Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Engrenagens 2

SOLUTION (16.1) Known: A helical spur gear with specified number of teeth, helix angle, \\u03b3, and pressure angle, \\u03c6, is given. Find: Determine: (a) the normal pressure angle, \\u03b8n, and the equivalent number of teeth, Ne. (b) the pressure angle and number of teeth on a spur gear of equivalent strength. Schematic and Given Data: Assumptions: 1. The helical gear teeth have standard tooth profiles. 2. The profile of the helical gear tooth in the plane normal to the tooth can be approximated by the standard profile of a spur gear tooth. Analysis: (a) From Eq. (16.2): \\u03b8n = tan^{-1}(tan \\u03c6 cos \\u03b3) = tan^{-1}(tan 20\\u00b0 cos 25\\u00b0), \\u03b8n = 18.2\\u00b0 Ne = N/cos\\u03b3 = 27/cos 25\\u00b0: Ne = 36.27 teeth (b) A spur gear of equal strength would have 36.27 teeth and \\u03c6 = 18.2\\u00b0 16-1 Comment: The increase in the number of teeth of the equivalent spur gear suggests that for the same pitch diameter, number of teeth and material, the helical gear teeth would have greater bending strength than spur gear teeth. SOLUTION (16.2) Known: The normal range of spur gear face widths is given as a function of P. It is desired that the helical gear face width be greater than or equal to twice the axial circular pitch. Find: (a) Determine the helix angle required to have a face width equal to 12/P and also equal to 2.0Pa. (b) Compare the helix angle found in (a) with the common range of helix angles, viz. 15 to 30 degrees. Schematic and Given Data: Assumption: The helical gear teeth have standard tooth profiles. Analysis: (a) Given that b = 2.0Pa = 12/P and Pa = p/tan \\u03b3 = p/sin\\u03b3; hence p/tan \\u03b3 = 2.0\\u03b0 = 12/P, or \\u03b3 = 27.64\\u00b0 (b) The helix angle is within the 15\\u00b0 to 30\\u00b0 \"commonly used range\" Comment: The range of helix angles possible under the conditions, b = 2.0Pa, 9/P \\u2264 b \\u2264 14/P and 15\\u00b0 \\u2264 \\u03b3 \\u2264 30\\u00b0 is 24.2\\u00b0 \\u2264 \\u03b3 \\u2264 30\\u00b0. SOLUTION (16.3) Known: For a pair of meshing helical gears mounted on parallel shafts, the normal circular pitch, gear center distance, speed ratio and the number of teeth on the pinion are given. Find: Determine the helix angle. Schematic and Given Data: Assumptions: 1. The helical gears are aligned and mounted to mesh along the pitch circles. 2. The gear teeth have standard involute profiles. Analysis: 1. The speed ratio = 2:1, and Np = 35. Hence, Ne = 70. mn = p/{\\u03c0} = 0.167 in. 3. Center distance, c = (dP + dG)/2 = mn(Np + Ng)/2 = mn(Np + Ng)/2 = 0.167(35 + 70) = 10.12 in. Comment: The helix angle of 30\\u00b0 is at the edge of the recommended range of 15 to 30 degrees. SOLUTION (16.4)\nKnown: A helical spur gear with specified number of teeth, helix angle, psi, and pressure angle, phi, is given.\n\nFind: Determine:\n(a) the normal pressure angle, phi_n, and the equivalent number of teeth, Ne.\n(b) the pressure angle and number of teeth on a spur gear of equivalent strength.\n\nSchematic and Given Data:\nAssumptions:\n1. The helical gear teeth have standard tooth profiles.\n2. The profile of the helical gear tooth in the plane normal to the tooth can be approximated by the standard profile of a spur gear tooth.\n\nAnalysis:\n(a) From Eq. (16.2): phi_n = tan^{-1}(tan(phi) cos(psi)) = tan^{-1}(tan(20°) cos(25°)), phi_n = 18.2°.\n(b) Ne = N/cos^3(psi) = 30/cos^2(25°); Ne = 40.29 teeth.