The state of plane stress shown where x = 6.8 ksi will occur at a critical point in an aluminum casting that is made of an alloy for which OUT=10 ksi and σUC= 25 ksi. Using Mohr's criterion, determine the shearing stress To for which failure should be expected. (Round the final answer to two decimal places.) The shearing stress To for which failure should be expected is ±[ ksi.
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- 3) An bronze rod is rigidly attached between a aluminum rod and a steel rod as shown. Axial loads are applied at the positions indicated. Find the maximum allowable value of P that will not exceed a stress in steel of 140MPA, in aluminum of 90MPA or in bronze of 100MPа. Aluminum A = 500mm? Bronze A = 200mm? Steel A = 150mm2 2P ЗР L st = 1.2m L br = 2m Lal = 3.5mAs shown, an aluminium alloy construction BCD with a circular cross section is fixed at end B and affected by a force of 150 N at the free end D. The diameter of the cross-section a-a is 20 mm. The yield strength of the material is 80 MPa: a) Determine the stresses at point A of the a-a cross-section. As indicated in the picture, draw the stress element in Cartesian coordinates and specify the stress values.(b) Calculate the factor of safety, n for Tresca, and the von Mises yield criterion to see if the structure would yield based on the stresses at point A.(c) In the major stress area, draw the yield loci of both criteria and indicate the operational stress state & why is the Rankine failure criterion inappropriate for aluminium alloys?A steel component is subjected to alternate cyclical loading. The steel follows Basquin's law for high cycle fatigue, o, x N = C, (where the stress amplitude is in MPa). Ignore the geometric detail and assume that Marin's modifying factors are all equal to 1. You are given the minimum stress ain = -213 MPa, the maximum stress omax = 213 MPa. The material data are Tensile strength oUTS = 539 MPa, Basquin's constant c, = 875 MPa, Basquin's exponent a = 0.085. a) Calculate the stress ratio R, the stress amplitude o, in MPa and the mean stress am in MPa. The answers are acceptable with a tolerance of 0.01 for R and of 1 MPa the stresses. R: MPa MPа b) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1 106 cycles) N :
- 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 = 0.3The state of plane stress shown occurs at a critical point of a steel machine component. As a result of several tensile tests, it has been found that the tensile yield strength is o, = 250 MPa for the grade of steel used. Determine the factor of safety with respect to yield using (a) the maximum shearing stress criterion, and (b) the maximum distortion energy criterion. ↑y 40 40 MPa 80 MPa 25 MPa ➜XSITUATION. As shown in the figure below, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assume that the rods were initially stress-free. Allowable stress in steel is 120 MPa and in bronze is 60 MPa. For this problem, a = 1.8m; b = 1.3m; c= 1.7m. a b I Steel: A 900 mm Bronze: A E 200 GPa E 83 GPa L-2.0m L=1.5m What is the value of P without exceeding the allowable stress of steel, in KN? 230 250 280 340 1200 mm
- SITUATION. As shown in the figure below, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assume that the rods were initially stress-free. Allowable stress in steel is 120 MPa and in bronze is 60 MPa. For this problem, a = 1.8m; b = 1.3m; c = 1.7m. b C NANTSELEINE Steel: A = 900 mm² E = 200 GPa L=1.5m Bronze: A=1200 mm² E=83 GPa L=2.0m What is the ratio of the elongation of steel to bronze? O 0.367 O 0.433 O 0.467 0.333Problem 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?An aluminium rod is rigidly attached between a steel rod and a bronze rod as shown in Figure 3. Axial loads are applied at the positions indicated. (a) Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminium of 90 MPa, or in bronze of 100 MPa. (b) Using the answer from question (a), determine the total elongation of the rods. Steel Aluminium A = 500 mm? Bronze A = 400 mm? E = 200 GPa A = 200 mm? F = 70 GPa E = 83 GPa 4P 2P 2.5 m 2.0 m 1.5 m Figure 3
- SITUATION. As shown in the figure below, a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assume that the rods were initially stress-free. Allowable stress in steel is 120 MPa and in bronze is 60 MPa. For this problem, a = 1.8m; b = 1.3m; c= 1.7m. a b C P Steel: A-900 mm² Bronze: A=1200 mm² E = 83 GPa L=2.0 m E = 200 GPa L=1.5m What is the value of P without exceeding the allowable stress of bronze, in kN? O 161 O 174 O 145 124 4A load applied to a machine component results in the state of plane stress ?x=80 MPa, ?y=100 MPa, ?xy=60 MPa. The component is made of a brittle high-strength steel that follows the maximum normal stress criterion with ?u=200 MPa. If increasing the load increases each stress component proportionally, determine the percentage increase that can be applied before the component fails.An aluminum alloy [E = 73 GPa; v = 0.33; a= 23.0 x 10-6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 250 mm, a cross-sectional area of A = 6900 mm², and a length of L = 5.9 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 AT = 50°C, the longitudinal normal strain in the plate is found to be 2400 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Ad. L Answer: (a) P = i (b) Δd = KN mm