Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Elementos de Fixaç o

SOLUTION (10.1)\nKnown: A double-threaded Acme screw of known major diameter is used in a jack having a plain thrust collar of known mean diameter. Coefficients of running friction are estimated as 0.10 for the collar and 0.13 for the screw.\nFind:\n(a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle of the screw.\n(b) Estimate the starting torque for raising and for lowering a 10,000 lb load.\n(c) If the screw is lifting a 10,000 lb load, determine the efficiency of the jack.\nSchematic and Given Data:\nAssumptions:\n1. The starting friction is about 1/3 higher than running friction.\n2. The screw is not exposed to vibration.\nAnalysis:\n1. From Table 10.3, there are 5 threads per inch.\np = 1 / 5 = 0.2 in.\nBecause of the double-threaded screw,\nL = 2p = 0.4 in.\nFrom Fig. 10.4a,\nThreaded depth = 0.5p = 0.10 in.\ndm = d - 0.5p = 0.90 in.\nFrom Eq. (10.1),\nα = tan⁻¹(L / rd) = tan⁻¹(0.4 / 0.90) = 8.05°\n2. For starting, increase the given coefficients of friction by 1/3:\nfc = 0.133, f = 0.173\nFrom Eq. (10.6),\nαn = tan⁻¹(tan α cos λ) = tan⁻¹(tan 14.5° cos 8.05°)\n= 14.36° From Eq. (10.4),\nT = Wdm / (2 πdm cos αa - rfL) + Wf e dc / 2\n= 10,000(0.90) [0.173π(0.90) + 0.4cos 14.36°]\n\t\t/ [π(0.90)cos 14.36° - 0.173(0.4)]\n= 10,000(0.133)(2.0)\n\t\t/ 2\n= 1477.5 + 1330 = 2807.5 lb-in. to raise the load\nFrom Eq. (10.5),\nT = Wdm / (2 πdm - Lcos α) + Wf e dc / 2\n= 10,000(0.90) [0.173π(0.90) - 0.4cos 14.36°]\n\t\t/ [π(0.90)cos 14.36° - 0.173(0.4)]\n= 10,000(0.133)(2.0)\n\t\t/ 2\n= 162.89 + 1330 = 1492.89 lb-in. to lower the load\nFrom Eq. (10.4), the friction free torque for raising the load is\nT = 10,000(0.90) (0.4)(cos 14.36°) / [π(0.90)cos 14.36°] = 636.6 lb-in\nEfficiency = 636.6/2264.5 = 28%\nWork input to the screw during one revolution = 2πT = 2π(2264.5) = 14228.8 lb-in\nWork output during one revolution = WL = (10000)(2)(0.2) = 4000 lb-in\nEfficiency = Work out/Work in = 4000/14228.8 = 28%\nComments:\n1. For a double threaded screw the work output during one revolution is WL where L = 2p.\n2. If a small thrust bearing were used so that the collar friction could be neglected, the efficiency would increase to 636.6/1264.5 = 50%. SOLUTION (10.2)\nKnown: A square-threaded, single thread power screw is used to raise a known load.\nThe screw has a mean diameter of 1 in. and four threads per inch. The collar mean diameter is 1.75 in. The coefficient of friction is estimated as 0.1 for both the thread and the collar.\nFind:\n(a) Determine the major diameter of the screw.\n(b) Estimate the screw torque required to raise the load.\n(c) If collar friction is eliminated, determine the minimum value of thread coefficient of friction needed to prevent the screw from overhauling.\nSchematic and Given Data:\nAssumption: The screw is not exposed to vibration.\nAnalysis:\n(a) From Fig. 10.4(c),\nd = dm + p / 2 = 1 + 0.25 / 2 = 1.125 in.\n(b) From Eq. (10.4a),\nT = Wdm / (2 πdm + L + Wf e dc / 2)\n= (13750)(1)(0.1)(1)(0.25)\n\t\t/ [π(1) - (0.1)(0.25)] / 2\n= 1245 lb-in + 1203 lb in. = 2448 lb in.\n(c) From Eq. (10.7), the screw is self-locking if\nf ≥ (L / πdm)0.25 / π(1) = 0.08 f \\geq 0.08\nTherefore, the minimum value of thread coefficient of friction needed to prevent\nthe screw from overhauling is 0.08.\n\nSOLUTION (10.3)\nKnown: A double-threaded Acme screw of known major diameter is used in a jack\nhaving a plain thrust collar of known mean diameter. Coefficients of running friction\nare estimated as 0.09 for the collar and 0.12 for the screw.\n\nFind:\n(a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle of\nthe screw.\n(b) Estimate the starting torque for raising and for lowering a 4000 N load.\n(c) If the screw is lifting a 4000 N load, determine the efficiency of the jack.\n\nSchematic and Given Data:\n\nLoad\n4,000 N\nDouble-threaded Acme screw\nd = 1 in.\nd_{c} = 50 mm\nf_{c} = 0.09\nf_{m} = 0.12\n\nAssumptions:\n1. The starting friction is about 1/3 higher than running friction.\n2. The screw is not exposed to vibration.\n\nAnalysis:\n1. From Table 10.3, there are 5 threads per inch.\np = \\frac{1}{5} = 0.2 in. = 0.0051 m\nBecause of the double-threaded screw,\nL = 2p = 0.4 in. = 0.0102 m,\nFrom Fig. 10.4a,\nThreaded depth = 0.5p = 0.10 in. = 0.00254 m\nd_{m} = d - 0.5p = 0.90 in. = 0.02286 m\nFrom Eq. (10.1),\n\\lambda = tan^{-1} \\left( \\frac{L}{\\pi d_m} \\right) = tan^{-1} \\left( \\frac{0.4}{0.907} \\right) = 8.05^{\\circ}\n10-4 2.\nFor starting, increase the coefficient of friction by 1/3:\nf c = 0.12, f = 0.16\nFrom Eq. (10.6),\n\\alpha = tan^{-1} (tan \\alpha cos \\lambda) = tan^{-1} (tan 14.5^{\\circ} cos 8.05^{\\circ})\n= 14.36^{\\circ}\nFrom Eq. (10.4),\nT = \\frac{W_{dm}}{2[\\pi d_m cos \\alpha + L cos \\alpha - f_L]} + \\frac{W_f d_c}{2}\n= \\frac{4000(0.02286)}{2[\\pi(0.02286)\\cdot 0.16\\cdot 0.02286 + 0.0102 cos 14^{\\circ}]}\n+ \\frac{4000(0.12)(0.05)}{2}\n= 7.175 + 12 = 19.175 N-m. to raise the load\nFrom Eq. (10.5), \nT = \\frac{W_{dm}}{2[\\pi d_m + L - L cos \\alpha]} + \\frac{W_f d_c}{2}\n= \\frac{4000(0.02286)}{2[\\pi(0.02286)\\cdot 0.16 + 0.0102 cos 14^{\\circ}]} + \\frac{4000(0.09)(0.05)}{2}\n= -5.819 + 12 = 6.18 N-m. to lower the load\n3.