Projetos e Elementos de Maquinas - Juvinall - Correias Exercicios Resolvidos

SOLUTION (19.1)\nKnown: The angle of wrap on the motor pulley is 160°, the slack-side tension is known. The belt friction factor is known. Centrifugal force is to be assumed negligible.\nFind: Determine the torque capacity of the motor pulley.\nSchematic and Given Data:\nAngle of wrap = 160°\nSlack Side\nP2 = 40 N\n\nAssumptions:\n1. The coefficient of friction is constant throughout the angle of wrap.\n2. The capacity of the belt drive is determined by the capacity of the small pulley.\n3. The belt withstands the load.\n4. Centrifugal force is negligible.\n\nAnalysis:\n1. For φ = 160° = 2.79 rad, P1/P2 = e^(fθ) = e^(0.3)(2.79) = 2.31\n2. For φ = 160°, T = (P1 - P2) r = (2.31 P2 - P2) r = 1.31 P2 r\n3. T = 1.31 P2 r = 1.31 (40 N)(50 mm) = 2620 N·mm\n\nComment: If the initial angle of wrap had been 150° rather than 160°, the torque capacity would be 2388 N·mm; i.e., the less angle of wrap on the driving pulley the lower the torque capacity of the pulley.\n19-1 SOLUTION (19.2)\nKnown: The angle of wrap on a pulley increases from 160° to 200° without change in slack-side tension.\nFind: Determine the percentage increase in torque capacity of the pulley.\nSchematic and Given Data:\nAngle of wrap = 160°\nSlack Side\nAngle of wrap = 200°\n\nAssumptions:\n1. The coefficient of friction is constant throughout the angle of wrap.\n2. The capacity of the belt drive is determined by the capacity of the small pulley.\n3. The belt withstands the load.\n4. Centrifugal force is negligible.\n\nAnalysis:\n1. For φ = 160° = 2.79 rad, P1/P2 = e^(fθ) = e^(0.3)(2.79) = 2.31\n2. For φ = 200° = 3.49 rad, P1/P2 = e^(fθ) = e^(0.3)(3.49) = 2.85\n3. For φ = 160°, T = (P1 - P2) r = (2.31 P2 - P2) r = 1.31 P2 r\n4. For φ = 200°, T = (P1 - P2) r = (2.85 P2 - P2) r = 1.85 P2 r\n5. Increased torque capacity = 1.85 - 1.31 = 0.41 = 41%\n19-2 Comment: If the initial angle of wrap had been 150° rather than 160°, then the increase in torque capacity would have been 55% rather than 41%.\nSOLUTION (19.3)\nKnown: The parameters c, r1, r2, and α are known for two pulleys and a belt.\nFind: Develop an equation for the belt length, L, as a function of c, r1, r2, and α.\nSchematic and Given Data:\n\nAssumptions:\n1. The effect of gravity on the belt shape is negligible.\n2. The belt has sufficient tension to prevent sagging of the top and bottom belt strands.\n\nAnalysis:\n1. L = l1 + l2 + l3\n l1 = c cos α\n l2 = 2πr2(180 + 2α)/360\n l3 = 2πr1(180 - 2α)/360\n2. L = 2c cos α + 2πr2(180 + 2α)/360 + 2πr1(180 - 2α)/360\n Thus, L = 2c cos α + π(2π/90)[r2(90 + α) + r1(90 - α)]\n19-3 SOLUTION (19.4)\nKnown: Figure P19.4 suggests an approximation length ABCD that will be equal to half the length of the belt.\n\nFind: From the half length approximation, develop an equation relating center distance, c, belt length, L, and pulley radii, r1, and r2.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The effect of gravity on the belt shape is negligible.\n2. The belt has sufficient tension to prevent sagging of the top and bottom belt strands.\n\nAnalysis:\n1. L/2 = BC + AB + CD\n2. BC = √(c² + (r2 - r1)²)\n AB = (1/2)πr1\n CD = (1/2)πr2\n3. L/2 = √(c² + (r2 - r1)²) + (1/2)πr1 + (1/2)πr2\n Thus, c² = L²/4 - 2[1/4Lπ(r1 + r2)] + π(r1 + r2)² - (r2 - r1)²\n or c² = [L² - 2Lπ(r1 + r2)] - (r2 - r1)²\n or c² = 1/4[L - π(r1 + r2)² - (r1 - r2)²]\n\n19-4 SOLUTION (19.5)\nKnown: The angles of wrap for two pulleys and a belt are given. Each pulley has a known radius.