Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - - Static Body Stresses

SOLUTION (4.1)\nKnown: A link is axially loaded in tension.\n\nFind: Identify the sections where the average compressive stress is equal to P/A. Also identify the sections where the maximum stress is equal to P/A.\n\nSchematic and Given Data:\n\nAssumptions:\n1. There are no residual stresses in the shaft.\n2. The load is applied exactly along the centroidal axis of the link.\n\nAnalysis:\n1. All sections have average stress = P/A.\n2. Sections C and D should have maximum stress approximately equal to P/A.\n\nComment: The maximum stress at a cross section is equal to P/A only if the load is uniformly distributed over the cross section.\n\nSOLUTION (4.2)\nKnown: A shaft is axially loaded in compression.\n\nFind: List the lettered sections where the average compressive stress is equal to P/A. Also list the sections where the maximum stress is equal to P/A.\n\nSchematic and Given Data:\n4-1 Assumptions:\n1. The shaft will not buckle.\n2. There are no residual stresses in the shaft.\n3. The load is applied exactly along the centroidal axis of the bar.\n\nAnalysis:\n1. All sections have average stress = P/A.\n2. Sections B, E, and F should have maximum stress approximately equal to P/A.\n\nComment: The maximum stress is equal to P/A only if the load is uniformly distributed over the cross section.\n\nSOLUTION (4.3)\nKnown: A force, P, in a bolted or pinned joint produces shear failure in the bolt or pin. The bolt or pin are made of steel of know diameter.\n\nFind: Select a steel from Appendix C-4a and determine the force, P, required to produce shear failure in a 0.375 in.-diameter bolt or pin: (a) with the configuration shown in Fig. 4.3, (b) with the configuration shown in Fig. 4.4.\n\nSchematic and Given Data:\n\nDecisions: Use as-rolled AISI 1040 with Su = 90.0 ksi.\n\nAssumptions:\n1. Bending is negligible.\n2. Plate interface friction is negligible.\n3. Bolt and pin materials are homogeneous.\n4. The loading is static.\n5. There are no residual stresses.\n6. The Sus = 0.62 Su is accurate for this analysis.\n\nAnalysis:\n1. For Fig. 4.3, since the bolt is in single shear,\n P = A*Su = π(0.1875 in)² (0.62)(90.0 ksi) = 6,163 lb.\n2. For Fig. 4.4, since the pin is in double shear,\n P = A*Su = 2 π(0.1875 in)² (0.62)(90.0 ksi) = 12,325 lb.\n\nComment: Sus =0.62Su is from Eq. 10.16 for the shear strength of steel bolts.\n4-2 SOLUTION (4.4)\nKnown: A force, P, is producing shear stress in a bolt or pin made of a ductile metal of known diameter and strength.\n\nFind: Determine the force, P, required to produce shear failure: (a) with the configuration shown in Fig. 4.3, (b) with the configuration shown in Fig. 4.4.\n\nSchematic and Given Data:\n\nAssumptions:\n1. Bending is negligible.\n2. Plate interface friction is negligible.\n3. Bolt and pin materials are homogeneous.\n4. The loading is static.\n5. There are no residual stresses.\n\nAnalysis:\n1. For Fig. 4.3, since the bolt is in single shear,\n P = A*Su = π(15 mm)²(200 MPa) = 141,372N = 141 kN.\n2. For Fig. 4.4, since the pin is in double shear,\n P = A*Su = 2 π(15 mm)²(200 MPa) = 282,743N = 283 kN.\n\nSOLUTION (4.5)\nKnown: A steel propeller shaft with a given diameter transmits a known power at a specified angular velocity.\n\nFind:\n(a) Determine the nominal shear stress at the surface.\n(b) Determine the outside diameter required to give the same outer surface stress if a hollow shaft of inside diameter 0.9 times the outside diameter is used.