 bpciarrow_forwardI need the answer as soon as possiblearrow_forwardPlease show work for practice problem 12arrow_forward
- I need the answer as soon as possiblearrow_forward5. The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy (initial length lo= 2.0 in.). After fracture, the total length was 3.014 in. and the diameter was 0.374 in. Load (Ib) Al (in.) 00000 3,000 6,000 0.00167 0.00333 7,500 0.00417 9,000 10,500 0.0090 0.040 12,000 0.26 12,400 11,400 0.50 (maximum load) 1.02 (fracture) a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity. c) Determine the yield strength at a strain offset of 0.002. d) Determine the tensile strength of this alloy. e) What is the approximate ductility, in percent elongation? f) Compute the modulus of resilience. g) Compute from the data and plot true stress versus true strain diagram.arrow_forwardA tensile test was performed on a metal specimen with a diameter of 1/2 inch and a gage length (the length over which the elongation is meas- ured) of 4 inches. The data were plotted on a load-displacement graph, P vs. AL. A best-fit line was drawn through the points, and the slope of the straight-line portion was calculated to be P/AL = 1392 kips/in. What is the modulus of elasticity? BIarrow_forward
- The answer is one of the options below please solve carefully and circle the correct option Please write clear .arrow_forwardFollowing experimental data are obtained from tensile test of a rectangular test specimen with original thickness of 2,5 mm, gauge width of 24 mm and gauge length of 101 mm: Load (N) Elongation (mm) 0 0 24372 0,183 23008 0,315 28357 5,777 35517 12,315 27555 17,978 23750 23,865 Based on the information above; draw stress-strain diagram of the material and answer the following questions. Question 1 ;Determine the elastic energy absorption capacity (in N.mm) of that specimen. Question 2; Determine the plastic energy absorption capacity (in N.mm) of that specimen.arrow_forwardStress 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…arrow_forward
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