A rod made of aluminum alloy, with a ga and yield strength of 150 MPa, was subj gauge length was changed to 100.1 mm mm, choose the nearest modulus of elas material. OE-74.5 Mpa; Nu = 0.43 OE 130 Mpa; Nu = 0.23 OE-74.5 GPa; Nu = 0.33 E = 11.5 Mpa; Nu = 0.49
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- The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular cross section of 0.2011in.2 in area and a gage length (the length over which the elongation is measured) of 2.000 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Determine the modulus of elasticity from the slope of the linear portion of the curve. d. Estimate the value of the proportional limit. e. Use the 0.2 offset method to determine the yield stress.The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen with a circular cross section. The diameter was 3 8 inch and the gage length (The length over which the elongation is measured) was 2 inches. a. Use the data in Table 1.5.2 to produce a table of stress and strain values. b. Plot the stress-strain data and draw a best-fit curve. c. Compute the, modulus of elasticity from the initial slope of the curve. d. Estimate the yield stress.An aluminum alloy bar with a rectangular cross section that has a width of 12.5 mm, thickness of 6.25 mm, and a gauge length of 50 mm was tested in tension to fracture according to ASTM E-8 method. The load and deformation data were as shown in Table P4.6. Using a spreadsheet program, obtain the following: a. A plot of the stress-strain relationship. Label the axes and show units. b. A plot of the linear portion of the stress-strain relationship. Determine the modulus of elasticity using the best fit approach. c. Proportional limit. d. Yield stress at an offset strain of 0.002 m/m. e. Tangent modulus at a stress of 450 MPa. f. Secant modulus at a stress of 450 MPa. TABLE P4.6 Load (kN) AL (mm) Load (kN) AL (mm) 33.5 1.486 3.3 0.025 35.3 2.189 14.0 0.115 37.8 3.390 25.0 0.220 39.8 4.829 29.0 0.406 40.8 5.961 30.6 0.705 41.6 7.386 31.7 0.981 41.2 8.047 32.7 1.245
- To stretch a 200 mm long tensile test specimen (a) of steel elastically by 0.08 mm. How many loads do I need to apply? (b) What is the maximum length of this sample without plastic deformation? How long can it be extended? (The modulus of elasticity of steel is 210 GPa, yield strength 580 MPa, tensile strength 920 MPa specified as.(b) (i) A tensile test specimen made from 0.4% C steel has a circular cross section of diameter d mm and a gauge length of 25 mm. When a load of 4500 N is applied during the test, the gauge length of the specimen extends to 25.02 mm. If the Young's Modulus of the steel is 199 GPa, calculate the diameter of the tensile test specimen used. 4An aluminum alloy cylinder with a diameter of 76 mm and a height of 150 mm. is subjected to a compressive load of 220 kN. Assume that the mate-rial is within the elastic region and a modulus of elasticity of 75 GPa. a. What will be the lateral strain if Poisson’s ratio is 0.33?b. What will be the diameter after load application?c. What will be the height after load application?
- Sample: Malleable Steel (AISI 4145) Original Diameter: 6.14mm Gauge Length: 55 mm Final Length: 68.12 mm Final Diameter: 3.54mm Load (kN) Deformation (mm) Stress (MPa) 12.95 Strain (%) 0.3838 0.54 0.7841 0.95 26.48 1.3899 1.63 1.9485 2.07 3.3090 3.10 4.5821 4.22 5.9359 5.03 7.0340 5.51 8.2413 6.22 10.9446 7.22 13.1951 8.18 12.8228 8.77 12.2583 9.11 12.5915 9.61 13.2536 10.33 13.6636 10.71 13.9772 11.35 14.5433 12.63 15.1155 13.88 15.4970 15.10 15.6484 17.09 15.4031 17.79 14.7655 18.73 13.6721 19.28 10.4617 19.88A round aluminum alloy bar with a 0.25-in. diameter and a 1-in. gauge length was tested in tension to fracture according to ASTM E-8 method. The load and deformation data were as shown in Table P4.8.Using a spreadsheet program, obtain the following: a. A plot of the stress–strain relationship. Label the axes and show units. b. A plot of the linear portion of the stress–strain relationship. Determine modulus of elasticity using the best fit approach. c. Proportional limit. d. Yield stress at an offset strain of 0.002 in/in. e. Initial tangent modulus. f. If the specimen is loaded to 3200 lb only and then unloaded, what is the permanent change in gauge length? g. When the applied load was 1239 lb, the diameter was measured as 0.249814 in. Determine Poisson’s ratio.A 1-cm-diameter, 30-cm-long titanium bar has a yield strength of 345 MPa, a modulus of elasticity of 110.3 * 10^3 MPa, and Poisson's ratio of 0.30. Determine the length and diameter of the bar when 2224N load is applied.
- Testing a round steel alloy bar with a diameter of 15 mm and a gauge length of 250 mm produced the stress-strain relationship shown in Figure P3.28. Determine a. the elastic modulus b. the proportional limit c. the yield strength at a strain offset of 0.002 d. the tensile strength e. the magnitude of the load required to produce an increase in length of 0.38 mm f. the final deformation, if the specimen is unloaded after being strained by the amount specified in (e) g. In designing a typical structure made of this material, would you expect the stress applied in (e) reasonable? Why? 750 500 250 0.01 0.02 0.03 0.04 Strain, m/m FIGURE P3.28 Stress, MPaThe 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.Please present FBD/s A tensile test is performed on a brass specimen 10mm in diameter using a gage lenght of 50mm. When the tensile load P reaches a valule of 20kN, the distance between the gage marks has increased by 0.122mm. What is the modulus of elasticitiy of the brass?