Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Fadiga

SOLUTION (8.1D)\nKnown: Appendix C-4a gives properties of various steel.\nFind: Select a steel with a rotating bending endurance limit for a standard R.R. Moore test specimen above 72 ksi.\nAssumption: We assume that the rotating bending endurance limit for a polished R.R. Moore test specimen is given by S'n = 0.5Su.\nDecision: Select 4140 normalized steel.\nAnalysis:\n1. From Appendix C-4a, 4140 normalized steel has a tensile strength of 148.0 ksi.\n2. S'a = 0.5Su = 0.5(148) = 74 ksi.\nComment: The endurance limit of 4140 steel, i.e. 74 ksi, is above 72 ksi.\n\nSOLUTION (8.2D)\nKnown: Appendix C-4a gives properties of various steel.\nFind: Select a steel with a 103-cycle fatigue strength for a standard R.R. Moore test specimen that is above 130 ksi.\nAssumption: We assume that the rotating bending 103-cycle fatigue life for a polished R.R. Moore test specimen is given by S for 103-cycles = 0.9Su.\nDecision: Select 4140 normalized steel.\nAnalysis:\n1. From Appendix C-4a, 4140 normalized steel has a tensile strength of 148.0 ksi.\n2. The 103-cycle fatigue life, S for 103-cycle = 0.9Su = (0.9)(148) = 133.2 ksi.\nComment: The 103-cycle life for 4140 is above 130 ksi. SOLUTION (8.3)\nKnown: Standard R.R. Moore test specimens are made of steels having known Brinell hardness.\nFind: Estimate the rotating bending endurance limit and also the 103 cycle fatigue strength.\nSchematic and Given Data:\nBhn = 100\nBhn = 300\nBhn = 500\nAssumptions:\n1. For steel, the tensile strength in psi is 500 times the Brinell hardness.\n2. The curve in Fig. 8.5 is an accurate representation of the S-N data for steel.\n3. For steel, the endurance limit in psi is 250 times the Brinell hardness.\n4. For steel, the endurance limit for 103 cycle is 90% of the ultimate strength.\nAnalysis:\n1. Su = 500 Bhn in psi.\n2. S'n = 0.25 Bhn in ksi.\n3. S for 103 cycle = 0.9Su.\n\nBhn\t Su (ksi)\t S'n (ksi)\t S for 10^3 cycle (ksi)\n100\t 50\t 25\t 45\n300\t 150\t 75\t 135\n500\t 250\t 100-125\t 225\n\nComments:\n1. The relationship S'n = 0.25 Bhn is accurate only to Brinell hardness values of about 400.\n2. For 10^3-cycle fatigue strength, actual stress is not as high as calculated values because of significant yielding.\n\nSOLUTION (8.4)\nKnown: Four known standard R.R. Moore specimens are given.\nFind: Estimate the long-life rotating bending fatigue strength (state whether it is for 10^8 or 5 x 10^8 cycles). Schematic and Given Data:\nWrought aluminum\t Su = 250 MPa\nWrought aluminum\t Su = 450 MPa\nAverage grade cast aluminum\nAverage grade forged magnesium\nAssumptions:\n1. The specimen is subjected to pure bending (i.e., zero transverse shear).\n2. Figs. 8.8, 8.9, and 8.10 can be used to estimate the fatigue strength of aluminum and magnesium.\nAnalysis:\n1. From Fig. 8.9, for the wrought aluminum having\nSu = 250 MPa, the fatigue strength at 5 x 10^8 cycles is 95 MPa.\n2. From Fig. 8.9, for the wrought aluminum having\nSu = 450 MPa, the fatigue strength at 5 x 10^8 cycles is 130 MPa.\n3. From Fig. 8.8, for the average grade cast aluminum, the fatigue strength at 5 x 10^6 cycles is 60 MPa for sand cast and 85 MPa for permanent mold cast.\n4. From Fig. 8.10, for average grade forged magnesium, the fatigue strength at 10^8 cycles is 130 MPa.\n\nSOLUTION (8.5)\nKnown: Standard R.R. Moore test specimens are made of steels having known ultimate tensile strengths.\nFind: Estimate the rotating bending endurance limit and also the 10^3 cycle fatigue strength.\nSchematic and Given Data:\nSu = 95 ksi\nS'n = 185 ksi\nSu = 240 ksi Assumptions:\n1. For steel, the tensile strength in psi is 500 times the Brinell hardness.\n2. The curve in Fig. 8.5 is an accurate representation of the S-N data for steel.\n3. For steel, the endurance limit in psi is 250 times the Brinell hardness.\n4. For steel, the endurance limit for 10^3 cycle is 90% of the ultimate strength.\n\nAnalysis:\n1. S'n = 0.5Su in ksi.