\n\nFind: Derive an equation relating α, r1, r2, and c.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The effect of gravity on the belt shape is negligible.\n2. The belt has sufficient tension to prevent sagging of the top and bottom belt strands.\n\nAnalysis: From inspection of the drawing, sin α = (r2 - r1)/c\n\nComment: For r1 = 7 mm, r2 = 17 mm and c = 34 mm as shown in the figure, α = sin⁻¹(10/34) = 17.1°. The angle of wrap on the smaller pulley is 145.8°.\n\n19-5 SOLUTION (19.6)\nKnown: A motor of given horsepower and speed drives an input pulley of known diameter and angle of wrap. The size 5 V belts have a known unit weight and angle β. The maximum belt tension is 150 lb and the coefficient of friction is 0.20.\n\nFind: Determine the number of belts required.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The maximum tension in the belt is limited to 150 lb.\n2. The coefficient of friction will be at least 0.20.\n3. Power is shared equally by each belt.\n\nAnalysis:\n1. We first calculate terms in Eq. (19.3a);\n with Eq. (19.2), Pe = mv² where V = 3.7 (π) 1750/60 = 339 in/sec\n Pe = 0.012 (339)² = 3.57 lb\n e(δ) sin β = e(0.2)(2.889) sin β = 6.45\n2. Substituting in Eq. (19.3a) and solving for P2;\n 150 - 3.57 = 6.45 or 146.4 = 6.45 P2 - 23.0\n Hence, P2 = 3.67 lb\n3. From Eq. (18.24), T = (P1 - P2) r = (150 - 26.3) 3.7/2 = 229 lb in.\n4. From Eq. (1.3), W per belt = 1750(229)/5252(12) = 6.36 hp/belt\n\n19-6 5. For 25 hp, 25/6.36 = 3.93, and 4 belts are required.\n\nComment: If a 30 hp motor was used then 5 belts would be required. As more and more belts are needed however, the effects of misalignment of the shafts (and consequent unequal sharing of the load) becomes important.\n\nSOLUTION (19.7)\nKnown: A pulley of given diameter drives a flat drum of given diameter. The center distance is known.\n\nFind: Determine the slip-limited capacity of the pulley relative to that of the drum.\n\nSchematic and Given Data:\n\nV-belt\n\n120 mm\ndiameter\n\n40 mm\ndiameter\npulley\n\nc = 120 mm\n\n2β = (2)(18°) = 36°\n\nAssumptions:\n1. The slack side of the belt does not sag.\n2. The friction coefficient is uniform and equal to 0.25 for both the pulley and drum.\n\nAnalysis:\n1. From Prob. 19.2, sin α = (r2 - r1) / c = 60 - 20 / 120\nα = 19.47°\n2. φ1 (pulley) = 180° - 2(19.47°) = 141.06° = 2.462 rad\nφ2 (drum) = 180° + 2(19.47°) = 218.94° = 3.821 rad\n3. The drum torque capacity:\n\n19-7 P1/P2 = e^(f0.25)(3.821) = 2.6, hence, P1 = 2.60 P2\nTorque = (P1 - P2) r = 1.60 P2 (60) = 96 P2\n\nThe pulley torque capacity:\nP1/P2 = e^(0.25)(sin 18°) = e^(0.25)(2.462/sin 18°) = 7.33\nTorque = (P1 - P2) r = 6.33 P2 (20) = 126.6 P2\n5. The pulley has 126.6 - 96 = 32% more capacity than the drum.\n\nComment: The above calculations used f = 0.25.\n\nSOLUTION (19.8)\nKnown: A pulley of given rotational speed, radius, and angle of wrap drives a V-belt with known friction, weight, and maximum tension.\n\nFind: Determine the maximum power transmitted by the pulley.\n\nSchematic and Given Data:\n\nV-belt\nβ = 18°\nn = 4000 rpm\nr = 100 mm\npulley radius\nφ = 170°\n\nBelt Maximum\nTension = 1300 N\nBelt Unit\nWeight = 1.75 N/m\n\nAssumptions:\n1. The friction coefficient is uniform throughout the contact area.\n2. The effect of centrifugal force is important.