\nFrom Eq. (10.4) with f_c = 0.09, f = 0.12\nT = \\frac{4000(0.02286)}{2[\\pi(0.02286)\\cdot 0.12 + 0.0102 cos 14^{\\circ}]} + \\frac{4000(0.09)(0.05)}{2}\n= 7.00 + 9 = 16.00 N-m\n4.\nFrom Eq. (10.4), the friction free torque for raising the load is\nT = \\frac{4000(0.02286)}{2[\\pi(0.02286) cos 14^{\\circ}]}\n= \\frac{4000(0.012)(0.0102) cos 14^{\\circ}}{2[\\pi(0.02286)cos 14^{\\circ}]} = 6.49 N-m\n5. Efficiency = \\frac{6.49}{16.00} = 40.6% \n6. Work input to the screw during one revolution is WL = 2\\pi(16.00) = 100.5 N-m\n7. Work output during one revolution is WL = (4000)(0.0051) = 40.8 N-m\n8. Efficiency = Work out/Work in = \\frac{40.8}{100.5} = 40.6%\n10-5 Comments:\n1. For a double threaded screw the work output during one revolution is WL where L = 2p.\n2. If a small thrust bearing were used so that the collar friction could be neglected,\nthe efficiency would increase to 6.497/7.00 = 92.7%.\n\nSOLUTION (10.4)\nKnown: A square-threaded, single thread power screw is used to raise a known load.\nThe screw has a mean diameter of 1 in. and four threads per inch. The collar mean\ndiameter is 1.5 in. The coefficient of friction is estimated as 0.1 for both the thread\nand the collar.\n\nFind:\n(a) Determine the major diameter of the screw.\n(b) Estimate the screw torque required to raise the load.\n(c) If collar friction is eliminated, determine the minimum value of thread coefficient\nof friction needed to prevent the screw from overhauling.\n\nSchematic and Given Data:\n\nLoad\n25,000 lb\nd_{g} = 1 in.\n4 threads/in.\nd_{c} = 1.5 in.\nf_{c} = 0.1\n\nAssumption: The screw is not exposed to vibration.\n\nAnalysis:\n(a) From Fig. 10.4(c):\nd = d_{m} + \\frac{P}{2} = 1 + \\frac{0.25}{2} = 1.125 in.\n\n(b) From Eq. (10.4a),\nT = \\frac{W_{dm}}{2[\\pi d_m + L]} + \\frac{W_f d_c}{2}\n10-6 (25000)(1) (0.1)\\pi(1) + 0.25\\n 2\\n = 2633 lb in. + 1875 lb in = 4138 lb in. \\n(c) From Eq. (10.7a), the screw is self-locking if\\n f \\u2265 \\n L \\u00b7 \\n Tdm = 0.25\\n \\pi(1) = 0.08\\n f \\u2265 0.08\\n Therefore, the minimum value of thread coefficient of friction needed to prevent the screw from overrhauling is 0.08. Analysis:\\n1. From Table 10.3, there are 4 threads per inch.\\n Dm = 1/4 = 0.25 in.\\n Because of the double-threaded screw,\\n L = 2p = 0.50 in.\\n From Fig. 10.4 (b),\\n Threaded depth = 0.3p = 0.075 in.\\n dm = d - 0.3p = 1.925 in.\\n From Eq. (10.1),\\n \\lambda = tan^{-1} \\left( \\frac{L}{Tdm} \\right) = tan^{-1} \\left( \\frac{0.5}{1.925} \\right) = 4.73º\\n 2. For starting, increase the coefficients of friction by 1/3:\\n f_e = 0.133, f = 0.147\\n From Eq. (10.6),\\n \\alpha = tan^{-1} \\left( \\tan \\alpha cos \\lambda \\right) = tan^{-1} \\left( \\tan 14.5º cos 4.73º \\right) = 14.45º\\n From Eq. (10.4),\\n T = \\frac{Wdm}{2} \\left( \\frac{\\pi Tdm - L cos \\alpha}{Tdm\\cos \\alpha + fL} \\right) + \\frac{W_fd_e}{2}\\n = \\frac{35000(1 925)}{2}\\n \\left( \\frac{0.147}{\\pi (1 925)\\cos 14.5º - 0.147(0.5)} + \\frac{35000(0.133)(2.75)}{2} \\right)\n = 799.9 + 640.1 = 1440 lb in. to raise the load\\n From Eq. (10.5),\\n T = \\frac{Wdm}{2}\\n = \\frac{35000(1 925)}{2} \\times \\left(\\frac{0.11(\\pi(1 925)} + 0.5cos 14.5º - 0.11(0.5)} \\right)\\n = 667.1 + 481.3 = 1148.4 lb in.\\n From Eq. (1.3),\\n \\dot{W}_{in} = \\frac{T_n}{5252} = \\frac{(1148.4/12)96}{5252} = 1.75 hp W_{out} = \\frac{(3500)(4)}{33000} = 0.424 hp\\n Therefore, efficiency = \\frac{0.424}{1.75} = 0.24 = 24%\\n 4. From Eq. (10.7), the screw is self-locking if\\n L \\cdot cos \\alpha_n = 0.5(cos 14.5º) = 0.08\\n Thus, if f = 0.11, the screw is self-locking and not overhauling. From Eq. (10.4a),\n\nT = Wdm \n2 \n[ f / (rdm + L) + Wf / (rdm - fL) ]\n= (125000)(1) \n2 [ (0.1)(2)(1) + 0.25 \n / (rdm)(1) - (0.1)(0.25)]\n= 1132 lb in. + 937 lb in. = 2069 lb in.\n\nFrom Eq. (10.7a), the screw is self-locking if\n\nf >= L / (rdm 0.25 / pi(d)) = 0.08\nf >= 0.08\n\nTherefore, the minimum value of thread coefficient of friction needed to prevent\nthe screw from overhauling is 0.08.\n\nSOLUTION (10.7)\nKnown: A double-threaded Acme stub screw of known major diameter is used in a\njack having a plain thrust collar of known mean diameter. Coefficients of running\nfriction are estimated as 0.10 for the collar and 0.11 for the screw.\n\nFind:\n(a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle of\nthe screw.\n(b) Estimate the starting torque for raising and for lowering a 5000 lb load.\n(c) If the screw is lifting a 5000 lb load at the rate of 4 ft/min, determine the screw\n\n(d) Also determine the efficiency of the jack under this steady state condition.