\n2. S for 10^3 cycle = 0.9Su\n\n Su (ksi) S'n (ksi) S for 10^3 cycle (ksi)\n 95 47.5 85.5\n 185 92.5 166.5\n 240 100-125 216\n\nComments:\n1. The relationship S'n = 0.5Su is accurate only to ultimate tensile strength values of 200 ksi. The endurance limit may or may not continue to increase for greater tensile strength values depending on the composition of the steel.\n\nSOLUTION (8.6)\nKnown: Four known standard R.R. Moore specimens are given.\n\nFind: Estimate the long-life rotating bending fatigue strength (state whether it is for 10^8 or 5 x 10^8 cycles).\n\nSchematic and Given Data:\nWrought aluminum\n Su = 29 ksi\nWrought aluminum\n Su = 73 ksi\nHigh grade\n cast aluminum\nHigh grade\n forged magnesium\n\nAssumptions:\n1. The specimen is subjected to pure bending (i.e., zero transverse shear).\n2. Figs. 8.8, 8.9, and 8.10 can be used to estimate the fatigue strength of aluminum and magnesium.\n8-4 Analysis:\n1. From Fig. 8.9, for the wrought aluminum having\n Su = 29 ksi, the fatigue strength at 5 x 10^8 cycles is 12 ksi.\n2. From Fig. 8.9, for the wrought aluminum having\n Su = 73 ksi, the fatigue strength at 5 x 10^8 cycles is 19 ksi.\n3. From Fig. 8.8, for the high grade cast aluminum, the fatigue strength\n at 5 x 10^8 cycles is 11 ksi for sand cast and 15 ksi for permanent mold cast.\n4. From Fig. 8.10, for high grade forged magnesium, the fatigue strength at 10^8 cycles is 22 ksi.\n\nSOLUTION (8.7)\nKnown: Standard R.R. Moore test specimens are made of steels having known ultimate tensile strengths.\n\nFind: Estimate the rotating bending endurance limit and also the 10^3 cycle fatigue strength.\n\nSchematic and Given Data:\n Su = 100 ksi\n Su = 160 ksi\n Su = 280 ksi\n\nAssumptions:\n1. For steel, the tensile strength in psi is 500 times the Brinell hardness.\n2. The curve in Fig. 8.5 is an accurate representation of the S-N data for steel.\n3. For steel, the endurance limit in psi is 250 times the Brinell hardness.\n4. For steel, the endurance limit for 10^3 cycle is 90% of the ultimate strength.\n\nAnalysis:\n1. S'n = 0.5Su in ksi.\n2. S for 10^3 cycle = 0.9Su\n\n Su (ksi) S'n (ksi) S for 10^3 cycle (ksi)\n 100 50 90\n 160 80 144\n 280 100-125 252\n\nComments:\n1. The relationship S'n = 0.5Su is accurate only to ultimate tensile strength values of 200 ksi. The endurance limit may or may not continue to increase for greater tensile strength values depending on the composition of the steel.\n8-5 2. For the 10^3-cycle fatigue strength, the actual stress is not as high as calculated values because of significant yielding.\n\nSOLUTION (8.8)\nKnown: Standard R.R. Moore test specimens are made of steels having known Brinell hardness.\n\nFind: Estimate the rotating bending endurance limit and also the 10^3 cycle strength.\n\nSchematic and Given Data:\n Bhn = 200\n Bhn = 350\n Bhn = 500\n\nAssumptions:\n1. For steel, the tensile strength in psi is 500 times the Brinell hardness.\n2. The curve in Fig. 8.5 is an accurate representation of the S-N data for steel.\n3. For steel, the endurance limit in psi is 250 times the Brinell hardness.\n4. For steel, the endurance limit for 10^3 cycle is 90% of the ultimate strength.\n\nAnalysis:\n1. Su = 500 Bhn in psi.\n2. S'n = 0.25 Bhn in ksi.\n3. S for 10^3 cycle = 0.9Su\n\n Bhn Su (ksi) S'n (ksi) S for 10^3 cycle (ksi)\n 200 100 50 90\n 350 175 87.5 157.5\n 500 250 100-125 225\n\nComments:\n1. The relationship S'n = 0.25 Bhn is accurate only to Brinell hardness values of about 400.\n2. For the 10^3-cycle fatigue strength, the actual stress is not as high as calculated values because of significant yielding.\n\nSOLUTION (8.9)\nKnown: Standard R.R. Moore specimens are subjected to loading.\n8-6 Find: Determine how the fatigue strengths change if the loading is reversed bending rather than rotating bending.\n\nSchematic and Given Data:\n\nM\nM\nNot Rotating\n\n\nNot Rotating\n\n\nRotating\n\n\nAssumption: The specimens are homogeneous throughout.