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. Stressand strain data obtained during the test are shown. Determine the proportional limit.Stress (ksi)4030-2010-0-0.00.0Upper scale0.0200.00222 ksiO 15 ksiE19 ksiO 10 ksi13 ksiLower scale-0.040 0.060 0.080 0.1000.004 0.006 0.008 0.010Strain (in./in.)0.1200.012

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Answered: A tensile test specimen of stainless…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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The following curve is a result of a tensile test. The gauge length was 2 in, and the specimen diameter was 0.505 in The fracture area was round with a diameter of 0.391 in. Remember to com 20000 Force (pounds force) 15000 10000 5000 0 16. The yield strength is 0 0.2% O 80,000 pl O 12.000 p 060,000 pl 85,000 p 0.1 0.2 Extension (inch) 0.3 0.4
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
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.)
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.
A 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? BI
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.
A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are shown. Determine the fracture stress. Stress (MPa) 50 40- 30- 20- 10- 0 0 0 Upper scale O 59 MPa O 24 MPa O 40 MPa O 50 MPa O 33 MPa 0.100 0.005 Lower scale 0.200 0.010 Strain (mm/mm) 0.300 0.015 0.400 0.020
Question 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 yield strength (in MPa) of the material. - Calculate the percent elongation of the specimen at yield point. (Use at least five decimal units) - Calculate the stiffness (in MPa) of the specimen material. - Calculate the ultimate strength (in MPa) of the material. - Calculate the percent elongation of the specimen at point of ultimate strength.
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 initial diameter of specimen is * O 25 022 20 O26
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…
hings The following results were obtained during a tensile test to destruction on a mild steel test piece of diameter 15.96mm and 80mm gauge length Extension for a load of 40KN = 0.08mm Maximum load applied during test = 93kN Diameter at fracture = 12.85mm %3D Final length between gauge points 106mm From these results, determine: i) The modulus of elasticity for mild steel ii) The tensile strength iii) Percentage of elongation iv) Percentage of reduction in area
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. 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.

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