8-21. The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. When the applied load on the specimen is 50 kN, the diameter is 12.67494 mm. Determine Poisson's ratio for the material. (MPa) 490 (mm/mm) 0.007
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- An element of material in plain strain is subjected to strains x = 0.0015, , y . = -0.0002, and xy = 0.0003. (a) Determine the strains for an element oriented at an angle = 20°. (b) Determine the principal strains of the element. Confirm the solution using Mohr’s circle for plane strain.An element of material in plain strain has the following strains: x = 0.001 and y = 0.0015. (a) Determine the strains for an element oriented at an angle = 250. (b) Find the principal strains of the element. Confirm the solution using Mohr’s circle for plane strain.The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm andagagelengthof50.8mm.IfaloadofP=60kNisapplied to the specimen, determine its new diameter and length. Take n = 0.35.
- Determine the elastic tensile load "F" that acts on mild steel specimen of Diameter 5 mm & Modulus of Elasticity 207 GPa. It is found that the specimen has undergone an extension of 1.2 mm due to the elastic load "F". Also, determine the length "L" of the specimen, if the strain-induced due to the load "F" on the specimen is 3 x 10-³. Solution Cross-sectional Area, A (in mm²) = Tensile Stress, (in N/mm²) = Tensile Load, (in N) = Length of the Specimen (in mm)The stress-strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 60 ksi, determine the approximate amount of elastic recovery and the increase in the gage length after it is unloaded.Tensile 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.
- The stress-strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support. o (ksi) 80 70 60 50 40 30 20 10 e (in./in.) 0 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0 .0.0005 0.0010.0015 0.002 0.0025 0.0030.0035The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 70 ksi, determine the approximate amount of elasticrecovery and the increase in the gage length after it is unloaded.Determine the elastic tensile load "F" that acts on mild steel specimen of Diameter 9 mm & Modulus of Elasticity 201 GPa. It is found that the specimen has undergone an extension of 1 mm due to the elastic load "F". Also, determine the length "L" of the specimen, if the strain- induced due to the load "F" on the specimen is 3.6 x 10-³. Solution Cross-sectional Area, A (in mm²) = 63.617 One possible correct answer is: 63.617251235193 Tensile Stress, (in N/mm²) 723.6 One possible correct answer is: 723.6 Tensile Load, (in N) = = 46033.261 One possible correct answer is: 46033.442993786 Length of the Specimen (in mm) 4.366 X One possible correct answer is: 277.77777777778
- The stress–strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material.The stress strain curve determined from a sample of material is shown in the figure. The original diameter was 14.3 mm and it changed to 14.28858 when a 43.5 kN load was applied. Calculate Poisson's ratio for the sample. o (MPa) 550 400 0.00275 Poisson's ratio v = 0.0Determine the elastic tensile load "F" that acts on mild steel specimen of Diameter 8 mm & Modulus of Elasticity 201 GPa. It is found that the specimen has undergone an extension of 1.6 mm due to the elastic load "F". Also, determine the length "L" of the specimen, if the strain-induced due to the load "F" on the specimen is 4.1 x 10-3. Solution Cross-sectional Area, A (in mm²) Tensile Load, F (in N) Length of the Specimen (in mm)