4. Based on Figure 1 shows, for a carbon steel, the partial tensile engineering stress-straincurve. Determine;(i)Yield strength (o)(ii)Ultimate Tensile Strength (oTs)(ii)Young Modulus (E)300200010° psi300MPa2000200200100010001001000.0000.0050.010Strain0.0150.0000.0200.0400.0600.080StrainFigure 1: Engineering stress-strain curve for alloy steelStress (MPa)StressStress (10° psi)

Answered: Image /qna-images/answer/05692b54-72cd-4755-ba97-e1fc3ab9b294.jpg

Answered: 4. Based on Figure 1 shows, 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

Problem 1.1MA

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Draw the tensile curves of the following materials, taking into account the appropriate elongation at break, modulus of elasticity and tensile stress, on the side stress-strain graph. a)Low carbon steel (Ecelik=200Gpa),(Syield=300MPa),(Stensile=400MPa) b)High carbon steel (Esteel=200GPa),(Stensile=700MPa) c)Aluminum (Ealuminum=70GPa), (Syield=200MPa),(Stensile=300MPa) d)Cast iron (Ecast iron=120GPa),(Stensile=200MPa)
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
a) Compression tests are generally performed on brittle materials-why?b) How a cast iron specimen fails under compression load?c) Define secant modulus?
A mild steel bar went through a tensile test conducted at the University of San Carlos Mechanical Engineering Laboratory. The following data was obtained from the test: i. Diameter of the steel bar = 30 mm ii. Gauge length of the bar = 200 mm iii. Load at the elastic limit = 250 kN iv. Extension at a load of 150 kN = 0.21 mm v. Maximum load = 380 kN vi. Total extension = 60 mm vii. Diameter of the rod at the failure = 22.5 mm Determine the Young’s modulus, the stress at the elastic limit, the percentage elongation, and the percentage decrease in area.
Analyse the differences between the 2 attached stress-strain curves for 0.9% carbon steel and 0.15% carbon steel.
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.
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
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 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 mild steel tensile specimen of initial length 44 mm and initial diameter 6.4 mm is subjected to a tensile test and the following data are obtained. - Yield Strength as 88 MPa - Maximum Strength as 212 MPa - Fracture Strength as 152 MPa - Percentage of Elongation as 63 % - Percentage of Reduction in area as 39% Determine the Final length, Final area, Yield load, Ultimate load
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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