calculate the lateral strain and the change in diameter.
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- 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.Compare the engineering and true secant elastic moduli for the natural rubber in Example Problem 6.2 at an engineering strain of 6.0. Assume that the deformation is all elastic.An applied torque results in a shear strain (y) of 3.75 x 103 in a solid circular steel shaft. At this strain, a portion of the shaft cross- section yields as shown in Figure 2. Determine the magnitude of the applied torque. Take G = 80 GPa and yy = 2.8125 x %3D 10-3 FIGURE 2: Plastic zone 50 mm 200 mm Elastic core
- An aluminium specimen with an initial gauge diameter d, = 10 mm and gauge length, 1, = 100 mm is %3D subjected to tension test. A tensile force P= 50 kN is applied at the ends of the specimen as shown, resulting in an elongation of 1 mm in gauge length. The Poisson's ratio (µ) of the specimen is Take shear modulus of material, G = 25 GPa. Consider engineering stress-strain conditions. PA steel wire with a diameter of d = 1/16” is bent around a cylindrical drum with a radius ofR = 36”.(a) Determine the maximum normal strain in the wire.(b) What is the minimum acceptable radius of the drum if the maximum normal strain mustremain below yield? Consider an elastic modulus of 30000 ksi and yield stress of 100ksi.(c) If R = 36”, what is the maximum acceptable diameter of the wire if the maximum normalstrain must remain below yield?During a tension test, measurements for the applied load and the corresponding elongation are taken at frequent intervals. These data points are then used to: a. calculate the stress caused by the applied load at each data point. b. calculate the strain in the specimen induced by the applied load at each data point. c. plot a stress-strain diagram. d. all of the above
- I5 In a tensile test experiment of carbon steel. a standard specimen (D= 0.505 in. Gauge length=2.0 in & total length 8 in) had a 0.2% offset yield strength of 80.000 psi and engineering strain of 0.010 at the yield strength point Calculate: The load (force) at this point. The instantaneous area of the specimen at this point. c. The true stress at this point. d. The true strain at this point. e. The total length of the specimen. if the load is released at this point.The specimen shown is made from a 25 mm diameter cylindrical steel rod with two 38 mm outer-diameter sleeves bonded to the rod as shown. Knowing that E = 200 GPa, determine (a) the load P so that the total deformation is 0.05 mm, (b) the corresponding deformation of the central portion BC. 38 mm diameter P' 25 mm diameter B 38 mm diameter C 50 mm 75 mm 50 mmA steel bar, whose cross section is 0.60 inch by 4.10 inches, was tested in tension. An axial load of P = 31,025 lb. produced a deformation of 0.115 inch over a gauge length of 2.10 inches and a decrease of 0.0080 inch in the 0.60-inch thickness of the bar. a. Determine the lateral strain. b. Determine the axial strain. c. Determine the Poisson’s ratio v. d. Determine the decrease in the 4.05-in. cross-sectional dimension (in inches).
- A cylindrical specimen of 1018 steel having a diameter of 6.35 mm and a gauge length of38.1 mm is pulled in tension. Use the load-elongation characteristics tabulated below. Determine the yield strength (σy) at a strain offset of 0.002.For a given steel cube section subjected to three dimensional state of principal stress. 75N/mm² ŠI 150 N/mm 30 N/mm² Consider yield stress = 250 N/mm², and µ = 0.3 The value of factor of safety according to maximum strain energy theory. -X30. An aluminum alloy rod has a circular cross section with a diameter of 8mm. The rod is subjected to a tensile load of 5kN. Assume that the material is in the elastic region and E = 69 GPa. If Poisson's Ratio is 0.33, what will be the lateral strain? E= 6/8 v = -E(lateral)/(axial)