\n\nDetermine if the screw will overhaul of a ball thrust bearing (of negligible friction)\nwere used in place of the plain thrust collar. Assumptions:\n1. The starting friction is about 1/3 higher than running friction.\n2. The screw is not exposed to vibration.\n\nAnalysis:\n1. From Table 10.3, there are 4 threads per inch.\np = 1/4 = 0.25 in.\nBecause of the double-threaded screw,\nL = 2p = 0.50 in.\nFrom Fig. 10.4 (b),\nThreaded depth = 0.3p = 0.075 in.\ndm = d - 0.3p = 1.925 in.\nFrom Eq. (10.1),\nλ = tan^{-1} (L / rdm) = tan^{-1} (0.5 / 1.925) = 4.73°\n\n2. For starting, increase the coefficients of friction by 1/3:\nfc = 0.133, ft = 0.147\nFrom Eq. (10.6),\nαn = tan^{-1} (tan α cos λ) = tan^{-1} (tan 14.5° cos 4.73°)\nFrom Eq. (10.4),\nT = Wdm \n2 \n[ f / (rdm - Lcos α) + Wf / (rdm cos α - fL)]\n= (5000(1.925)(0.147)(1.925) + 0.5cos 14.45° \\\n * 5000(0.133)(2.5))\n\n= [π(1.925)cos 14.45° - 0.147(0.5)] \n\n= 1142.7 + 831.3 = 1974 lb in. to raise the load\nFrom Eq. (10.5),\nT = Wdm \n2 \n[ f / (rdm - Lcos α) + Wf / (rdm cos α + fL)]\n\n= (5000(1.925)(0.11)(1.925) + 0.5cos 14.45° \n\n= 328.5 + 831.3 = 1160 lb in. to lower the load.\n\n4(12) in.\n\n0.5 in. / rev = 96 rpm\nFrom Eq. (10.4), with f = 0.1, ft = 0.11,\nT = (5000/(1.925)) * [0.11(1.925) + 0.5cos 14.45°] +\n5000(0.1)(2.5)\n\n= 953 + 625 = 1578 lb in. From Eq. (1.3),\n\nWin = (1578/12)/96 = 2.40 hp\n\n(5000)^4\n\nWout = 33000 = 0.606 hp\n\nTherefore, efficiency = 0.606 = 25%\n\n4. From Eq. (10.7), the screw is self-locking if\n\nf >= Lcos α / (rd) = 0.5(cos 14.45°) \nrd = π(1.925) = 0.08\n\nThus, if ft = 0.11, the screw is self-locking and not overhauling.\n\nSOLUTION (10.8)\nKnown: A jack uses a single square-thread screw to raise a known load. The major\ndiameter and pitch of the screw and the thrust collar mean diameter are known.\nRunning friction coefficients are estimated.\n\nFind:\n(a) Determine the thread depth and helix angle.\n(b) Estimate the starting torque for raising and lowering the load.\n(c) Estimate the efficiency of the jack for raising the load.\n(d) Estimate the power required to drive the screw at a constant 1 revolution per\nsecond. Analysis:\n(a) From Fig. 10.4(c),\nThread depth = p/2 = 6/2 = 3 mm\nFrom Eq. (10.1),\n\\lambda = tan^{-1}(L / \\pi d_m) where d_m = d - p/2 = 33 mm\n\\lambda = tan^{-1}(6 / \\pi(33)) = 3.31°\n(b) For starting, increase the coefficients of friction by 1/3, then\nf = 0.20, fc = 0.16\nFrom Eq. (10.4a),\nT = Wdm / 2 + L / \\pi d_m + Wf e dc / 2\n= (50,000)(0.033)\n= 2 (50,000)(0.16)(0.080)\n/ \\pi(0.033)(0.20)(0.006) + 2\n= 215 + 320 = 535 N*m to raise the load\nFrom Eq. (10.5a),\nT = Wdm / 2 + L / \\pi d_m + Wf e dc / 2\n= (50,000)(0.033)(0.20)(0.033)\n= 116 + 320 = 436 N*m to lower the load\n(c) From Eq. (10.4a), with f = 0.15, fc = 0.12,\nT = (50,000)(0.033)\n= 173 + 240 = 413 N*m\nWork input to the screw during one revolution\n= 2\\piT = 2\\pi(413) = 2595 N*m\nWork output during one revolution\n= W * p = (50,000)(0.006) = 300 N*m\nEfficiency = Work_out Work_in = 300 2595 = 11.6%\n10-13 Analysis:\n1. From section 10.3.1, and considering that service conditions may be conducive to relatively high friction, estimate f = fc = 0.15 for running friction.\n2. From Table 10.3, p = 0.1 in., and with a single thread, L = 0.1 in.\n3. From Fig. 10.4(a),\n\\bar{d_m} = d = p / 2 = 0.5 - 0.05 = 0.45 in.\n\\alpha = 14.50°\nFrom Eq. (10.1),\n\\lambda = tan^{-1}(L / \\pi d_m) = tan^{-1}(0.1 / \\pi(0.45)) = 4.05°\nFrom Eq. (10.6),\n\\alpha_n = tan^{-1}(tan \\alpha cos \\lambda) = tan^{-1}(tan 14.5° cos 4.05°) = 14.47°\n(Note: with h = 40°, it is obvious that \\alpha_n = \\alpha and well within the accuracy of assumed friction coefficients)\n4. From Eq. (10.4),\nT = Wdm / 2 + L / \\pi d_m cos \\alpha + Wf e dc / 2\n= (200)(0.45)\n= (0.15)(0.45) + 0.1(cos 14.47°) + (200)(0.15)(0.625)\n= 10.27 + 9.37 = 19.64 lb in. Use T = 20 lb in.\nAt the end of a 5-in. handle, the clamping force required = 20 / 5 = 4 lb\n10-15 Check:\nTorque during load raising with f = fc = 0\nT = (50,000)(0.033) 0 + 0.006 + 0\n= 47.8 N*m\nEfficiency = T(with zero frictions) T(actual) = 47.8 413 . 11.6%\nCheck (partial):\nTorque during load raising if collar friction is eliminated = 173 N*m\nEfficiency (screw only) = 47.8 173 = 28%\n(d) From Eq. (1.2),\nW = nT 9549 = 60(413)\n9549 = 2.6 kW\nSOLUTION (10.9)\nKnown: An ordinary C-clamp uses a 1/2 in. Acme thread and a collar of 5/8 in. mean diameter.\nFind: Estimate the force required at the end of a 5-in. handle to develop a 200 lb clamping force.\nSchematic and Given Data:\nAssumptions:\n1. Coefficients of running friction are estimated as 0.15 for both the collar and the screw.\n2. The screw has a single thread.\n10-14 Schematic and Given Data:\n\nAssumption: Collar friction can be neglected.\n\nAnalysis:\n1. From Table 10.3,\np = L = 1/(1.75) = 0.571 in.\n2. From Fig. 10.4(d),\nd_m = d - P/2 = 3 - 0.571/2 = 2.71 in.