8-14. A tension test was performed on an aluminum 2014- T6 alloy specimen. The resulting stress-strain diagram is shown in the figure. Estimate (a) the proportional limit, (b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method.
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- 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 STRESSA ul ? KBIS M 1:11 O photo_2021-0... -> 200 stress 150 (MPa) (00 50 0-2 strain From the tensile stress-strain behavior for the brass specimen d the diamcter is 12-8 mm Calculate the folluding:- 1- The Yield stress. 2- Modulus of elasticity. 3 - The Tensile strength . IIStress 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…
- Q-3 The following creep data were taken on an aluminum alloy at 400°C (750°F) and a constant stress of 25 MPa (3660 psi). Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Time Time (min) Strain (min) Strain 0.00 18 0.82 0.22 20 0.88 4 0.34 22 0.95 0.41 24 1.03 8 0.48 26 1.12 10 0.55 28 1.22 12 0.62 30 1.36 14 0.68 32 1.53 16 0.75 34 1.77200 stress 150 (MPa) 100 50 0.1 0-2 strain From the tensile stress-strain behavior for the brass specimen the diameter is 12-8 mm Calcalate the folluwing:- %3B 1- The Yield stress. 2-Modulus of elasticity. 3 - The Tensile strength.. From the tensile stress-strain behaviour for a casted iron as shown in the following curve. Please answer the following questions (1) Estimate the modulus of elasticity, (2) Calculate the change in length of a specimen with original length of 100mm that is subjected to a tensile stress of 300MPa. (Hint: Using the Hooke's law to get the strain.) 600 500 400 - 300 - 200 - 100 - 2 4 6. 8 10 12 14 16 Strain Strength (MPa)
- Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025Tensile test specimens are extracted from the "X" and "y" directions of a rolled sheet of metal. "x" is the rolling direction, "y" is transverse to the rolling direction, and "z" is in the thickness direction. Both specimens were pulled to a longitudinal strain = 0.15 strain. For the sample in the x-direction, the width strain was measured to be ew= -0.0923 at that instant. For the sample in the y-direction, the width strain was measured to be gw=-0.1000 at that instant. The yield strength of the x-direction specimen was 50 kpsi and the yield strength of the y-direction specimen was 52 kpsi. Determine the strain ratio for the x direction tensile test specimen. Determine the strain ratio for the y-direction tensile test specimen. Determine the expected yield strength in the z-direction. Give your answer in units of kpsi (just the number). If the sheet is plastically deformed in equal biaxial tension (a, = 0, to the point where & = 0.15, calculate the strain, 6, that would be expected.A tension test was performed on an aluminum 2014- T6 alloy specimen. The resulting stress-strain diagram is shown in the figure. Estimate (a) the proportional limit, (b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method. a (MPa) 420 350 280 210 140 70 (mm/mm) 0.06 0.02 0.002 0.08 0.008 004 0.10 0.004 0.006 0.010
- Sketch a representative stress-strain digrama for a ductile material showing the following on it (label the axis and provid units). 1- YIELD STRENGTH. 2- TENSILE STRENGTH. 3- FRACTURE POINT. 4- HARDENING PART. 5- SOFTENING PART. 6- A METHOD TO CALCULATE THE STIFFNESS. 7- A METHOD TO CALCULATE THE DUCTILITY.Part A The stress-strain diagram for a steel alloy having an original diameter of 0.40 in. and a gage length of 9 in. is shown in the figure below. (Figure 1) Determine the modulus of elasticity for the material. Express your answer to three significant figures and include appropriate units. HA ? Eapprox = Value Units Submit Request Answer Figure < 1 of 1 Part B a (ksi) 80 Determine the load on the specimen that causes yielding. 70 Express your answer to three significant figures and include appropriate units. 60 50 40 HA 30 20 Py = Value Units 10 € (in./in.) 0.04 0,08 0.12 0,16 0,20 0,24 0,28 O 0.0005 0.001 0.0015 0.002 0.0025 0.0030.0035 Request Answer SubmitIn our previous topic regarding strain, repeat the problem on the strain using thefollowing parameters:F = 150000 +/- 150 lbfSide length = 1.2 +/- 0.025 inModulus of Elasticity = 25000 +/- 11.5 ksiDetermine the relative error. Is it above or below the accepted 5% error?