\n\nAnalysis:\n1. From Eq. (19.2), Pc = mV² = (1.75 N/m)(4000 / 9.81 m/s²)(4000 / 60) = 313 N\n2. From Eq. (19.3), 1300 - 313 = e^(0.25)(sin 18°)((170/π)(17°/180) = 6.82. Hence, P2 = 458 N\n3. T = (P1 - P2) r = (1300 - 458)(0.100) = 84 N·m\n4. From Eq. (1.2), W = 4000(84) / 9549 = 35.18 kW\n\n19-8 SOLUTION (19.9)\nKnown: A pulley of given rotational speed, radius, and angle of wrap drives a V-belt with known friction, weight, and maximum tension but (a) two V-belts are needed, (b) a single V-belt with twice the maximum tensile capacity is used.\n\nFind: Determine the power transmitted by the pulley for cases (a) and (b).\n\nSchematic and Given Data:\n\nCase (a) Two V-belts, Case (b) One V-belt with twice the cross sectional area\n\nV-belt\nβ = 18°\nn = 4000 rpm\nr = 100 mm\npulley radius\nφ = 170°\n\nBelt Maximum\nTension = 1300 N\nBelt Unit\nWeight = 1.75 N/m\n\nAssumptions:\n1. If two belts are used, each will have the same capacity and the load will be shared equally between them.\n2. If a belt with twice the cross section is used, there will be no loss in the tensile strength capacity because of a size effect.\n\nAnalysis:\n1. For case (a) each of the two identical belts could transmit 35 kW, hence total power capacity is 70 kW.\n2. For case (b) doubling the section would double m', giving Pc = 626 N.\nDoubling Pc and P1 with no change in e^(f/sin β) would double P2. Thus, power capacity is doubled in cases (a) and (b).\n\nSOLUTION (19.10)\nKnown: A pulley of given rotational speed, diameter, and angle of wrap drives a V-belt with known friction, weight, and maximum tension.\n\nFind: Determine the maximum power transmitted by the smaller pulley.\n\n19-9 Schematic and Given Data:\nV-belt\nβ = 18°\nn = 3500 rpm\nd1 = 6 in.\nd2 = 12 in.\nφ = 170°\nDriven pulley\n\nBelt Maximum\nTension = 250 lb\nBelt Unit\nWeight = 0.012 lb/in.\n\nAssumptions:\n1. The friction coefficient is uniform throughout the contact area.\n2. The effect of centrifugal force is important.\n\nAnalysis:\n1. From Eq. (19.2), Pc = m·v² = 0.012 lb/in (3500 6 in/s)² = 37.5 lb\n2. From Eq. (19.3), P1 - Pe = P2 - Pe = e^{φ}\nwhere f = f_s for the V-belt\nsinβ\n\n250 - 37.5 = e(0.2/sin 18° × (170°/180°) = 6.82\nP2 = 68.7 lb\n3. T = (P1 - P2) r = (250 - 68.7) (3) = 544 lb in.\n4. From Eq. (1.3), Ẇ = nT lb-ft = 3500(544 lb in.) (1 ft/12 in.) = 30.2 hp\n\nSOLUTION (19.11)\nKnown: A V-belt pulley of given diameter and rotational speed drives a second pulley at known rotational speed. The V-belt has a given unit weight and coefficient of friction.\n19-10 Find:\n(a) Determine values of P1 and P2.\n(b) Determine the loads applied by the belt to each shaft.\n(c) Determine initial belt tension when the drive is not operating.\n(d) Determine values of P1 and P2 when the drive is operating at normal speed but transmitting only 6 kW.\n\nSchematic and Given Data:\nSingle V-belt, β = 18°\nUnit weight = 2.2 N/m\nPower transmitted = 12 kW\nInitial belt tension just adequate to prevent slippage\n? mm diameter pulley\n\nDriving pulley\nn = 1750 rpm\nf = 0.20\nc = 400 mm\n\nAssumption: Initial belt tension is marginally adequate to prevent slippage.