Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Failure Theories

SOLUTION (6.1D) Known: A poem appears in Chapter Six titled: \"The One-Hoss Shay\". Find: Write a one paragraph summary of the poem \"The One-Hoss Shay\". Analysis: A \"chaise\" should be designed and built in a logical way such that it has uniform strength throughout and after a hundred years, looks okay, feels okay, and shows little trace of age, has no loss of strength and the strength is uniform throughout, the same as before, and then breaks down and wears out \"all at once and nothing first\", and the modern version would require that the heap or mound (and the modern version would require that the heap or mound) be immediately ready for recycling). SOLUTION (6.2) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: 7075-T651 Aluminum Su = 78 ksi Sy = 70 ksi Kic = 60 ksi in^0.5 2w = 6 in. t = 0.035 in. c = 1 in. P = ? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. 6-1 Analysis: 1. From Eq. (6.2), sig = Kic = 60 / 1.8c = 47.14 ksi 2. Since the area equals 2wt, P = sig(2wt) = 47.14(6)(0.035) = 9,899 lb Comment: P/A stress based on the net area, t(2w - 2c), is 56.57 ksi which is less than Sy = 70 ksi; hence, the second assumption is satisfied. SOLUTION (6.3) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: D6AC Steel at -40 °F Su = 227 ksi Sy = 197 ksi Kic = 100 ksi in^0.5 2w = 6 in. t = 0.06 in. 2c = 1 in. P = ? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. Comment: P/A stress based on the net area, t(2w - 2c), is 94.28 ksi which is less than Sy = 197 ksi; hence, the second assumption is satisfied. SOLUTION (6.4) Known: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load. Find: Estimate the highest tensile load that the plate will support. Schematic and Given Data: D6AC Steel at room temperature Su = 220 ksi Sy = 190 ksi Kic = 115 ksi in^0.5 2w = 6 in. t = 0.06 in. 2c = 1 in. P = ? Assumptions: 1. The crack length is a small fraction of the plate width. 2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength. 3. Yielding has occurred within one small volume of the material at the crack root. 4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material. Analysis: 1. From Eq. (6.2), sig = Kic / 1.8c 1.8 = 90.353 ksi 2. Since the area equals 2wt, P = sig(2wt) = 90.353(6)(0.06) = 32,527 lb Comment: The P/A stress based on the net area, t(2w - 2c), is 108.4 ksi which is less than Sy = 190 ksi. Hence the second assumption is satisfied. SOLUTION (6.5)\nKnown: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load.\nFind: Estimate the highest tensile load that the plate will support.\nSchematic and Given Data:\n4340 Steel at room temperature\nS_u = 260 ksi\nS_y = 217 ksi\nK_{IC} = 115 ksi in 0.5\n2w = 6 in.\nt = 0.06 in.\n2c = 1 in.\nP = ?\nAssumptions:\n1. The crack length is a small fraction of the plate width.\n2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength.\n3. Yielding has occurred within one small volume of the material at the crack root.\n4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K_I equals or exceeds the fracture toughness K_{IC} for the material.\nAnalysis:\n1. From Eq. (6.2), \\sigma_g = K_{IC} / (1.8\\sqrt{c}) = 115 / (1.8*0.5) = 90.353 ksi\n2. Since the area equals 2wt, P = \\sigma_g(2wt) = 90.353(6)(0.06) = 32,527 lb\nComment: The P/A stress based on the net area, (t(2w - 2c), is 108.4 ksi which is less than S_y = 217 ksi. Hence the second assumption is satisfied.\n6-4 SOLUTION (6.6)\nKnown: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load.\nFind: Estimate the highest tensile load that the plate will support.\nSchematic and Given Data:\nTi-6Al-4V (annealed)\ntitanium alloy\nS_u = 130 ksi\nS_y = 120 ksi\nK_{IC} = 110 ksi in 0.5\n2w = 6 in.\nt = 0.06 in.\n2c = 1 in.\nP = ?\nAssumptions:\n1. The crack length is a small fraction of the plate width.\n2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength.\n3. Yielding has occurred within one small volume of the material at the crack root.\n4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K_I equals or exceeds the fracture toughness K_{IC} for the material.\nAnalysis:\n1. From Eq. (6.2), \\sigma_g = K_{IC} / (1.8\\sqrt{c}) = 110 / (1.8*0.5) = 86.42 ksi\n2. Since the area equals 2wt, P = \\sigma_g(2wt) = 86.420(6)(0.06) = 31,112 lb\nComment: The P/A stress based on the net area, (t(2w - 2c), is 103.6 ksi which is less than S_y = 120 ksi. Hence the second assumption is satisfied. SOLUTION (6.7)\nKnown: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load.\nFind: Estimate the highest tensile load that the plate will support.