\nand α = 2.5°\n3. Since cos α = 0.999, use Eq. (10.4a):\nT = W_dm/2 * (πd_m + L)/(πd_m - r_L) + W_f* d_c/2\n= (52,000)(2.71)/2 * (0.1)/(π(2.71) - (0.571)) + 0\n= 11,851 lb in. = 988 lb ft (during load raising)\n4. To raise the gate 36 in./min with L = 0.571 in.\nrequires 36/0.571 = 63.05 = 63 rpm\n5. From Fig. (1.3),\nẆ = (63.05)(988) / 5252 = 11.9 hp say 12 hp\nTherefore, 12 hp are required to drive each screw.\n10-16 6. Check:\nWork output per gate = 52,000 lb (3 ft/min)\n= 156,000 lb ft/min = 4.73 hp\nλ = tan⁻¹(L/πd_m) = tan⁻¹(0.571/π(2.71)) = 3.84°\nFrom Fig. 10.8,\nEfficiency = 40 %\nThus, W_required = 4.73 / 0.4 = 11.83 hp\n\nSOLUTION (10.11)\nKnown: The power screws in Problems (a) 10.4, (b) 10.7, (c) 10.8, (d) 10.9, and (e) 10.10 are given.\nFind: Calculate the nominal values of torsional, axial, thread bearing, and thread shear stresses.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The length of thread engagement is 1.5 times the outside diameter of the screw.\n2. For torsional stress, assume effective diameter = d_r, then τ = 16T / πd_r³.\n3. For axial stress, where A_t is not given,\nassume σ = 4W / πd_r².\n4. For bearing stress, assume d_i ≈ d_r,\nthen σ = 4W / π(d² - d_r²).\n10-17 5. For thread shear stress, assume d_i = d_r, and define j as the fraction of nut thickness (in this problem, 1.5d) involved in screw thread shear fracture surface B of Fig. 10.11. Then τ = W / (A * πd(i * 1.5d));\n\nAnalysis:\n(a) Assume starting friction = 1/3 running friction, or f = 0.133.\nFrom Eq. (10.4a),\nT = W_dm/2 * (πd_m + L)/(πd_m - r_L) + W_f * d_c / 2\n= 25,000(1) * [0.133(1) + 0.25]/\n2 * π(1) + 25,000(0.133)(1.5) / 2\n= 2686 + 2494 = 5180 lb in.\n(b) From Prob. 10.7, T = 1973 lb in.\nFrom Fig. 10.4(b):\nj = (0.5)p + 0.3p tan 14.5°/p = 0.58 in.\n(c) From Prob. 10.8,\nT = 535 Nm\n(d) Using starting friction = f = 4/3 (0.15) = 0.20 in place of running friction,\nf = 0.15 in Prob. 10.9 solution:\nτ = (200)(0.45)/(2)\n[ (0.2)π(0.45) + 0.1(cos 14.47°)]/π(0.45)(cos 14.47°)\n= 9.48 + 12.50 = 21.98 ≈ 22 lb in.\nFrom Fig. 10.4(a);\n(e) Using starting friction of f = 4/3 (0.1) = 0.133,\n\tτ = (52,000)(2.71)/2 * [0.133(2.71) + 0.571] / [π(2.71) - (0.133)(0.571)]\n= 14,223 lb in.\nFrom Fig. 10.4(d):\nd_r = d - p = 3 - 1/1.75 = 2.43 in.\n10-18 Prob Part W T Load Torque d af j Torsion stress_t Axial stress_a Bearing stress_b Shear stress_s\n(a) 25,000 5180 1.125\" 0.875\" 0.5 3.94 ksi 41.6 ksi 63.6 ksi 10.8 ksi\n(b) 5000 1993 2\" 1.85\" 0.58 1.6 ksi 1.9 ksi 11.0 ksi 0.49 ksi\n(c) 50 KN 535 No36 mm 30 mm 0.5 101 MPa 70.7 MPa 160.7 MPa 19.6 MPa\n(d) 22 0.5\" 0.43 1.81 ksi 2.8 ksi 0.34 ksi\n(e) 52000,144.23\" 2.43\" 2.5 ksi 11.2 ksi 2.9 ksi\nComment: In the above table, the units of load, W, are lb, and the units of torque, T, are lb-in, unless specified otherwise.\n\nSOLUTION (10.12D)\nKnown: Several different types of fasteners are used that prevent or resist unauthorized removal.\n\nFind: Examine and sketch several fasteners used in vending machines, computers, and other items that prevent or resist unauthorized removal.\n\nAnalysis:\n1. The article, Twenty Tamper-proof Fasteners, by Federico Strasser appearing in Product Engineering Design Manual edited by Douglas C. Greenwood, McGraw-Hill, 1959, p. 72-73, illustrates twenty tamper-proof fasteners.\n2. The abstract of the article states: Ways to prevent or indicate unauthorized removal of fasteners in vending machines, instruments, radios, TV sets and other units. Included are positively retained fasteners to prevent loss where retrieval would be difficult.\n\nThe author provides the following descriptions:\n1a. Wax or other suitable material fills recess above screw. Wax flush with plate hides screw position if surface is painted.\n1b. Cupped sleeve riveted in screw hole provides cavity for wax when plate is too thin for recessing.\n1c. Pin prevents rotation of square cup which would allow screw to be removed without disturbing wax.\n2a. Lead seal crimped over twisted ends of wire passing through screw allows only limited slackening of nut.\n2b. Two or more screws strung through heads with wire are protected against unauthorized removal by only one seal. Code or other signet can be embossed on seals during crimping.\n3. Sheet-metal disk pressed into groove can only be removed with difficulty and discourages tampering.\n\n10-19 4a. Spanner-head screws are available in all standard heads and sizes from U.S. manufacturers. Special driver is required for each screw size except 1/4-in. diameter and above.\n4b. Left-handed screw thread is sometimes sufficient to prevent unauthorized loosening.\n4c. Special head lets screw be driven but not unscrewed.\n5. Tapped cover and casing allows screw with reduced shank diameter to be completely unscrewed from casing yet retained positively in cover. For thin sheet-metal covers, split ring on reduced shank is preferable. Snap ring in groove or transverse pin are effective on unreduced shank. Simple and cheap method is fiber washer pushed over thread.