\n\n(b) A spur gear of equal strength would have 40.29 teeth and phi = 18.2°. Comment: The increase in the number of teeth of the equivalent spur gear also reflects the fact that for the same pitch diameter and number of teeth, helical gear teeth can have greater bending strength than spur gear teeth.\nSOLUTION (16.5)\nKnown: The normal range of spur gear face widths is given as a function of P. It is desired that the helical gear face width be greater than or equal to twice the axial circular pitch.\n\nFind:\n(a) Determine the helix angle required to have a face width equal to 13/P and also equal to 2.2Pa.\n(b) Compare the helix angle found in (a) with the common range of helix angles, viz. 15 to 30 degrees.\n\nSchematic and Given Data:\nAssumption: The helical gear teeth have standard tooth profiles.\n\nAnalysis:\n(a) Given that b = 2.2P2, b = 13/P and Pa = tan(psi) = pi/(P tan(psi)); hence P tan(psi) = P, or psi = 28°.\n(b) The helix angle is within the 15° to 30° \"commonly used range.\"\n\nComment: The range of helix angles possible under the conditions, b = 2.2Pa, 9/P <= b <= 14/P and 15° <= psi <= 30° is 26.3° <= psi <= 30°. SOLUTION (16.6)\nKnown: For a pair of meshing helical gears mounted on parallel shafts, the normal circular pitch, gear center distance, speed ratio and the number of teeth on the pinion are given.\n\nFind: Determine the helix angle.\n\nSchematic and Given Data:\nAssumptions:\n1. The helical gears are aligned and mounted to mesh along the pitch circles.\n2. The gear teeth have standard involute profiles.\n\nAnalysis:\n1. The speed ratio = 2:1, and Np = 35. Hence, Ng = 70.\n2. mp = P/pi = 0.167 in.\n3. Center distance,\nc = (dg + dp)/2 = m(Np + Ng)/2 = mg(Np + Ng)/2cos(psi).\n4. 9 = 0.167(35 + 70)/(2cos(psi)).\n5. Hence, cos(psi) = 0.974; psi = 13.05° = 13° 03′.\n\nComment: The helix angle of 13° 03′ is not in the recommended range of 15 to 30 degrees. From Eq. (16.1), p = p n / cos ψ. From Eq. (15.6), m = p / π (and m n = p n / π), hence m = m n / cos ψ. \nFrom Eq. (15.4), d = m N. \nIn order to use the same arm, d sun + d planet must have the same value for the two gear sets.\nHence, m(N planet + N sun) must be equal for the two sets.\nFor the original set, m = 4 mm / cos 0.42 \n4 / cos 0.42(48 + 24) = 4 / cos ψ(42 + 27).\nψ = 0.5053 rad = 28.95°\nTherefore, the helix angle for optional gears must be 0.5053 rad or 28.95°.\nComments:\n (1) In this problem the helix angle was fixed to conform with the requirement of keeping the same arm and normal module for the optional gearing as well as the original gearing. If instead, the module m was required to be kept the same, there would be a choice in the values for normal module and helix angle, since, m = m n / cos ψ.\n (2) It is possible to check if the same arm could be used when, there are three equidistant planets on the rim. For the original gearing the number of teeth on the ring between planets is 120/3 = 40, and on the sun it is 24/3 = 8. Thus three equidistant planets can mesh with the sun and ring if the planet has an even number of teeth (see section 15.13). This is indeed the case since the number of teeth on the planet is equal to 48. Using a similar argument for the optional gearing, three equidistant planets can be checked and found to mesh properly.\nSOLUTION (16.8)\nKnown: A double reduction helical gear arrangement is given. Number of teeth and module in the normal plane are specified. The helix angle of the high speed gears is specified.