I need the answer as soon as possible Q3) The figure below simplified a load-elongation diagram of a tensile test for low carbonsteel test specimen with (10 mm) diameter and (60 mm) gauge length, determine:i.ii. Ultimate stress.Yield stress.iii. Fracture stress.1601401201008060402026.Elongation (mm)101214(N) peon

Solution: Yield point is the point on the stress vs strain or the load vs elongation curve where a…

Answered: Q3) The figure below simplified a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A tensile test was performed on a metal specimen with a diameter of 1/2 inch and a gage length (the length over which the elongation is meas- ured) of 4 inches. The data were plotted on a load-displacement graph, P vs. AL. A best-fit line was drawn through the points, and the slope of the straight-line portion was calculated to be P/AL = 1392 kips/in. What is the modulus of elasticity? BI
Question A cylindrical specimen of metal having a diameter of 12.88 mm and a gauge length of 63.50 mm is tested using a tensile testing machine. The elongation measurement are recorded in Table 3. nPlot the stress-strain curve on the graph paper provided based on data in Table 3 Based on the stress-strain curved plotted in (i): Compute the modulus of elasticity Determine the yield strength at a strain offset of 0.002. Determine the tensile strength Determine the ductility in percent elongation and percent area reduction Table 3: Load-elongation readings for a metal specimen Jadual 3: Bacaan beban pemanjangan untuk statu spesimen logam) Load (N) Elongation (mm) 1380 0.03 2780 0.06 5630 0.12 7430 0.2 8140 0.25 9870 0.64 12850 1.91 14100 3.18 14340 4.45 13830 5.72 12500 6.99 Fracture
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
1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 0.0019 -0.0057 13.092 0.0762 21.819 0.1270 0.0095 30.547 32.729 0.1778 0.7620 34.911 3.0480 30.01 mm O F Rigid Plate Cylindrical Alloy -Steel casement Figure k Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.
The following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.25 inches and the gage length was 4 inches. The stress scale of the stress-strain diagram is given with the factor a = 10 ksi. Estimate: (a) The modulus of elasticity. (b) The ultimate strength. (c) The yield strength (0.2% offset). (d) The percent elongation at fracture. 2013 Michael Swanbom STRESS VS. STRAIN BY NC SA 7a bat Sat 2at at 0.05 STRAIN 0.01 0.04 0.06 0.08 0.02 0.03 0.07 0.09 STRESS
Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. What tension force is required to cause yielding in a bar of the same material but with a diameter of 20 mm? P, kN AL mm 7.22 0.0839 14.34 0.1636 21.06 26.8 31.7 34.1 35.0 0.241 0.308 0.380 0.484 0.614 36.0 0.924 36.5 1.279 36.9 37.2 1.622 1.994
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
The results of a tensile test are: Diameter of the specimen Gauge length Load at limit of Proportionality Extension at the limit of Proportionality Maximum Load : 10mm :40 mm : 70kN :0.08mm. :100 kN Calculate stress at limit of proportionality and young's modulus.
1.9 (A). During a tensile test on a specimen the following results were obtained: Load (kN) 50 55 30 0.094 40 0.127 Extension (mm) Load (kN) Extension (mm) 15 0.05 70 5.08 75 7.62 Diameter of gauge length = 19 mm Diameter at fracture = 16.49 mm 80 12.7 0.157 82 16.0 1.778 80 19.05 60 2.79 Gauge length 100 mm Gauge length at fracture = 121 mm 65 3.81 70 22.9 Plot the complete load extension graph and the straight line portion to an enlarged scale. Hence determine: (a) the modulus of elasticity; (d) the nominal stress at fracture; (b) the percentage elongation; (e) the actual stress at fracture; (c) the percentage reduction in area; (f) the tensile strength. [116 GN/m²; 21%; 24.7%i 247 MN/m²; 328 MN/m2; 289 MN/m²]
Question Following experimental data are obtained from tensile test of a rectangular test specimen with original thickness of 2,5 mm, gauge width of 24 mm and gauge length of 101 mm: Load (N) Elongation (mm) 0 0 24372 0,183 23008 0,315 28357 5,777 35517 12,315 27555 17,978 23750 23,865 Based on the information above; draw stress-strain diagram of the material and answer the following questions. - Calculate the yield strength (in MPa) of the material. - Calculate the percent elongation of the specimen at yield point. (Use at least five decimal units) - Calculate the stiffness (in MPa) of the specimen material. - Calculate the ultimate strength (in MPa) of the material. - Calculate the percent elongation of the specimen at point of ultimate strength.
Specimen (Rebar) Original Length Length when it was clamped Diameter (mm) Internal Diameter (mm) External 1 300 mm 163mm 10 15.030 2 300 mm 180 mm 10.020 14.386 3 300 mm 197mm 9.40 15.028 Dimension of Specimen Specimen A0 P max Tensile strength Modulus of elasaticity Percent Elongation 1 474 N 179.381 2.776 MPa 19.76 MPa 13.4% 2 475 N 159.070 2.895 MPa 18.74 MPa 16.67% 3 485 N 179.381 2.706 MPa 20.058 Mpa 13.4% Result of testing Observation What is the possible discussion of these tablesP.S THIS IS NOT GRADED, ITS JUST EXAMPLES
1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves were generated, both curves generated from the same set of data. Use the graphs to fill in the mechanical properties of the material tested in the box below. Don't forget units! Stress vs Strain Stress, psi Stress, psi 80000 70000 60000 50000 40000 30000 20000 10000 0 0.00 80000 70000 60000 50000 40000 30000 20000 10000 0.02 0 0.000 0.002 0.04 0.004 0.06 0.006 0.08 0.10 Strain Stress vs Strain 0.008 0.12 Elastic Modulus, E: 0.2% Offset Yield Strength, oo: Tensile Strength, ou: Breaking Strength, of: % Elongation: 0.14 0.010 0.012 0.014 Strain 0.16 0.18 0.016 0.018 0.20 0.020

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Q3) The figure below simplified a load-elongation diagram of a tensile test for low carbon steel test specimen with (10 mm) diameter and (60 mm) gauge length, determine: i. Yield stress. ii. Ultimate stress. iii. Fracture stress. 160 140 120 100 80 60 40 20 10 12 14 Elongation (mm) (Nx) peo