TutorialMechanical properties1. Based on data collected as Table 1, plot graph Stress vs strain, determine YS, TS and E.|Table 1: Data load and length for tensile testGage length(mm)LoadGage lengthLoad(N)(mm)(N)35,22035,72040,54050.800050.977850.810250.82034,89051.00329,77951.816014,67050.830548,39053.340019,56024,45027,62029,39032,68033.95034,58050.840659,03065,87069,42069,670 (maximum)55.880050.850850.861058.420060.960050.871161.468050.901668,15063.500050.927060,810 (fracture)66.0400 (after fracture)50.9524Original, specimen diameter: 12.7 mm.

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Answered: 1. Based on data collected as Table 1,…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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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.
Draw a typical stress vs strain tensile test curve for the following materials (two seperate graphs) and label the axis. A ductile metallic test specimen that is stretched to failure displaying a characteristic yield point and show the following parts on the curve. 1- Yield point 2- Ultimate Tensile Strength 3- Breaking point 4- Elastic Region 5- Plastic Region 6- Necking region
Was there a discrepancy between the values of Young's Modulus found in the tensile test and the bending test? If so, why does this occur?
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 tensile test specimen has a starting gage length 50 mm and a cross-sectional area 200 mm2. During the test, the specimen yields under a load of 30,000 N (this is the 0.2% offset) at a gage length of 52 mm.The maximum load of 63,000 N is reached at a gage length of 57 mm just before necking begins. Final fracture occurs at a gage length of 63.5 mm. Determine (a) yield strength, (b) modulus of elasticity, (c) tensile strength, (d) engineering strain at maximum load, and (e) percent elongation.
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.
For the following inspection opening, calculate the correct values for A, A1, A2, A3, A4, A5 The design pressure is 900 kPa.g and the design strengths for the metals of shell and branch are 133MPa A= a. 3545.24 b. 3354.63 c. 3454.36 d. 3445.42 A1= a. 1377.45 b. 1673.71 c. 1773.82 d. 1534.53 A2= a. 721.63 b. 733.32 c. 718.59 d. 738.93 A3 = a. 600 b. 500 c. 450 d. 550 A4 = a. 211 b. 214 c. 220 d. 216 A5= a. 346 b. 356 C. 366 d. 336
A) An engineer works for a company that has many gray cast iron rods of different diameters and would like to design an equivalent rod that could handle a force of P= 10.0 kipkip without fracturing. What is the minimum diameter that this new rod must have to support P? B)A tensile test is being conducted on a steel-rod specimen with a gauge length of L0 = 4.0 in and an initial diameter of d0 = 0.75 in. If the final length of the rod at fracture is Lf= 5.53 in , find the percent elongation of the rod at fracture.
Suppose Polymethylmethacrylate (PMMA) and Polydimethylsiloxane (PDMS) have undergone tensile tests of similar settings at normal temperature and pressure conditions. With aid of a sketch, contrast their stress-strain responses.
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.
What is self-heating in strain gauges? Why is it caused and what are its implications on end result?
How is the nominal or engineering strain found directly from the strain gage?

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