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- Results of a tensile test of a 0.505-in. diameter aluminum alloy test bar are shown below. Initial length = 2.0 in. The proportional limit occurred with a load = 7000 lbs. What is the length after deformation of another bar (same aluminum alloy material) of initial length of 50 in. at a stress level of 30,000 psi? Load (Ib) (A) (in.) 0.000 1000 0.001 3000 0.003 5000 0.005 7000 0.007 7500 0.030 7900 0.080 8000 (maximum load) 0.120 7950 0.160 7600 (fracture) 0.205 O A. 50.1 in. O B. 50.3 in. O C.50.15 in. O D. 51.5 in.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.From the shear stress - shear strain diagram shown below, all the following are true except: Shear Stress (MPa) 500 450 400 350 300 250 200 150 100 50 0. 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 Shear Strain (rad) Shear Stress-Strain Curve for Brass Select one: O a. Shear stress and Shear strain are 430 MPa and 1.72 rad at fracture. O b. Shear stress and Shear strain are 250 MPa and 0.05 rad at Yielding. Oc. The elastic zone ends at a shear stress of 400 MPa and shear strain of 1 rad Od. The ultimate Shear stress and Shear strain are 430 MPa and 1.72 rad.
- 5. The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy (initial length lo= 2.0 in.). After fracture, the total length was 3.014 in. and the diameter was 0.374 in. Load (Ib) Al (in.) 00000 3,000 6,000 0.00167 0.00333 7,500 0.00417 9,000 10,500 0.0090 0.040 12,000 0.26 12,400 11,400 0.50 (maximum load) 1.02 (fracture) a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity. c) Determine the yield strength at a strain offset of 0.002. d) Determine the tensile strength of this alloy. e) What is the approximate ductility, in percent elongation? f) Compute the modulus of resilience. g) Compute from the data and plot true stress versus true strain diagram.A 11 in. inner diameter, 0.35 " wall thickness pipe is under a pressure of 2.5 ksi where strain gages installed along axial and circumferential directions register strains of 180 and 900 micro-strains (x10^-6), respectively. A) What is the Poisson's Ratio of this material and it's Elastic Modulus? B) In a uniaxial test, the pipe's material is observed to yield at a longitudinal strain of 0.1 in/in. Assuming a factor of safety of 2, the pipe can withstand impact energy of _______ lb - in per foot without suffering permanent deformation. C) If the pipe is depressurized and then subjected to a torque of 50 lblb - ft.ft., it will experience a shear strain of ________ rad.A 11 in. inner diameter, 0.35 " wall thickness pipe is under a pressure of 2.5 ksi where strain gages installed along axial and circumferential directions register strains of 180 and 900 micro-strains (x10^-6), respectively. A) What is the Poisson's Ratio of this material and it's Elastic Modulus? B) In a uniaxial test, the pipe's material is observed to yield at a longitudinal strain of 0.1 in/in.in/in. Assuming a factor of safety of 2, the pipe can withstand impact energy of _______ lblb - in.in. per foot without suffering permanent deformation. C) If the pipe is depressurized and then subjected to a torque of 50 lblb - ft.ft., it will experience a shear strain of ________ rad.
- 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? BI1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of ResilienceTable B2: Stress-strain data for uniaxial compression test on Sample Normal stress Uniaxial strain (%) Lateral strain (%) (MPa) 0.0 0.0000 0.0000 5.0 0.0319 -0.0150 10.0 0.0720 -0.0275 15.0 0.1025 -0.0425 20.0 0.1450 -0.0600 25.0 0.1755 -0.0725 30.0 0.2150 -0.0875 35.0 0.2455 -0.1025 41.0 0.2815 -0.1125 46.0 0.3125 -0.1209 52.0 0.3515 -0.1285 58.0 0.4000 -0.1325 (f) Based on the plotted curve, calculate the average/tangent Young's modulus and Poisson's ratio for this sample, at 50 % UCS?
- 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 1.994Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…Problem :- The following data was obtained tensile test on a specimen of 6 mm diameter and gauge length Lo=30 mm: load(KN) 0 40 80 95 90 135 150.2 145 130 120 elongation(mm) 0 0.022 0.13 0.25 0.55 2.01 3.88 4.12 5.22 6 Plot engineering and True (stress-strain) curves, then determine the (oy, ɛy), (ou, ɛu), (of, , ef), ductility, modulus of elasticity. Then select on the drawing ((strain hardening (n)) , (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region)?