\n\nAnalysis: In reverse bending, maximum stresses are limited to the top and bottom of the specimen. In rotating bending, maximum stresses are produced all around the circumference of the specimen. Therefore, statistically the fatigue strengths are generally slightly higher for the reverse bending case, but (conservatively) it is usually assumed that reverse bending and rotating bending are the same.\n\nSOLUTION (8.10)\nKnown: Standard R.R. Moore specimens are subjected to loading.\nFind: Determine how the fatigue strengths change if the loading is reversed axial loading rather than rotating loading. Analysis: The fatigue strength would be lower by roughly 20%.\nComment: Please see the discussion in the text.\n\nSOLUTION (8.11)\nKnown: Standard R.R. Moore specimens are subjected to loading.\nFind: Determine how the fatigue strengths change if the loading is reversed bending rather than rotating bending.\n\nSchematic and Given Data:\n\nM\nM\nNot Rotating\n\n\nNot Rotating\n\n\nRotating\n\n\nAssumption: The specimens are homogeneous throughout.\n\nAnalysis: In reverse bending, maximum stresses are limited to the top and bottom of the specimen. In rotating bending, maximum stresses are produced all around the circumference of the specimen. Therefore, statistically the fatigue strengths are generally slightly higher for the reverse bending case, but (conservatively) it is usually assumed that reverse bending and rotating bending are the same.\n\nSOLUTION (8.12)\nKnown: Standard R.R. Moore specimens are subjected to loading.\nFind: Determine how the fatigue strengths change if the loading is reversed axial loading rather than rotating loading. Schematic and Given Data:\n\nAssumption: The concentricity of load and geometric axes reflects an intermediate condition between precision ground parts and non-precision as-cast or as-forged parts.\n\nAnalysis: The fatigue strength would be lower by roughly 20%.\nComment: Please see the discussion in the textbook.\n\nSOLUTION (8.13)\nKnown: Standard R.R. Moore test specimens are made of steels having known Brinell hardness.\nFind: Estimate the endurance limit and also the 10^3 cycle fatigue strength for reversed torsional loading.\n\nSchematic and Given Data:\n\nBhn = 100\nBhn = 300\nBhn = 500\n\nAssumption: Figs. 8.5 and 8.11 can be used to estimate the endurance limit and 10^3 cycle fatigue strength for reversed torsional loading.\n\nAnalysis:\n1. Su = 500 Bhn in psi.\n2. Sn' = 0.25 Bhn in ksi.\n3. Sn = 0.58 Sn'\n4. S for 10^3 cycle = 0.9 Sus where Sus = 0.8 Su for steel 5. Bhn S_u (ksi) S_n (ksi) S for 10^3 cycle (ksi)\n100 50 25(0.58) = 14.5 36\n300 150 75(0.58) = 43.5 108\n500 250 (100-125)0.58 = 58 to 72 180\n\nComment: The relationship S_n' = 0.25 Bhn is accurate only to Brinell hardness values of about 400.\n\nSOLUTION (8.14)\nKnown: Four known standard R.R. Moore specimens are given.\n\nFind: Estimate the long-life fatigue strength for reversed torsional loading. (State whether it is for 10^8 or 5 x 10^8 cycles.)\n\nSchematic and Given Data:\n\nWrought aluminum\nS_u = 250 MPa\n\nWrought aluminum\nS_n = 450 MPa\n\nAverage grade\ncast aluminum\n\nAverage grade\nforged magnesium\n\nAssumption: Figs. 8.8, 8.9, and 8.10 can be used to estimate long life fatigue strength for reversed torsional loading.\n\nAnalysis:\n1. From Fig. 8.9, for the wrought aluminum having S_u = 250 MPa, the rotating bending fatigue strength at 5 x 10^8 cycles is S_n' = 95 MPa. Since, for reversed torsional loading S_n = 0.58 S_n' ; S_n = 0.58(95) = 55 MPa\n2. From Fig. 8.9, for the wrought aluminum having S_u = 450 MPa, the rotating bending fatigue strength at 5 x 10^8 cycles is S_n' = 130 MPa. Thus, for reversed torsional loading, S_n = 0.58(130) = 75 MPa\n3. From Fig. 8.8, for average grade cast aluminum, the rotating bending fatigue strength at 5 x 10^8 cycles is 60 MPa for sand cast and 85 MPa for permanent mold cast. Thus, for reversed torsional loading,\nS_n = 0.58(60) = 35 MPa for sand cast\nS_n = 0.58(85) = 49 MPa for permanent mold cast\n4. From Fig. 8.10, for average grade forged magnesium, the rotating bending fatigue strength at 10^8 cycles is 130 MPa. Thus, for reversed torsional loading,\nS_n = 0.58(130) = 75 MPa\n\n8-10