5. 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 Al (Ib) (in.) 00000 3,000 0.00167 6,000 7,500 9,000 10,500 12,000 12,400 11,400 0.00333 0.00417 0.0090 0.040 0.26 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.

5. 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 Al (Ib) (in.) 00000 3,000 0.00167 6,000 7,500 9,000 10,500 12,000 12,400 11,400 0.00333 0.00417 0.0090 0.040 0.26 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.

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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6. The following data were collected from a 12-mm diameter test specimen of magnesium (lo = 30.00 mm). After fracture, the total length was 32.61 mm and the diameter was 11.74 mm. Load (N) Al (mm) 0.0000 5,000 10,000 15,000 20,000 25,000 26,500 0.0296 0.0592 0.0888 0.15 0.51 0.90 1.50 (maximum load) 2.10 2.79 (fracture) 27,000 26,500 25,000 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.
A test specimen data made of polyvinyl chloride with 10.16-mm-diameter were collected (l. = 50.80 mm) and shown in Table 1. After fracture, the total length was extended to 53.08 mm and the diameter was 9.98 mm. Load (kN) Al (mm) 0 0.00 1.35 0.19 2.65 0.40 4.00 0.60 5.35 0.81 6.65 1.17 7.35 1.78 (maximum load) 7.10 2.39 6.30 3.05 (fracture) Table 1: The results of the tensile test for polyvinyl chloride. Plot the engineering stress-strain curve (list the stress and strain value in a Table) and evaluate: (i) (ii) the 0.2% offset yield strength; the modulus of elasticity; (iii) the percentage of reduction in area; (iv) the engineering stress at fracture; (v) the modulus of resilience.
ASAP
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 True stress at fracture is. * O 463.441 N/mm2, O 525.441 N/mm2, O 254.441 N/mm2, 498.441 N/mm2,
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 The % reduction in area is * O 15.819% 18.819 O 17.819 O 20.819
The table below shows the deformation data that resulted from applying a pure tensile load to a brass alloy rod with an initial length of 30 mm and a diameter of 10 mm. This rod was subjected to necking and beyond necking deformation. The data at fracture is shown in the last row. Applied Load (N) Length (mm) Diameter (mm) 75,100 33.8 68, 250 34.9 50,200 35.7 6.7 Solve for the TRUE STRESS/ES and TRUE STRAINS. 8.3 7.8
7. The following data were collected from a 20-mm diameter test specimen of a ductile cast iron (lo = 40.00 mm). After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Load Al (N) (mm) 0.0000 25,000 0.0185 50,000 75,000 90,000 105,000 120,000 131,000 0.0370 0.0555 0.20 0.60 1.56 4.00 (maximum load) 7.52 (fracture) 125,000 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.
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
Based upon the information in the table below, which material is the stiffest? Elastic Strain at Fracture Fracture Yield Strength (MPa) Tensile Modulus Strength (MPa) 120 550 340 Fractures before yielding 850 Strength (MPa) 105 500 265 650 720 Material 100 415 310 0.40 0.15 0.23 (GPa) 150 310 210 350 210 700 0.14 OE OB A. OD
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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
Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mm and a 50 mm gage length was tested to fracture. Load and deformation data obtained during the test are given: Determine: (a) the modulus of elasticity. (b) the proportional limit. Load (kN) 0.00 Change in length (mm) (c) the ultimate strength. 0.00 (d) the yield strength (0.20% offset). 0.00 25.79 0.00 (e) the fracture stress. 0.05 51.59 0.10 (f) the true fracture stress if the final diameter 66.55 73.36 80.18 87.77 0.20 of the specimen at the location of the fracture 0.30 0.45 was 8.89 mm 0.63 95.35 0.81 101.42 105.98 112.05 1.01 1,18 1.41 only solve d, e,f 118.13 123.46 128.03 131.85 133.39 134.95 135.73 1.66 1.94 2.27 2.60 2,82 3.15 3.35 3.68 136.53 136.56 3.93 135.84 4.26 135.14 4,66 133.66 4.94 130.67 5.22 5.47 128.44 124.69 5.70 120.92 5.82 117.91 5.95 fracture 116.52