Calculate the (total tensile strength) of the mitotic tensile strength test for the information in the table below: Time Dmm Load Kn sec 1 60 16 14 2 54 15.5 14 57 16 15.5 52 15 18 53 15 14 59 17.5 19 50 9 17 56 19 21 58 10 21.4 55 11 25 34 5 6 7 8 9 10

Calculate the (total tensile strength) of the mitotic tensile strength test for the information in the table below: Time Dmm Load Kn sec 1 60 16 14 2 54 15.5 14 57 16 15.5 52 15 18 53 15 14 59 17.5 19 50 9 17 56 19 21 58 10 21.4 55 11 25 34 5 6 7 8 9 10

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

See similar textbooks

Similar questions

For a piece of copper alloy, a standard stress test was applied to it, the following data was collected, from which a stress-strain diagram must be generated, where as data it is known that the initial diameter of the element is 0.505in. The analysis must include the following: 1.Modulus of Elasticity and modulus of resilience. 2.Percentage of elongation. 3.Percentage of area reduction. 4.Real and engineering stress at fracture. It is known that after the specimen fractures, its dimensions in terms of length and diameter are 3.014 and 0.374in, respectively.
In a tensile test on a specimen of black mild steel of 12 mm diameter, the following results were obtained for a gauge length of 60 mm. Load W(kN) 5 10 15 20 25 30 35 40 Extension x (10-3 mm) 14 27.2 41 54 67.6 81.2 96 112 When tested to destruction. Maximum load = 65 kN; load at fracture = 50 kN, diameter at fracture = 7.5 mm, total extension on gauge length = 17 mm. Find young's modulus, specific modulus, ultimate tensile stress, breaking stress, true stress at fracture, limit of proportionality, percentage elongation, percentage reduction in area. The relative density of the steel is 7.8. Draw the straight line graph.
In a tensile test on a specimen of black mild steel of 12 mm diameter, the following results were obtained for a gauge length of 60 mm. Load W(kN) 5 10 15 20 25 30 35 40 Extension x (10-3 mm) 14 27.2 41 54 67.6 81.2 96 112 When tested to destruction. Maximum load = 65 kN; load at fracture = 50 kN, diameter at fracture = 7.5 mm, total extension on gauge length = 17 mm. Find young's modulus, specific modulus, ultimate tensile stress, breaking stress, true stress at fracture, limit of proportionality, percentage elongation, percentage reduction in area. The relative density of the steel is 7.8. Draw the straight line graph. Answer: breaking stress,true stress at fracture and limit of proportionality.
The data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13mm and a gage length of 50mm. At fracture, the elongation between the gage marks was 3.0mm and the minimum diameter was 10.7mm. Plot the conventional stress-strain curve for the steel and determine the propotional limit, modulus of elasticity (i.e the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 50mm, and percent reduction area. TENSILE-TEST DATA Load(kN) Elongation(mm) 5 0.005 10 0.015 30 0.048 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracture
The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 157 kN. - Length of the specimen is 23 mm. - The yield strength is 89 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 18 kN, iv) Young's Modulus if the elongation is 1.3 mm at 18 kN and (v) Final diameter if the percentage of reduction in area is 25 %. Solution: Initial Cross-sectional Area (in mm2) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm2) Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)
The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 150 kN. - Length of the specimen is 24 mm. - The yield strength is 78 kN/mm2. - The percentage of elongation is 42 %. Determine the following iii) Stress under an elastic load of 14 kN, iv) Young's Modulus if the elongation is 1.7 mm at 14 kN and (v) Final diameter if the percentage of reduction in area is 26 %. I wont answer for this three questions only: 1)Final Area of the Specimen at Fracture (in mm) 2)Final Diameter of the Specimen after Fracture (in mm) 3)Initial Cross-sectional Area (in mm2)
The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 142 kN. - Length of the specimen is 23 mm. - The yield strength is 82 kN/mm2. - The percentage of elongation is 48 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 15 kN, iv) Young's Modulus if the elongation is 1.6 mm at 15 kN and (v) Final diameter if the percentage of reduction in area is 20 %. Solution Initial Cross-sectional Area (in mm2) Answer for part 1 The Diameter of the Specimen (in mm) Answer for part 2 Final Length of the Specimen (in mm) Answer for part 3 Stress at the elastic load (in N/mm2) Answer for part 4 Young's Modulus of the Specimen (in N/mm2) Answer for part 5 Final Area of the Specimen at Fracture (in mm) Answer for part 6 Final Diameter of the Specimen after Fracture (in mm)
The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 151 kN. - Length of the specimen is 23 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 17 kN, iv) Young's Modulus if the elongation is 2.4 mm at 17 kN and (v) Final diameter if the percentage of reduction in area is 24 %. Find: 1)Stress at the elastic load (in N/mm2) 2)Young's Modulus of the Specimen (in N/mm2) 3)Final Area of the Specimen at Fracture (in mm) 4)Final Diameter of the Specimen after Fracture (in mm)
The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 151 kN. - Length of the specimen is 23 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 17 kN, iv) Young's Modulus if the elongation is 2.4 mm at 17 kN and (v) Final diameter if the percentage of reduction in area is 24 %. THE QUESTION IS : FIND THE Final Diameter of the Specimen after Fracture (in mm)????
Select All the correct statements from below. a. If the averaged distance between partial dislocations for material A observed under TEM is smaller than material B, the estimated stacking fault energy of material A is lower than material B. b. K_IC is a material property that doesn't change with testing conditions c. The main difference between high cycle fatigue and low cycle fatigue is the difference in the number of cycles to failure d. A negative R value in fatigue test indicate the minimum stress is compressive in nature
Three identical fatigue specimen ( denoted A ,B and C ) are fabricated from none ferrous alloy .Each one is subjected to one of the maximum- minimum stress cycles listed in the following table, the frequency is the same for all the three tests.. Which of the following samples will have the longest life and the shortest life Specimen Stress,max ( MPa) Stress , min (MPa) A +450 -150 B +300 -300 C +500 -200
Determine the indentation diameter and the surface area of indentation of a copper material subjected to a Brinell hardness test with a test force of 8.2 kN using a hardened steel ball indentor of 9 mm. Take the Brinell Hardness Number as 902. Solution: i) Indentation Diameter (in mm) = ii) Surface Area of Indentation (in mm^2) =