\n(c) Compare the weights of the solid and hollow shafts.\n4-3 Schematic and Given Data:\n\nAssumptions:\n1. Bending and axial loads are negligible.\n2. The bar is straight and round.\n3. The material is homogeneous, and perfectly elastic within the stress range involved.\n4. The effect of stress raisers is negligible.\n\nAnalysis:\n1. From Eq. (1.3), T = 5252 W\nn\nT = 5252(2500) = 6565 lb ft = 78,780 lb in.\n\n2. From Eq. (4.4), \\tau = 16 T\n\\pi d^3\n\\tau = 16 (78,780)\n\\pi (2)^3 = 50,153 psi = 50 ksi\n\n3. For a hollow shaft,\nd_o^3 - (0.9)(d_i)^3 = 2^3\n0.344 d_o^3 = 8\nd_o = 2.86 in.\n\n4. Wt. hollow / Area hollow = 2.86^2 - (0.9 x 2.86)^2\nWt. solid / Area solid = 2^2 = 0.39\nComment: It is more economical to use a hollow shaft when pure shear stress is involved.\n\n4-4 SOLUTION (4.6D)\nKnown: A steel driveshaft transmits a known power at a specified angular velocity. \n\nFind: Select the diameter for a steel driveshaft and:\n(a) determine the nominal shear stress at the surface.\n(b) determine the outside diameter required to give the same outer surface stress if a hollow shaft of inside diameter 0.9 times the outside diameter is used.\n(c) compare the weights of the solid and hollow shafts.\n\nSchematic and Given Data:\n\nDecision: Select a diameter of 3 in.\n\nAssumptions:\n1. Bending and axial loads are negligible.\n2. The bar is straight and round.\n3. The material is homogeneous, and perfectly elastic within the stress range involved.\n4. The effect of stress raisers is negligible.\n\nAnalysis:\n1. From Eq. (1.3), T = 5252 W\nn\nT = 5252(250 hp)\n5000 rpm = 262.6 lb ft = 3151.2 lb in.\n\n2. From Eq. (4.4), \\tau = 16 T\n\\pi d^3\n\\tau = 16 (3151.2 lb in.)\n\\pi (3 in.)^3 = 594.4 psi\n\n4-5 3. For a hollow shaft,\nd_o^3 - (0.9)(d_i)^3 = 3^3\n0.344 d_o^3 = 27\nd_o = 4.282 in.\n\n4. Wt. hollow / Area hollow = 4.282^2 - (0.9 x 4.282)^2\nWt. solid / Area solid = 3^2 = 0.39\nComment: It is more economical to use a hollow shaft when pure shear stress is involved.\n\nSOLUTION (4.7)\nKnown: A steel propeller shaft with a given diameter transmits a known power at a specified angular velocity.\n\nFind:\n(a) Determine the nominal shear stress at the surface.\n(b) Determine the outside diameter required to give the same outer surface stress if a hollow shaft of inside diameter 0.8 times the outside diameter is used.\n(c) Compare the weights of the solid and hollow shafts.\n\nSchematic and Given Data:\n\nAssumptions:\n1. Bending and axial loads are negligible.\n2. The bar is straight and round.\n3. The material is homogeneous, and perfectly elastic within the stress range involved.\n4. The effect of stress raisers is negligible. Analysis:\n1. From Eq. (1.2), T = 9549 W\n T = 9549(700) = 4456.2 N*m\n2. From Eq. (4.4), \\tau = \\frac{16 T}{\\pi d^3}\n \\tau = \\frac{16 (4456.2)}{\\pi (0.03)^3} = 840.56 MPa\n3. For a hollow shaft,\n d_o^3 - (0.8)^3 d_i^3 = (0.03)^3\n 0.5904 d_o^3 = 2.7 x 10^-5\n d_o = 35.8 mm\n4. W_t hollow = \\frac{35.8^4 - (0.8 x 35.8)^4}{Area hollow} \n W_t solid = \\frac{Area solid}{30^2} = 0.51\nComment: It is more economical to use a hollow shaft when pure shear stress is involved.\n\nSOLUTION (4.8D)\nKnown: A 40 mm steel shaft transmits 500 kw.\nFind: Select a steel from Appendix C-4a and a rotational speed for the 40 mm shaft, and then:\n(a) determine the nominal shear stress at the surface within the \\tau < 0.2s limit.