\nFind: Determine: \n (a) the total speed reduction provided by the four gears, \n (b) the helix angle of the low speed gears, \n (c) the helix angle of the low speed gears if the low speed gears had 24 and 34 teeth respectively but with the same module in the normal plane. Schematic and Given Data:\n 40 Teeth\nOutput\n c\nInput\n 24 Teeth\n 60 Teeth\n 5 mm\n 20 Teeth\n 3.5 mm\nNote: The helix angles are different for the gear sets.\nAssumptions: \n 1. The helical gears are aligned and mounted to mesh along the pitch circles.\n 2. The gear teeth have standard involute profiles.\nAnalysis: \n (a) Speed ratio = 60/24 = 40/20: speed ratio = 5. \n (b) For equal center distance, c = d p + d e / 2 = m(N p + N g) = m n / cos ψ (N p + N g) must be the same for the two gear pairs.\n Therefore, 3.5 / cos 0.44 (24 + 60) = 5 / cos ψ(20 + 40); and the helix angle of the low speed gears, ψ = 0.3944 rad. \n (c) Similarly, 5 / cos ψ(24 + 34) = -3.5 / cos 0.44(24 + 60) ψ = 0.4680 rad.\nComments: \n (1) The helix angle for the low speed gears is within the common range of helix angles (15° to 30°). \n (2) The total speed ratio of the unit changes in part (c) to 60/24: 34/24 = 3.54.\n16-9 SOLUTION (16.7)\nKnown: For a simple planetary gear train the number of teeth on the sun and ring helical gears is given. Module of teeth and helix angle are also specified.\nFind: Determine the helix angle of the optional helical gearing using the same arm and gear helix angle, but with a 27 tooth sun and 111 tooth ring.\nSchematic and Given Data:\nNumber of Teeth\n Sun Ring\nNormal Module Helix Angle\nOriginal set: 24 120 4 mm 0.42 rad\nOptional set: 27 111 4 mm ?\nAssumptions:\n 1. The sun, planet and ring gears are aligned and mounted to mesh along their pitch circles.\n 2. All the gears have standard involute tooth profiles.\nAnalysis: \n 1. Since the sun, planets, and ring gear all mesh together, all must have the same module and helix angle.\n 2. From the planetary gear arrangement, d ring = d sun + 2d planet, or N ring = N sun + 2N planet.\n 3. Hence, for the original gears, N planet = 48 teeth, and for the replacement gears, N planet = 42 teeth.\n16-7 (3) Unlike spur gears, in part (c) it is possible to change the number of teeth on the helical gears while keeping the center distance and normal plane module the same because in helical gears the module is a function of the normal plane module as well as the helix angle. (4) To assure that each gear tooth in each gear pair meshes with every other tooth in that pair, an odd number of gear teeth could be used for the larger gears in both gear sets. SOLUTION (16.9D) Known: A pair of spur gears of specified numbers of teeth, face width and module is to be replaced with helical gears of the same module in the normal plane. Find: Determine a suitable combination of numbers of teeth and helix angle for the helical gears. Schematic and Given Data: Spur Gears: Helical Gears: Np = 20 Np = ? Ng = 60 Ng = ? b = 50 mm b = 50 mm m = 4.0 mm mn = 4.0 mm y = 0 y = ? Decisions: 1. The ratio Ng/Np must remain 60/20 = 3, but with mn remaining 4.0 mm, fewer teeth will be required to preserve the same center distance. 2. Try Np = 18, Ng = 54. Assumptions: 1. Spur and helical gear sets are aligned and mounted to mesh along their pitch circles. 2. All gears have teeth of standard involute tooth profile. Design Analysis: 1. For equal center distance, c = m(Np + Ng) = mn/cos(y) (Np + Ng) must be the same for the spur and helical gear pairs: 4.0 cos(0(20 + 60) = 4.0 cos(y)(18 + 54). Therefore, y = 25.84° = 0.4510 rad and lies in the commonly used range 15° to 30°.