\n6. Open-ended slot in sliding cover allows screw end to be staked or burred so screw cannot be removed, once assembled.\n7. Nut is retained on screw by staking or similar method, but if removal of nut is occasionally necessary, coaxial binding-head screw can be used. Where screw end must be flush with nut, pin through nut tangential to undercut screw limits nut movement. Rotatable nut or screw should have sufficient lateral freedom to accommodate slight differences in location when two or more screws are used.\n\nSOLUTION (10.13D)\nKnown: There are various types of commercially available locknuts.\n\nFind: Give examples of locknuts for (1) pins, keys, tabs, safety wire, (2) deformed threads, (3) secondary spring elements, (4) frictional interference, and (5) free spinning until seated.\n\nAnalysis:\n1. The article, The Fundamentals of Selecting Locknuts, by C.C. Feroni, in Product Design Engineering Manual, edited by Douglas C. Greenwood, McGraw-Hill, 1959, p. 47-49, lists and illustrates the basic classes of commercial locknuts.\n2. The author provides the following discussion for each basic class.\n\nCLASS I - Pins, Keys, Tabs, Safety Wire\nAssembly is more costly than for other types, and there is the possibility of forgetting the locking member at assembly. Also, the very vibration which the locknuts are intended to withstand may cause fatigue failure of the pin, tab or wire. On the other hand, these types of nuts are most satisfactory where some relative motion exists between the respective part of the joint.\n\nCLASS II - Deformed Threads\nWidely used for moderate service conditions because of low original cost. Several factors limit their performance: (1) Locking friction is a result of high pressure in a few localized places; (2) Performance is usually erratic; (3) Cyclic loading and vibration can materially wear the interference points. Often merely loading the nut causes local yielding and a loss of locking torque. Also, deformed threads do not lend themselves to frequent reuse.\n\n10-20 CLASS III - Secondary Spring Elements\nThese often give an attractive balance between cost and performance. Initial cost is usually low and reliability is adequate for many applications. Limitations: (1) Spring member may fail from vibrations; (2) Locking effectiveness is reduced or even lost if bolt stretches or mating surfaces wear; (3) Most nuts of this type tend to score the surfaces on which they bear.\n\nCLASS IV - Frictional Interference\nHighest performance, but more expensive than any of the other classes. Locking action comes from plastic deformation of elements of the nut itself. These elements are either non-metallic inserts, or slotted collars. With either, spring-back tendency which produces action is uniform and does not depend solely upon bolt tension. Also these types need not be seated to lock. Reusability is high. Operating temperature is limited.\n\nCLASS V - Free Spinning\nThese include types that are free to spin until seated. Advantages: Easy and inexpensive installation; easily removed and replaced. Disadvantages: An opening of joint or loss of bolt tension converts them into free running plain nuts. Also, since the clamping force is in addition to bolt stress and is usually confined to the lower threads where load is already high, combined stress in these threads lowers the fatigue and impact resistance.\n\nSOLUTION (10.14D)\nKnown: There are various types of commercially available locknuts.\n\nFind: Develop a list of ten factors that should be considered in selecting the class of locknut that should be used.\n\nAnalysis:\n1. Can the risk be run that occasionally something may accidentally be forgotten in assembly? Class I nuts depend on a secondary element, such as a lock washer or a cotter pin. Not only might this be overlooked in assembly, but lock wires or cotter pins can break under severe vibration. Without the secondary element, these nuts are free to spin as plain nuts. There seems to be little justification for the popularity that still exists for this class. Probably it is partially inertia in resisting a change from practice and partially a reluctance to trust what seems like less positive methods. Actually many types often perform better.\n2. Can the nut be seated tight against the work? Class V types lock only when they are tightly seated against one of the surfaces being joined. If any relaxation occurs, they are free to turn off. Such relaxation can occur due to creep of the bolt, wear or corrosion of mating surfaces.\n\n10-21 3. Is locking pressure spread evenly, or concentrated on a few threads? If vibration is severe, the friction which prevents a nut from loosening should be spread over as large an area as possible. This requires expensive, precise manufacture. Some nuts lock with interference between a few threads or load some threads more than others.