\n(b) determine the outside diameter required to give the same outer surface stress if a hollow shaft of inside diameter 0.8 times the outside diameter is used.\n(c) compare the weights of the solid and hollow shafts.\n\nSchematic and Given Data:\nRotation\n500 kW\nn = 1250 to 2000 rpm\n40 mm\n Decisions: Select normalized AISI 4340 with S_u = 1279.0 MPa. Select 2000 rpm for the rotational speed.\n\nAssumptions:\n1. Bending and axial loads are negligible.\n2. The bar is straight and round.\n3. The material is homogeneous, and perfectly elastic within the stress range involved.\n4. The effect of stress raisers is negligible.\n5. \\tau \\leq 0.25s is a limit on acceptable steel selections.\n\nAnalysis:\n1. From Eq. (1.2), T = 9549 W\n T = \\frac{9549(500)}{2000} = 2387.3 N*m\n2. From Eq. (4.4), \\tau = \\frac{16 T}{\\pi d^3}\n \\tau = \\frac{16 (2387.3)}{\\pi (0.04)^3} = 190.0 MPa\n3. \\tau \\leq 0.25s\n 190.0 MPa \\leq 0.2(1279 MPa)\n 190.0 MPa < 255.8 MPa\n4. For a hollow shaft,\n d_o^3 - (0.8)^3 d_i^3 = (0.04)^3\n 0.5904 d_o^3 = 6.4 x 10^-5\n d_o = 47.7 mm\n5. W_t hollow = \\frac{47.7^4 - (0.8 x 47.7)^4}{Area hollow}\n W_t solid = \\frac{Area solid}{40^2} = 0.51\nComments:\n1. The steel selected has \\tau \\leq 0.25S_u.\n2. The weight of the hollow shaft is 51% the weight of the solid shaft.\n\nSOLUTION (4.9)\nKnown: A steel propeller shaft with a given diameter transmits a known power at a specified angular velocity.\nFind:\n(a) Determine the nominal shear stress at the surface.\n(b) Determine the outside diameter required to give the same outer surface stress if a hollow shaft of inside diameter 0.85 times the outside diameter is used.\n(c) Compare the weights of the solid and hollow shafts.\n Schematic and Given Data:\nRotation\n3200 hp, n = 2000 rpm\n2.5 in.\nAssumptions:\n1. Bending and axial loads are negligible.\n2. The bar is straight and round.\n3. The material is homogeneous, and perfectly elastic within the stress range involved.\n4. The effect of stress raisers is negligible.\n\nAnalysis:\n1. From Eq. (1.3), T = 5252 W\n T = \\frac{5252(3200)}{2000} = 8403 lb ft = 100,838 lb in.\n2. From Eq. (4.4), \\tau = \\frac{16 T}{\\pi d^3}\n \\tau = \\frac{16 (100,838)}{\\pi (2.5)^3} = 32,868 psi = 33 ksi\n3. For a hollow shaft,\n d_i^3 - (0.85)^4 d_o^3 = 2.53\n 0.478 d_o^3 = 15.625\n d_o = 3.20 in.\n4. W_t hollow = \\frac{3.20^2 - (0.85 x 3.20)^2}{Area hollow}\n W_t solid = \\frac{Area solid}{2.5^2} = 0.46\nComment: It is more economical to use a hollow shaft when pure shear stress is involved. SOLUTION (4.10)\nKnown: The maximum shear stress is given for a hollow shaft of known geometry subjected to pure torsion.\n\nFind: Determine the torque that produces the given maximum shear stress.\n\nSchematic and Given Data:\n\nd_i = 20 mm\nd_o = 24 mm\n\t\\tau_{max} = 100 MPa\n\nAssumptions:\n1. The shaft is straight.\n2. The material is homogeneous and perfectly elastic.\n3. There are no stress raisers.\n\nAnalysis:\n1. From Eq. (4.3), \\tau = \\frac{Tr}{J} = \\frac{16Td_o}{\\pi(d_o^4 - d_i^4)}\n2. T = \\frac{\\pi(100)(24^4 - 20^4)}{16(24)} = 0.141 \\times 10^6 N \\cdot mm = 141 N \\cdot m\n\nSOLUTION (4.11)\nKnown: The same value of torque is applied to both a solid square shaft (b x b) and a solid round shaft of radius r.\n\nFind:\n(a) Determine the ratio b/f for both square and round shafts so as to produce equal maximum shear values.\n(b) Compare the weight of the two shafts for the square and round shafts.\n(c) Compare the ratio of strength-to-weight for the square and round shafts.\n\n4-10