*84 The stress-strain diagram for a metal alloy having anoriginal diameter of 12 mm and a gauge length of 50 mm isgiven in the figure. Determine approximately the modulus ofelasticity for the material, the load on the specimen that causesyielding, and the ultimate load the specimen will support.a (MPa)600500400300200100-e(mm/mm)o 004 008 a12 0.16 0.20 024 0.28O 0005 0.001 0.0015 0002 0.0025 0.003 0.0035

Answered: *84 The stress-strain diagram for a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.350 in. Stress (ksi) 120 Upper scale Lower scale H 0.0 0.020 0.040 0.060 0.080 0.100 0.120 0.0 0.002 0.004 0.006 0.008 0.010 0.012 Strain (in./in.) 90 60 30 0- 300 ksi O 126 ksi EA O 237 ksi O 202 ksi O 261 ksi
The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gage length of 2 in. and a diameter of 0.5 in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is Gal = 3.811032 ksi.
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…
A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the ultimate strength. Stress (ksi) 60 50- 40- 30- 20 10- 0 0 0 Upper scale O 66 ksi 28 ksi O 72 ksi O 53 ksi O 62 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (in./in.)
A steel column is a cylinder 6.8 m long and 0.35 m in diameter. What is the change in length of the column when it experiences a tensile stress resulting from the application of a 3600 N force? The elastic modulus for steel is 20 × 1010 Nm2/kg2. ΔL = m
5. Consider a cylindrical specimen of a steel alloy (Figure 6.22) 10 mm in diameter and 75 mm long that is pulled in tension. Determine its elongation when a load of 20,000 N is applied. Stress (MPa) 600 500 400 300 200 100 0 0.00 0.04 Stress (MPa) 500 400 300 200 100 0.000 0.002 0.004 0.006 0.08 Strain Strain 0.12 0.16 0.20 Figure 6.22 Tensile stress-strain behavior for a steel alloy.
The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression vy/3E where y = 2.3 lb/in.³ is the specific weight of the material, y = 0.8 in. is the distance from the free (i.e., bottom) end of the bar, L = 8 in. is the length of the bar, and E = 23000 ksi is a material constant. Determine, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar. (c) the maximum normal strain in the bar.
The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure below (Figure 1). The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.5 mm. Take v=0.35 Figure σ (MPa) 490 0.007 € (mm/mm) Part A If a load of P = 40 kN is applied to the specimen, determine its new diameter. Express your answer to four significant figures and include appropriate units. d= Submit Part B L= O Value Submit μA Request Answer Determine its length. Express your answer to four significant figures and include appropriate units. Provide Feedback Value Units Request Answer ? Units ?
The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression yy/3E where y = 3.0 lb/in.³ is the specific weight of the material, y = 6.3 in. is the distance from the free (i.e., bottom) end of the bar, L = 21 in. is the length of the bar, and E = 27000 ksi is a material constant. Determine, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar (c) the maximum normal strain in the bar. Answer: (a) d = i (b) avg (c) Emax i x10-6 in. με με
The stress-strain diagram for a steel alloy having an original diameter of 0.3 in. and a gage length of 5 in. is shown in the figure below. (Figure 1) Figure σ (ksi) 80 70 60 50 40 30 20 10 0 0 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0 0.0005 0.001 0.0015 0.002 0.0025 0.0030.0035 e (in./in.) < 1 of 1 ▼ Part A Determine the modulus of resilience. Express your answer to three significant figures. (ur)approx= Submit Part B (14)approx= Determine the modulus of toughness. Express your answer to three significant figures. Submit ■ AΣ¢ † vec Provide Feedback Request Answer |VL]ΑΣΦΠ = Ivec Request Answer [C] ? MIC ? in-lb in³ in-kip in³
The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression yy/3E where y = 2.1 lb/in.³ is the specific weight of the material, y = 2.0 in. is the distance from the free (i.e., bottom) end of the bar, L = 5 in. is the length of the bar, and E = 27000 ksi is a material constant. Determine, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar (c) the maximum normal strain in the bar. Answer: (a) 8 = i (b) avg (c) Emax = i x10 in. με με

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