\n4. Are all nuts of a given type equally reliable? Some classes of nuts are not uniform in their locking ability. The amount of distortion of shape which causes locking may vary from nut to nut. Also, these types and constants of the normal variations of bold diameter within normal tolerance limits. Thus, on bolts with diameters near the low limit, some nuts may not lock at all.\n5. Will the joint be exposed to high temperatures? Some types of Class I locknuts have plastic or other non-metallic inserts to obtain locking action, and are not recommended for use above 250°F.\n6. Will the nut be frequently removed and reused? Often, bolted joints must be broken periodically for inspection, repairs, access, or maintenance. The types of nuts which jam a few threads together for locking can damage bolt threads to the extent that the joint cannot be remade unless bolts are replaced.\n7. Is speed of assembly important? It obviously takes more assembly time for nuts which require extra motions to lock, such as insertion of cotter pins or lock wires. With large volume production, this is an important factor.\n8. Is ease of assembly important? When a locknut is required for a relatively inaccessible location, the free spinning type, Class V, may be preferred.\n9. Will the nut damage the bolt or the work surfaces? When a joint design is such that maximum tightening is required, stress that results from jammed-on nuts can cause should be.\n10. Is there relative motion between parts bolted together? If this is so, a castellated nut with pin or lock wire may be best, since repeated rotary motion might loosen other types.\n\nSOLUTION (10.15D)\nKnown: The web site http://www.nutty.com/ lists different kinds of (a) nuts, (b) bolts, and (c) washers.\n\nFind: Review the web site and list the different kinds of (a) nuts, (b) bolts, and (c) washers. Comment on how to evaluate the products promoted on a web site.\n\nAnalysis:\n(a) nuts\n hex nut, nylon insert lock nut, grade 5 hex nut, jam nylon insert lock nut, grade 8 hex nut, grade 8 torque lock nut, jam nut, grade 8 flange torque lock nut, heavy hex nut, 2-way lock nut, machine screw nut, left hand nut, acorn (cap) nut, tee nut, acme nut, rod coupling nut, slotted nut, reducer rod coupling nut, castle nut, k-lock (keps) nut, 2H heavy hex nut, flange lock nut, wing nut, square nut, wood insert nut, cage nut.\n(b) bolts\n grade 2 hex head cap screw, carriage bolt, grade 5 hex head cap screw, lag bolt, grade 8 hex head cap screw, tap (full thread) bolt, grade 8 flange frame bolt, grade 5 shaker screen bolt, square head bolt, grade 8 shaker screen bolt, grade 5\n10-22 carriage bolt, grade 8 plow bolt, hanger bolt, step bolt, elevator bolt, A325 structural bolt, L - anchor bolt, A490 structural bolt, J - bolt.\n(c) washers\n USS flat washer, grade 8 USS flat washer, SAE flat washer, grade 8 SAE flat washer, shim washer, fender washer, split lock washer, grade 8 split lock washer, finishing (cup) washer, galvanized/rubber bonding, dock washer, square bevel washer, internal tooth lock washer, external tooth lock washer, high collar split lock washer.\n\nSOLUTION (10.16D)\nKnown: The web site http://www.boltscience.com/ provides information related to bolt technology.\n\nFind: Review the site information related to bolted joint technology and answer the following questions:\n(a) Is vibration the most frequent cause of bolt/nut loosening? If not, what is the most frequent cause of loosening?\n(b) What are three common causes of relative motion in threads?\n(c) Can conventional spring lock washers be used to prevent self loosening when bolts without lock washers would loosen because of relative motion?\n(d) What is prevailing torque?\n(e) What are direct tension indicators?\n\nAnalysis: The web site provides the following answers:\n(a) It is widely believed that vibration causes bolt loosening. By far the most frequent cause of loosening is side sliding of the nut or bolt head relative to the joint, resulting in relative motion occurring in the threads. If this does not occur, then the bolts will not loosen, even if the joint is subjected to severe vibration.\n(b) There are three common causes of the relative motion occurring in the threads:\n 1. Bending of parts which results in forces being induced at the friction surface. If slip occurs, the head and threads will slip which can lead to loosening.\n 2. Differential thermal effects caused as a result of either differences in temperature or differences in clamped materials.\n 3. Applied forces on the joint can lead to shifting of the joint surfaces leading to bolt loosening.\n\n(c) Conventional spring lock washers are no longer specified, because it has been shown that they actually aid self loosening rather than prevent it.\n\n(d) The prevailing torque is the torque required to run a nut down a thread on certain types of nuts designed to resist vibration loosening. The resistance can be provided by a plastic insert or a non circular head. Threads coated with an adhesive also exhibit a prevailing torque.