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

SOLUTION (12.1) Known: A torsion bar has a known length and diameter.\nFind: \n(a) Estimate the change in shear stress when one end of the rod rotates through 70° relative to the other end.\n(b) Estimate the change in torque.\nSchematic and Given Data:\nAssumptions:\n1. The bar remains straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\nAnalysis:\n1. From Table 5.1, for a torsional case (Case 2);\nθ = TL /JG; therefore, T = θJG / L.\n2. θ = (80/180)π = 1.39 rad\nJ = πd^4/32 = π(0.312)^4/32 = 0.00093 in.^4\nFrom (Appendix C-1), G = 11.5 × 10^6 psi\n3. Therefore, T = 1.39 × 0.00093 × 11.5 × 10^6 / 50 = 297.32 lb in.\n4. From Eq. (4.4), for a solid round rod,\nτ = 16T/πd^3 = (16)(297.32) / π(0.312)^3 = 49.9 ksi\n12-1 SOLUTION (12.2) Known: A torsion bar has a known length and diameter.\nFind:\n(a) Estimate the change in shear stress when one end of the rod rotates through 75° relative to the other end.\n(b) Estimate the change in torque.\nSchematic and Given Data:\nAssumptions:\n1. The bar remains straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\nAnalysis:\n1. From Table 5.1, for a torsional case (Case 2);\nθ = TL /JG; therefore, T = θJG / L.\n2. θ = (75/180)π = 1.31 rad\nJ = πd^4/32 = π(0.312)^4/32 = 0.00093 in.^4\nFrom (Appendix C-1), G = 11.5 × 10^6 psi\n3. Therefore, T = 1.31 × 0.00093 × 11.5 × 10^6 / 42.5 = 329.66 lb in.\n4. From Eq. (4.4), for a solid round rod,\nτ = 16T/πd^3 = (16)(329.66) / π(0.312)^3 = 55.3 ksi\n12-2 SOLUTION (12.3) Known: A torsion bar spring serves as a counterbalance for a trap door with a given weight. The maximum allowable torsional stress for the spring is 50 ksi.\nFind:\n(a) Determine the length and diameter of a solid steel torsion bar that would counterbalance 80% of the door weight when closed, and provide a 6 lb-ft torque holding the door against the stop.\n(b) Make a graph showing gravity torque, spring torque, and net torque all plotted against door opening angle.\nSchematic and Given Data:\nAssumptions:\n1. The bar is straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\n3. The cross section considered is sufficiently remote from points of load application and from stress raisers.\nAnalysis:\n1. From Eq. (4.4), for a solid round rod τ = 16T/πd^3 or d = (3√(16T/πτ))\nWhen the door is closed, the bar will counterbalance 80% of the door weight.\nThus, Tc = (0.80)(60 lb)(24 in.) = 1152 lb in.\nSince a maximum allowable torsional stress is 50 ksi,\nd = (3√(16(1152)/π(50,000))) = 0.49 in.\n2. From Table 5.1, for a torsional case (Case 2) θ = TL /JG and K = T / (θJG / L)\n12-3 where J = πd^4 / 32 = 0.005660 in.4\nG = 11.5 x 10^6 psi (Appendix C-1)\nK = ΔT = T_e - T_c = (1152 - 72) lb in. = 562.5 lb in./rad\nΔθ = 1.92 rad\nThus, L = JG/K = (0.005660)(11.5 x 10^6) / 562.5 = 115.7 in.\n3. Let clockwise torque be positive.\n1600\n1200\n800\n400\n0\n-400\n-800\n-1200\n-1440 cos θ\nGravity torque = 1440 cos θ\nGravity + Spring torque\nSpring torque\n\nDoor opening angle, θ\n30°\n60°\n90°\n110°\n120°\n-72\n1152\n\nSOLUTION (12.4)\nKnown: A torsion bar spring serves as a counterbalance for a trap door with a given weight. The trap door must be designed to open only 60°. The maximum allowable torsional stress for the spring is 50 ksi.\nFind:\n(a) Determine the length and diameter of torsion bar that would counterbalance 80% of the door weight when closed, and provide a net torque of 12 lb-ft when open 60°.\n(b) Make a graph showing gravity torque, spring torque, and net torque all plotted against door opening angle.\n12-4 Schematic and Given Data:\nAssumptions:\n1. The bar is straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\n3. The cross section considered is sufficiently remote from points of load application and from stress raisers.\nAnalysis:\n1. From EQ. (4.4), for a solid round rod τ = 16T / πd^3 or d = (16T / π)^(1/3)\nwhen the door is closed, the bar will counter-balance 80% of the door weight. Thus, T_e = (0.80)(60 lb)(24 in) = 1152 lb in. Since a maximum allowable torsional stress is 50 ksi, d = (16(1152)/π(50,000))^(1/3) = 0.49 in.\n\n2. From Table 5.1, for a torsional case (case 2), θ = π / (JG) and K = T / (θJG / L)\nwhere J = πd^4 / 32 = 0.005660 in.4\nG = 11.5 x 10^6 psi (Appendix C-1)\nFor 60° opening, gravity torque = 60(24) cos 60° = 720 lb in.\nThus, spring torque required, T_e = 720 + 144 = 864 lb in.