A tensile force of 5 kN produce an elastic reduction in diameter of 4 x 10-4 mm on a metal alloy specimen that has a diameter of 6 mm. Determine: *. The modulus of elasticity of the alloy (Unit: GPa) !. The Poisson' s ratio of this alloy
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- An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0×10-6/°C] bar is subjected to a tensile load P. The bar has a depth of d = 260 mm, a cross-sectional area of A = 14720 mm2, and a length of L = 5.5 m. The initial longitudinal normal strain in the bar is zero. After load P is applied and the temperature of the bar has been increased by AT = 46°C, the longitudinal normal strain is found to be 1680 µɛ. % D Calculate the change in bar depth d after the load P has been applied and the temperature has been increased. L P Answer: Ad = i mmThe following information about the o-e curve is given for a steel alloy. E = 0.001527 at o = 300 MPa and ɛ = 0.003054 for o = 600 MPa. (a) Draw the stress-strain diagram and calculate the E for this alloy.Problem 1 An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Aluminum Steel A=800mm2 A=1000mm2 Bronze A=700mm? 4P 2P 500mm 600mm 700mm 1. What is the maximum value of P that will not exceed the axial stress bronze of 105 MPa? 2. What is the maximum value of P that will not exceed the axial stress in aluminum of 90 MPa? 3. If P=10KN, what is the axial force to be carried by the aluminum in KN? 4. If P is 5KN, what is the axial stress of steel?
- Q1/ Consider the brass alloy for which the stress-strain behavior is shown in the figure below. A cylindrical specimen of this material 10.0 mm in diameter and 101.6 mm long is pulled in tension with a force of 10,000 N. If it is known that this alloy has a value for Poisson's ratio of 0.35, compute (a) the specimen elongation, and (b) the reduction in specimen diameter. 500 70 Tensile strength 450 MPa (65,000 psi) 60 400 50 10 psi 300 MPa 40 40 30 200 Yield strength 30 200 250 MPa (36,000 psi) 20 100 20 10 100 10 0.005 0.10 0.20 0.30 0.40 Strain Stress (MPa) Stress (10 psi)S Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion tests until fracture. Determine the following for the two metals (show all calculations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 m/m. c. Ultimate strength d. Modulus of resilience e. Toughness f. Which metal is more ductile? Why? 900 Metal A 600 Metal B 300 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Strain, m/m FIGURE P1.16 Stress, MPaAn aluminum alloy rod has a circular cross section with a diameter of 7mm. 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 0.38, what will be the lateral strain? E=6/8 v= -€(lateral)/e(axial) V ominorbooomes
- The following figure shows the tensile stress-strain curve for a plain-carbon steel. 600 80 500 MPa 600 10 psi 60 80 400 400 60- 300 40 40 200 200 20 20 100 0.005 0.05 0.10 0.15 Strain (a) What is this alloy's tensile strength? MPа (b) What is its modulus of elasticity? GPa (c) What is the yield strength? i MPa Stress (MPa) Stress (10 psi)Question No.2 Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion tests until fracture. Determine the following for the two metals (show all calculations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 m/m. c. Ultimate strength d. Modulus of resilience e. Toughness I. Which metal is more ductile? Why? 900 -Metal A E 600 Metal B 300 0.00 a02 004 a.06 0.08 0.10 0.12 014 Strain, matm FIGURE P1.16 Strees, MPaCalculate the stress and strain at each force interval. Plot a graph of the stress-strain curve. Estimate the yield point of the steel and note its location on the curve. Estimate the ultimate strength of the steel and note its location on the curve.
- Strain = 600 Stress = Strain = 500 Stress = 400 500 300 400 300 200 200 100 100 0.000 0.002 0.004 0.006 Strain 0.00 0.04 0.08 0.12 0.16 0.20 Strain Stress (MPa)An aluminum alloy [E = 67 GPa; ν = 0.33; α = 23.0 × 10–6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 225 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 4.1 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by ΔT = 63°C, the longitudinal normal strain in the plate is found to be 2900 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Δd.3. The distribution of stress in an aluminum machine component is given (in megapascals) by Ox = y + z? Oy = x + z Oz = 3x + y Txy = 3z2 Tyz = x Txz = %3D Calculate the state of strain at a point positioned at (1,2,4). Use E=70 GPa and v = 0.3