\n\n(e) Direct Tension Indicators (DTIs) is a term sometimes used to describe load indicating washers. Projections on the face of the washer (usually on the face abutting the bolt head or nut) that deform under loading as the bolt is tensioned. An indication of the tension in the bolt can be made by measuring the gap\n10-23 between the washer face and the nut or bolt head. The smaller the gap - the greater the tension in the bolt. Load indicating washers are commonly used in civil rather than mechanical engineering applications.\n\nComment: The task of developing a procedure for evaluating the integrity of information provided by a web site is left as an exercise for a bushy-tailed student. We suggest starting by defining the word \"integrity\". As with most all problem solving, we first need to understand and/or define the problem.\n\nSOLUTION (10.17)\nKnown: The bolt shown is made from cold drawn steel. The load fluctuates continuously between 0 and 8000 lb.\n\nFind:\n(a) The minimum required value of initial load to prevent loss of compression of the plates.\n(b) The minimum force in the plates for the fluctuating load when the preload is 8500 lb.\n\nSchematic and Given Data:\n\nF_e\n F_c = 0 to 8,000 lb\n k_e = 6k\n\nAssumption: The bolt, nut, and plate materials do not yield.\n\nAnalysis:\n1. Compression of the plates is lost when F_c = 0 when maximum load is applied. From Eq. (10.13)\n F_i = F_e + [ k_c / (k_b + k_c) ] F_e\n = 0 + [ 6k / (k_b + 6k) ] 8,000 = 6,857 lb\n2. Minimum force in plates occurs when fluctuating load is maximum. From Eq. (10.13)\n F_c = F_i [ - k_c / (k_b + k_c) ] F_e\n10-24 SOLUTION (10.18)\nKnown: The bolt shown is made from cold drawn steel. The load fluctuates continuously between 0 and 8000 lb.\nFind:\n(a) The minimum required value of initial load to prevent loss of compression of the plates.\n(b) The minimum force in the plates for the fluctuating load when the preload is 8500 lb.\nSchematic and Given Data:\nAssumption: The bolt, nut, and plate materials do not yield.\nAnalysis:\n1. Compression of the plates is lost when Fc = 0 when maximum load is applied.\nFrom Eq. (10.13),\nFi = Fc + (kc / (kb + kc)) Fe\n= 0 + (4 / 5) 8,000 = 6,400 lb\n\n2. Minimum force in plates occurs when fluctuating load is maximum.\nFrom Eq. (10.13),\nFc = Fi * (kc / (kb + kc)) Fe\n= 8,500 * (4 / (kb + 4kb)) 8,000 = 8,500 - 6,400 = 2,100 lb SOLUTION (10.19)\nKnown: In a given assembly, two parts are clamped together by a bolt. The ratio of the clamped member stiffness and the bolt stiffness is given. The initial bolt tension and the range of the fluctuating external load are also given.\nFind: Draw a graph (plotting force vs. time) showing three or four external load fluctuations, and corresponding curves showing the fluctuations in total bolt load and total joint clamping force.\nSchematic and Given Data:\nAssumption: The bolt size and material are such that the bolt load remains within the elastic range.\nAnalysis:\n1. The total bolt load when an external load is applied is, from Eq. (10.13),\nFb = Fi + (kb / (kb + kc)) Fe = 1100 + (1 / (1 + 6)) (6000)\n= 1957 lb\nFc = Fi - (kc / (kb + kn)) Fe = 1100 - (6 / (7)) (6000)\n= -4043 lb, since -4043 < 0, Fc = 0 lb and Fb = 6000 lb\n\n2. When Fc = 0, separation takes place: 1100 - (6 / 7) Fe = Fc = 0 and thus\nFe = (7 / 6) (1100) = 1283 lb\n\n3. With no external load: Fb = Fc = Fi SOLUTION (10.20)\nKnown: In a given assembly, two parts are clamped together by a bolt. The ratio of the clamped member stiffness and the bolt stiffness is given. The initial bolt tension and the range of the fluctuating external load are also given.\nFind: Draw a graph (plotting force vs. time) showing three or four external load fluctuations, and corresponding curves showing the fluctuations in total bolt load and total joint clamping force.\nSchematic and Given Data:\nAssumption: The bolt size and material are such that the bolt load remains within the elastic range.\nAnalysis:\n1. Using Eq. (10.13) for Fe = 6000 lb, Fb = Fi + k_b \\frac{F_e}{k_b + k_c} = 1100 + \\frac{1}{1 + 3} (6000) = 2600 lb\nFc = Fi - k_e \\frac{F_e}{k_e + k_b} = 1100 - \\frac{3}{4} (6000) = -3400 lb\nSince -3400 < 0, Fc = 0 and Fb = 6000 lb\n2. When Fc = 0, separation takes place: 1100 - \\frac{3}{4} F_e = Fc = 0 and thus\nFc = \\frac{4}{3} (1100) = 1467 lb\n3. For Fc = 0 lb, Fb = Fc = Fi\n4. Schematic and Given Data:\nAssumption: The bolt size and material are such that the bolt load remains within the elastic range.\nAnalysis:\n1. The total bolt load when an external load is applied is, from Eq. (10.13),\nFb = Fi + k_b \\frac{F_e}{k_b + k_c} = (5000)(10) + \\frac{1}{1 + 4} (20,000) = 54,000 N\nFc = Fi - k_e \\frac{F_e}{k_e + k_f} = 50,000 - \\frac{4}{5} (20,000) = 34,000 N\n2. SOLUTION (10.22)\nKnown: The cylinder head of a piston-type air compressor is held in place by ten bolts. Total joint stiffness is four times total bolt stiffness. Each bolt is tightened to an initial tension of 5000 N. The total external force acting to separate the joint fluctuates between 10,000 and 20,000 N.