\nK = ΔT / Δθ = (1152 - 864) lb in. = 274.3 lb in./rad\n1.05 rad\nTherefore, L = JG / K = (0.005660)(11.5 x 10^6) / 274.3 = 237 in.\n12-5 3. Let clockwise torque be positive.\nGravity torque = 1440 cos θ\n720\n200\n400\n600\n800\n1000\n1200\n1400\n1600\n\nDoor opening angle, θ\n20°\n40°\n60°\n80°\n\nSOLUTION (12.5)\nModified Problem 12.3\nKnown: A torsion bar spring serves as a counterbalance for a trap door with a given weight. The maximum allowable torsional stress for the spring is 350 MPa.\nFind:\n(a) Determine the length and diameter of a solid steel torsion bar that would counterbalance 80% of the door weight when closed, and provide a 8 N·m torque holding the door against the stop.\n(b) Make a graph showing gravity torque, spring torque, and net torque all plotted against door opening angle.\nSchematic and Given Data:\nDoor stop\nR_D\n\n600 mm\n250 N\nT_e\nT_e = 8 N·m\nθ = 110°\n250 N\nR\n\nOpened\nClosed\n12-6 Assumptions:\n1. The bar is straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\n3. The cross section considered is sufficiently remote from points of load application and from stress raisers.\n\nAnalysis:\n1. From Eq. (4.4), for a solid round rod τ = 16T/πd^3 or d = (16T/πτ)^(1/3)\nWhen the door is closed, the bar will counterbalance 80% of the door weight. Thus, Tc = (0.80)(250 N)(0.60 m) = 120 N.m.\nSince a maximum allowable torsional stress is 350 MPa,\nd = ∛(16/16)(250 x 10^6) = 0.01204 m or d = 12.04 mm\n2. From Table 5.1, for a torsional case (Case 2) θ = TL/JG and K = T/θJG/L\n\nK = ΔT/Δθ = (120 - 8)/1.92 = 58.3 N.m/rad\n\nThus, L = JG/K = (2.063 x 10^-9)(79 x 10^9)/58.3 = 2.79 m\n\nLet clockwise torque be positive.\n\n200\nGravity torque = 150 cos θ\n150\n\nGravity +\nSpring torque\n\n-50\n\n-100\n\n-150\n\n-120\n\n12-7 Find:\n(a) Determine the length and diameter of torsion bar that would counterbalance 80% of the door weight when closed, and provide a net torque of 16 N.m when open (60°).\n(b) Make a graph showing gravity torque, spring torque, and net torque all plotted against door opening angle.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The bar is straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\n3. The cross section considered is sufficiently remote from points of load application and from stress raisers.\n\nAnalysis:\n1. From Modified Problem 12.4, d = 12.04 mm\n2. From Table 5.1, for a torsional case (case 2), θ = TL/JG and K = T/θJG/L\n\nwhere J = πd^4/32 = 2.063 x 10^-9 m4\nG = 79 x 10^9 Pa\nFor 60° opening, gravity torque = 250(0.6) cos 60° = 75 N.m\nThus, spring torque required, To = 75 + 16 = 91 N.m\n\nK = ΔT/Δθ = (120 - 91)/1.05 = 27.62 N.m/rad\n\nTherefore, L = JG/K = (2.063 x 10^-9)(79 x 10^9)/27.62 = 5.90 m\n\n12-8 3. Let clockwise torque be positive.\n\nSOLUTION (12.6)\nKnown: A torsion bar has a known length and diameter.\n\nFind:\n(a) Estimate the change in shear stress when one end of the rod rotates through 70° relative to the other end.\n(b) Estimate the change in torque.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The bar remains straight and the torque is applied about the longitudinal axis.\n2. The material is homogeneous and perfectly elastic within the stress range involved.\n\n12-9 Analysis:\n1. From Table 5.1, for a torsional case (Case 2):\n θ = TL / J therefore, T = θJG / L\n2. θ = (70/180)π = 1.22 rad\n J = πd^4/32 = π(0.312)^4/32 = 0.00093 in.^4\n From (Appendix C-1), G = 11.5 x 10^6 psi\n3. Therefore, T = 1.22 x 0.00093 x 11.5 x 10^6\n = 290.43 lb in.\n4. From Eq. (4.4), for a solid round rod,\n τ = 16T/πd^3 = (16)(290.43) / π(0.312)^3 = 48.7 ksi\n\nSOLUTION (12.7D)\nKnown: The web site http://www.indspring.com.\n\nFind:\n(a) List the mandatory specifications required to order a compression spring.\n(b) List other information or data important in ordering a compression spring.\n\nAnalysis: The web site provides the following information:\n(a) Mandatory specifications\n 1. OUTSIDE DIAMETER\n 2. INSIDE DIAMETER\n 3. Solid and Rate\n 4. Maximum solid height\n 5. Direction of helix\n 6. Type of ends\n\n(b) Advisory data\n 1. FREE LENGTH\n 2. Wire diameter in. (mm)\n 3. Mean coil diameter in. (mm)\n 4. No. of active coils\n 5. Total no. of coils\n\nSpecial Information\n 1. Type of Material\n 2. Finish\n 3. Squareness (free); within degrees\n 4. Frequency of compression, cycles/sec., and working range, in. (mm) to in. (mm) of length\n 5. Operating temp. degrees F (degrees C)\n 6. End use or application\n 7. Other\n\n12-10