And **The modulus of elasticity** (MPa)I need a clear step by step answer please :) 16.6815.0213.3511.6910.028.356.695.023.351.690.021.69713.5428.465Strain (6)3.3895.0816.77310, 15711.85015.23416.920According to the engineering stress strain curve above.Determine the offset strength oo.005 (in MPa):Answer:Stress (Mpa)

Stress is the ratio of force which is applied on a material to the cross sectional area of the…

Engineering

Mechanical Engineering

And **The modulus of elasticity** (MPa) I need a clear step by step answer please :)

And The modulus of elasticity (MPa) I need a clear step by step answer please :)

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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(SI Units) In a tensile test on a steel specimen, true strain = 0.11 at a stress of 245 MPa. When true stress = 340 MPa, true strain = 0.31. Determine the strength coefficient and the strain-hardening exponent in the flow curve equation.
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
For some metal alloy, a true stress 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 418 MPa (60630 psi) is applied if the original length is 500 mm (19.69in) ? Assume a value of 0.22 for the strain- hardening exponent, n
Annealed low-carbon steel has a flow curve with strength coefficient = 80,000 lb/in2 and strain-hardening exponent = 0.25. A tensile test specimen with gage length = 2.0 in is stretched to a length = 3.5 in. Determine the flow stress and average flow stress that the metal experienced during this deformation. use correct stratgy and correct E 0.55962
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)
To determine the nominal or engineering stress and strain experienced by a specimen of a material while it is subjected to a tension test, and to be able to read important values from a conventional stress-strain diagram obtained from the test.A tension test is being conducted on a steel-rod specimen with a gauge length of L0=2 in and initial diameter of d0=0.5 in. Data were collected to form the conventional stress-strain diagram as shown. From the diagram, f = 73.0 ksi , e = 101.0 ksi g=83.0ksi, and h=0.15in/in Part A - Nominal or engineering strain in the rod Assuming that the strain remains constant throughout the region between the gauge points, determine the nominal strain ε experienced by the rod if it is elongated to L = 2.5 in . Express the nominal strain in inches per inch to three significant figures. Part B - Nominal or engineering stress in the specimen Assuming that the stress is constant over the cross-sectional area and if the tension force used is P = 8.0 kips ,…
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.
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The following data were obtained from the tensile test of Aluminum alloy. The initial diameter of testspecimen was 0.505 inch and gauge length was 2.0 inch. Plot the stress strain diagram and determine(a) Proportional Limit (b) Modulus of Elasticity (c) Yield Stress at 0.2% offset (d) Ultimate Stress and(e) Nominal Rupture Stress.
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22) In stress and strain diagram for a perfectly ductile material, the yield strength, and ultimate tensile strength are well defined Select one: True False