\nSchematic and Given Data:\nLead beryllium copper\nS_u = 98 ksi\nS_y = 117 ksi\nK_{IC} = 70 ksi in 0.5\n2w = 8 in.\nt = 0.05 in.\n2c = 1.5 in.\nP = ?\nAssumptions:\n1. The crack length is a small fraction of the plate width.\n2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength.\n3. Yielding has occurred within one small volume of the material at the crack root.\n4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K_I equals or exceeds the fracture toughness K_{IC} for the material.\nAnalysis:\n1. From Eq. (6.2), \\sigma_g = K_{IC} / (1.8\\sqrt{c}) = 70 / (1.8*0.75) = 44.9 ksi\n2. Since the area equals 2wt, P = \\sigma_g(2wt) = 44,900(8)(0.05) = 17,960 lb\nComment: The P/A stress based on the net area, (t(2w - 2c), is 55.26 ksi which is less than S_y = 117 ksi. Hence the second assumption is satisfied. SOLUTION (6.8)\nKnown: A thin plate of known material is loaded in tension and has a central crack of given length perpendicular to the direction of the applied load.\n\nFind: Estimate the highest tensile load that the plate will support.\n\nSchematic and Given Data:\n\nP\n\nLead brass\nSu = 55 ksi\nSy = 42 ksi\nKc = 35 ksi in^0.5\n\n2w = 8 in.\nt = 0.05 in.\n2c = 1.5 in.\nP = ?\n\nAssumptions:\n1. The crack length is a small fraction of the plate width.\n2. The tensile stress based on the net area (minus the area of the crack) is less than the yield strength.\n3. Yielding has occurred within one small volume of the material at the crack root.\n4. Crack propagation to total fracture occurs instantaneously when the limiting value of the stress intensity factor K1 equals or exceeds the fracture toughness KIC for the material.\n\nAnalysis:\n1. From Eq. (6.2), σg = KIC / (1.87c) = 35 / (1.8*0.75) = 22.45 ksi\n\n2. Since the area equals 2wt, P = σg(2wt) = 22,450(8)(0.05) = 8980 lb\n\nComment: The P/A stress based on the net area, (t(2w - 2c), is 27.6 ksi which is less than Sy = 42 ksi. Hence the second assumption is satisfied.\n SOLUTION (6.9)\nKnown: A thick plate having a central crack is loaded in tension to a known gross-area stress.\n\nFind: Determine the critical crack depth, acr, at which rapid fracture will occur for 7075-T651 aluminum.\n\nSchematic and Given Data:\n\nP\n\n7075-T651 Aluminum\n2w = 6 in.\nt = 1 in.\na/2c = 0.25\nσg = 0.73 Sy\n\nAssumptions:\n1. The 7075-T651 aluminum is at room temperature.\n2. The crack length is critical when the value of the stress intensity factor K exceeds KIC.\n\nAnalysis:\n1. From Eq. (6.4), and setting K = KIC:\n\nacr = [KIC/σg(0.39 - 0.053(σg/Sy)²)]²\nSince σg = 0.73 Sy\n\nacr = (KIC/(0.73 Sy))²[0.39 - 0.053(0.73)²]\nacr = 0.68(KIC/Sy)²\n6-8 2. Using Table 6.1 to find Sy and Kic for 7075-T651 Aluminum:\nKIC = 27 ksi√in., Sy = 70 ksi, acr = 0.101 in.\n\nComments:\n1. Eq. (6.4) is appropriate if 2a/t > 6, a/2c > about 0.25, a/c > 3, a/t < 0.5, and σg < 0.8 Sy. For this problem, these conditions are satisfied.\n2. An important design requirement of internally pressurized members is that a crack be able to propagate through the full wall thickness (thereby causing a leak that can be readily detected) without becoming unstable and leading to total fracture.\n\nSOLUTION (6.10)\nKnown: A thick plate having a central crack is loaded in tension to a known gross-area stress.\n\nFind: Determine the critical crack depth, acr, at which rapid fracture will occur for D6AC steel at room temperature.\n\nSchematic and Given Data:\n\nP\n\nD6AC Steel at room temperature\n2w = 6 in.\nt = 1 in.\na/2c = 0.25\nσg = 0.73 Sy\n\n\nAssumption: The crack length is critical when the value of the stress intensity factor K exceeds Kic.\nAnalysis:\n1. From Eq. (6.4), and setting K = KIC:\n\nacr = [KIC/σg(0.39 - 0.053(σg/Sy)²)]²\n6-9\n Since σg = 0.73 Sy\n\nαcr = (KIC / (0.73 Sy)) [0.39 - 0.053(0.73)²]\n\nαcr = 0.68[KIC / Sy]²\n\n2. Using Table 6.1 to find Sy and KIC for D6AC steel at room temperature: KIC = 70 ksi/in., Sy = 190 ksi, αcr = 0.092 in.\n\nComments:\n1. Eq. (6.4) is appropriate if 2a/t > 6, a/2c = about 0.25, w/c > 3, and a/t < 0.5, and σg < 0.8 Sy. For this problem, these conditions are satisfied.\n2. An important design requirement of internally pressurized members is that a crack be able to propagate through the full wall thickness (thereby causing a leak that can be readily detected) without becoming unstable and leading to total fracture.\n\nSOLUTION (6.11)\nKnown: A thick plate having a central crack is loaded in tension to a known gross-area stress.\n\nFind: Determine the critical crack depth, acr, at which rapid fracture will occur for D6AC steel at -40 °F.\n\nSchematic and Given Data:\nD6AC Steel at -40 °F\n2w = 6 in.\nt = 1 in.\na/2c = 0.25\nσg = 0.73 Sy