What change in volume would a 100 mm cube of steel suffer at a depth of 5 km in sea water? Take E=2.05 x 10 5N/mm2 and N = 0.82 x 10 5N/mm2
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What change in volume would a 100 mm cube of steel suffer at a depth of 5 km in sea water?
Take E=2.05 x 10 5N/mm2 and
N = 0.82 x 10 5N/mm2
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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.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.A tensile test was performed on a metal specimen having a circular cross section with a diameter of 1 2 inch. The gage length (the length over which the elongation is measured) is 2 inches. For a load 13.5 kips, the elongation was 4.6610 3 inches. If the load is assumed to be within the linear elastic rang: of the material, determine the modulus of elasticity.
- What change in volume would a 100 mm cube of steel suffer at a depth of 5 km in sea water? Take E=2.05 x 10 5N/mm2 and N = 0.82 x 10 5N/mm2Note: Modulus of elasticity of steel is 2.0x106 kg/cm2Actual weight of steel bars = 1.520 kg/m, Standard weight of steel bars (16mm )= 1.579 kg/m, Load at yield point = 3480 kgf ,Load at breaking point = 7540 kgf, Maximum load = 7. 9260 kgf, Unit weight of steel = 7854 kg/m^3. What is the cross-sectional area in m^2? Compute it's yield strength,ultimate strength and breaking strength in MPA?
- A 4 m long steel plate having a cross section of 27 mm by 368 mm is use as a hanger and subjected to a tensile load of 2476 kN. The proportional limit of the steel is assumed to be 200 MPa. If the Poisson’s ratio is 0.87, determine the dimensional change along 27 mm.Determine the maximum tensile strength of the steel bar described by the following data Initial diameter = 12 mm Final diameter = 6.5 mm Initial length = 60 cm Final length = 86 cm Ultimate load = 5000 kgfUse modulus of elasticity 30000000 psi
- Calculate the plane stresses involved where two pieces of steel are welded together. Our sign convention defines forces causing the object to be in tension positive and forces causing the object to be in compression negative. As we can see from the picture our σx = −4.5MPa and σy = 10MPa. Steel has an average Young’s modulus of elasticity of E = 200 GPa and an average Poisson’s ratio of ν = .285 Because it is not defined we will assume that initially there is no shear stress, τxy = 0. Solve only for: Compression in the x-direction(deformation along the weld)=(-Blank 4)? x 10^-5One of the following values is a possible for possion ratio for steel ? a. Zero b. one c. 0.3 d 0.9The rectangular block shown in Figure is subjected to tension within the elastic range. The increase in the length of a is 2 * 10-3in. and the contraction of b is 3.25 * 10-4 in. If the original lengths of a and b were 2 in. and 1 in., respectively, what is Poisson’s ratio for the material of the specimen?