8-10 A 0.505-in.-diameter metal bar with a 2-in.gage length lo is subjected to a tensile test.The following measurements are made inthe plastic region:Change inForceGage lengthDiameter(lb)(in.) (A/)(in.)27,5000.21030.480027,0000.44280.456625,7000.69970.4343Determine the strain-hardening exponentfor the metal. Is the metal most likely to beFCC, BCC, or HCP? Explain.

Answer:We are given a metal bar with an initial diameter of 0.505 inches and an initial gage length…

Answered: 8-10 A 0.505-in.-diameter metal bar…

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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In tensile test a plain carbon steel specimen has a (40mm) gauge length and the Final area (A final) of specimen after tensile test was 264.327. The load which caused fracture was (122.5 KN). After fracture, the final length was 47.516mm Choose the correct answers for the following sentences: Initial area (A final) of specimen * O568.254 O 313.994 325.256 152.368
A test rod of rust-resistant steel, with a diam. of 12.8 mm and a measuring length of 50.8 mm is subjected to a tensile test. The data from the test is shown in the table below.
in a Torsion test done. the following data were collected: the radius of the specimen was 3mm and the gauge length was 76.2mm. During the elastic zone: T1=0. 01N.m. 81=14.760, 12-17.13N.m, and 02 36.910, The modulus of rigidity G will be equal to: Select one O a 44.283 GPa. O b. 26.522 GPa. O C 462.91 MPa. O d. 12.94 GPa.
1. A cylindrical specimen of a metal object with a radius of 3.8E-2 m and a gauge length of 20.31 mm is pulled in tension. Use the load-elongation characteristics tabulated below to answer the following questions; Load Length (N) (mm) 20.31 20.38 12 20.45 18 20.51 24 20.55 30 20.61 36 20.64 42.8 20.68 48.9 20.72 55.2 20.76 60 20.79 65.8 20.84 70.7 20.87 77.3 20.93 83.1 20.98 88.6 22.15 91.58 23.24 91.68 25.41 0 6 Change in length Length (x10-2 m) (m) Computed strain Computed stress (N/m²) (a) Compute the cross-sectional area of brass. (3
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The data illustrated below have been determined while the tension test of the stainless steel sample by amateur engineers. The gage length of the stainless steel sample was 50.8 mm (2.0in.) and, the original diameter of the steel sample was 12.8 mm (0.505in.). Load (Ib) Elongation (in.) Load (Ib) Elongation (in.) 14000 0.020 2 310 0.0022 14400 0.025 4 640 0.0044 14 500 0.060 6950 0.0066 14600 0.080 9 290 0.0088 14800 0.100 11 600 0.0110 14 600 0.120 13000 0.0150 13600 Fracture With the help of Excel, plot the stress-strain diagram for this test conducted, and define the following mechanical properties: a) Young modulus; b) Proportional limit; c) Yield stress @0.2% offset; d) ultimate stress; e) nominal rupture stress.
1.9 (A). During a tensile test on a specimen the following results were obtained: Load (kN) 50 55 30 0.094 40 0.127 Extension (mm) Load (kN) Extension (mm) 15 0.05 70 5.08 75 7.62 Diameter of gauge length = 19 mm Diameter at fracture = 16.49 mm 80 12.7 0.157 82 16.0 1.778 80 19.05 60 2.79 Gauge length 100 mm Gauge length at fracture = 121 mm 65 3.81 70 22.9 Plot the complete load extension graph and the straight line portion to an enlarged scale. Hence determine: (a) the modulus of elasticity; (d) the nominal stress at fracture; (b) the percentage elongation; (e) the actual stress at fracture; (c) the percentage reduction in area; (f) the tensile strength. [116 GN/m²; 21%; 24.7%i 247 MN/m²; 328 MN/m2; 289 MN/m²]
Results of a tensile test of a 0.505-in. diameter aluminum alloy test bar are shown below. Initial length = 2.0 in. The proportional limit occurred with a load = 7000 lbs. What is the Modulus of Elasticity? Load (Ib) (AI) (in.) 0.000 1000 0.001 3000 0.003 5000 0.005 7000 0.007 7500 0.030 7900 0.080 8000 (maximum load) 0.120 7950 0.160 7600 (fracture) 0.205 O A. 30 x 10° psi O B. 1 x 100 psi O C.3x 106 psi OD. 10 х 10° psi
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I want answers to all four questions if possible. Thanks for help :) Following 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. - Calculate the fracture strength (in MPa) of the material. - Calculate the percent elongation of the specimen at fracture point. - Determine the modulus of resilience (in N.mm/mm3) of the material. (Use at least five decimal units) - Determine the toughness index number (in N.mm/mm3) of the material.
The following data was obtained as a result of tensile testing of a standard 0.505 inch diameter test specimen of magnesium. After fracture, the gage length is 2.245 inch and the diameter is 0.466 inch. a). Calculate the engineering stress and strain values to fill in the blank boxes and plot the data. Load(lb) Gage Length (in) Stress (kpsi) Strain 0 2 1000 2.00154 2000 2.00308 3000 2.00462 4000 2.00615 5000 2.00769 5500 2.014 6000 2.05 6200 (max) 2.13 6000 (fracture) 2.255 b). Calculate the modulus of elasticity c). If another identical sample of the same material is pulled only to 6000 pounds and is unloaded from there, determine the gage length of the sample after unloading.

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