\nFind: Draw a graph (plotting force vs. time) showing three or four external load fluctuations, and draw corresponding curves showing the fluctuations in total bolt load and total joint clamping force.\nSchematic and Given Data:\nAssumption: The bolt size and material are such that the bolt load remains within the elastic range.\nAnalysis:\n1. Using Eq. (10.13) for F_e = 20,000 N,\nFb = Fi + k_b \\frac{F_e}{k_e + k_d} = ((5000)(10)) + \\frac{1}{1 + 4} (20,000) = 54,000 N\nFc = Fi - k_e \\frac{F_e}{k_e + k_b} = 50,000 - \\frac{4}{5} (20,000) = 34,000 N\n2. For F_e = 10,000 N,\nFb = 50,000 + \\frac{1}{1 + 4} (10,000) = 52,000 N\nFc = 50,000 - \\frac{4}{5} (10,000) = 42,000 N SOLUTION (10.23)\nKnown: Two parts of a machine are held together by bolts that are initially tightened to provide a total initial clamping force of 10,000 N. The elasticities are such that k_c = 2k_b.\n\nFind:\n(a) Determine the external separating force that would cause the clamping force to be reduced to 1000 N.\n(b) If this separating force is repeatedly applied and removed, determine values of mean and alternating force acting on the bolts.\n\nSchematic and Given Data:\n\nAssumption: The stress on the bolt is within the elastic limit.\n\n10-31 Analysis:\n(a) From Eq. (10.13),\nF_C = F_I - \\left( \\frac{k_c}{k_c + k_b} \\right) F_e: \n1000 = 10,000 - \\left( \\frac{2}{2 + 1} \\right) F_e: 9000 = \\frac{2}{3} F_e\nHence, F_e = 13,500 N\n\n(b) Load off:\nF_b = F_I = 10,000 N\nLoad on:\nF_b = 10,000 + \\frac{1}{3}(13,500) = 14,500 N\n\nF_m = \\frac{10,000 + 14,500}{2} = 12,250 N\nF_a = \\frac{14,500 - 10,000}{2} = 2250 N\n\nSOLUTION (10.24)\nKnown: Two parts of a machine are held together by bolts that are initially tightened to provide a total initial clamping force of 2000 lb. The elasticities are such that k_c = 5k_b.\nFind:\n(a) Determine the external separating force that would cause the clamping force to be reduced to 500 lb.\n(b) If this separating force is repeatedly applied and removed, determine values of mean and alternating force acting on the bolts.\n\nSchematic and Given Data:\n\n10-32 Assumption: The stress on the bolt is within the elastic limit.\n\nAnalysis:\n(a) From Eq. (10.13),\nF_C = F_I - \\left( \\frac{k_c}{k_c + k_b} \\right) F_e:\n500 = 2000 - \\left( \\frac{5}{5 + 1} \\right) F_e: 1500 = \\frac{5}{6} F_e\nHence, F_e = 1800 lb\n(b) Load off:\nF_b = F_I = 2000 lb\nLoad on:\nF_b = 2000 + \\frac{1}{6}(1800) = 2300 lb\n\nF_m = \\frac{2000 + 2300}{2} = 2150 lb\nF_a = \\frac{2300 - 2000}{2} = 150 lb\n\nSOLUTION (10.25)\nKnown: Two parts of a machine are held together by bolts that are initially tightened to provide a total initial clamping force of 2000 N. The elasticities are such that k_c = 6k_b.\nFind:\n(a) Determine the external separating force that would cause the clamping force to be reduced to 1000 N.\n(b) If this separating force is repeatedly applied and removed, determine values of mean and alternating force acting on the bolts.\n\nSchematic and Given Data:\n\n10-33 Assumption: The bolt stress is less than the elastic limit of the bolt material.\n\nAnalysis:\n(a) From Eq. (10.13),\n F_c = F_i - k_e\n ------\n k_c + k_a :\n 500 = 2000 - (6 / (6 + 1)) F_e : 1500 = (6 / 7) F_e\nHence, F_e = 1750 lb\n(b) Load off;\n F_b = F_i = 2000 lb\nLoad on;\n F_b = 2000 + (1 / 7)(1750) = 2250 lb\n F_m = (2000 + 2250) / 2 = 2125 lb\n F_a = (2250 - 2000) / 2 = 125 lb\n\nSOLUTION (10.26)\nKnown: Drawing 1 and 2 are identical except for placement of the spring washer. The bolt and the clamped members are \"infinitely\" rigid in comparison with the spring washer. In each case the bolt is initially tightened to a force of 10,000 N before two known external loads are applied.\n\nFind:\n(a) For both arrangements, draw block A as a free-body in equilibrium.\n(b) For both arrangements, draw a bolt force-vs-time plot for the case involving repeated application and removal of the external loads.\n\nSchematic and Given Data:\n(1)\n1000 N\n1000 N\n(2)\n1000 N\n1000 N\nAssumption: The bolt stress is within the elastic limit of the bolt material.\n\n10-34 Analysis:\n(a) (1) 10,000 N\n (2) 8000 N\n\n 1000 N 12,000 N 1000 N\n 1000 N 10,000 N 1000 N\n(b)\n 12 kN\n 10 kN\n ----------\n0 Time\n (1)\n0 Time\n (2)\n\nComment: Note that in neither case does the 10 kN force of the flexible spring washer change.\n\nSOLUTION (10.27)\nKnown: Drawing 1 and 2 are identical except for placement of the spring washer. The bolt and the clamped members are \"infinitely\" rigid in comparison with the spring washer. In each case the bolt is initially tightened to a force of 10,000 N before two known external loads are applied.\n\nFind: Plot F_b and F_c versus F_e for drawings 1 and 2.\n\nSchematic and Given Data:\n(1)\n1000 N\n1000 N\n(2)\n1000 N\n1000 N\n\n10-35