\n\nAnalysis:\n1. P1 and P2\nThe larger pulley diameter = 180 (1750) / (1050) = 300 mm\n2. From Prob. 19.5, sin α = r2 - r1 / c = 150 - 90 / 400\n\n3. For smaller pulley, φ = 180° - 2α; α = 8.63°\n4. From Eq. (19.2), Pc = m ω²\n\n5. From Eq. (19.3): P1 - 61.0\nP2 - 61.0\n6. From Eq. (1.2): T = 9549 Ẇ / n\n19-11 7. T = (P1 - P2) r : 65.48 = (P1 - P2) 0.090; hence, P1 - P2 = 727.54 N ----- (2)\n8. Simultaneous solution of equation (1) and equation (2) gives P2 = 198.7, P1 = 926.2. Rounding off: P1 = 926 N, P2 = 199 N\n(b) Shaft loads\n1. Driving shaft torque,\nT = (926.2 N -198.7 N) (0.090 m) = 65.5 N m\n(which checks with calculated torque of 65.48 N m)\nDriven shaft torque: T = (926.2 N -198.7 N) (0.150 m)\nT = 109.2 N m\n3. Radial load = √(P1 + P2 cos 2α)² + (P2 sin 2α)² =\n√(926.2 + 198.7 cos 17.26°)² + (198.7 sin 17.26°)² = 1117.6 N\n4. Radial load applied to each shaft = 1118 N\n5. Assume P1 & P2 remain = 926.2 + 198.7 = 1124.9 N\n(c) Initial belt tension (no rotation)\nP = 1124.9 N = 562.5 N\n\n2. D1 and P2 for 6 kW\n1. The torque is reduced to 65.48 = 32.74 N m\nHence, (P1 - P2) 0.090 m = 32.74 N m; P1 - P2 = 363.78 N\n2. Assuming P1 + P2 remains = 1124.9 N, P1 = 744 N, P2 = 381 N\n\nSOLUTION (19.12)\nKnow: A V-belt pulley of known diameter and rotational speed transmits a given power to a driven pulley of known diameter. This is accomplished with a belt of specified unit weight, angle B, and coefficient of friction.\nFind:\n(a) Determine values of P1 and P2.\n(b) Determine the loads applied by the belt to each shaft.\n(c) Determine values of P1 and P2 if the power is reduced to a known value.\n19-12 Schematic and Given Data:\nCase (a) and (b) Power = 12 hp.\nCase (c) Power = 3 hp\nSingle V-belt, \\u03b2 = 18\\u00b0\nUnit weight = 0.012 lb/in.\nDriving pulley\n6 in.dia.\nn = 1750 rpm\nf = 0.20\nDriven pulley\n12 in.dia.\nn = ? rpm\nc = 20 in.\nAssumption: The initial belt tension is just adequate to prevent slippage (given).\n\nAnalysis:\n(a) P1 and P2 for 12 hp\n1. From Pro. 19.4, sin \\u03b1 = r2 - r2 - 6 - 3.\nHence, \\u03b1 = 8.63\\u00b0\n2. For the smaller pulley, \\u03d5 = 180\\u00b0 - 2\\u03b1 = 162.7\\u00b0 = 2.84 rad\n3. From Eq. (19.2), Pc = m \\u03c9 r2^2\nPc = (0.012 lb/in.)/(386 in/s2) (1750 x 2\\u03c0 rad/s)2 (3 in.)2 = 9.40 lb\n4. From Eq. (19.3a), P1 = 9.40\nP2 = 9.40 = (0.20/sin 18\\u00b0)(2.84) = 6.285 ----- (1)\n5. From Eq. (1.3), T = 5252 (\\u03c9)/(n) = 36.0 lb ft = 432.2 lb in.\n6. T = (P1 - P2) r; 432 = (P1 - P2)(3), hence P1 - P2 = 144.05 lb ----- (2)\n7. Simultaneous solution of equation (1) and equation (2) gives: P1 = 36.67 lb (37 lb) and P1 = 180.72 lb (181 lb)\n\n(b) Shaft loads\n1. Driving shaft torque = T = (180.72 - 36.67)(3/12) = 36 lb ft\n2. Driven shaft torque = T = (180.72 - 36.67)(6/12) = 72 lb ft\n19-13 3. Radial load = \\u221a((P1 + P2 cos 2\\u03b2)^2 + (P2 sin 2\\u03b2)^2\n= \\u221a((180.72 + 36.67 cos 17.26\\u00b0) + (36.67 sin 17.26\\u00b0)2 = 216 lb (applied to each shaft)\n\n(c) P1 and P2 for 3 hp\n1. Torque is reduced to 432.2 - 216.1 lb in. Hence,\n(P1 - P2)(3) = 216.1; P1 - P2 = 72.0 lb\n2. With the assumption that average tension remains unchanged; i.e., no adjustments are made, we have P1 + P2 = 180.72 + 36.67 = 217.39 lb\n3. Simultaneous solution gives: P1 = 144.7 lb, P2 = 72.7 lb, rounding off, P1 = 145 lb, P2 = 73 lb\n\nSOLUTION (19.13D)\nKnown: A web site address is given as http://www.grainger.com.\n\nFind: Select an A-type V-belt with a length of 32 in. List the manufacturer, description, and price.\nAnalysis: A product search of the web site gives:\nItem #: IA095\nMfg. Name: Browning\nDescription: A Type V-Belt-32\", 12\" Top Width-5/16\" Thick-RMA #A30\nPrice: $8.22 Schematic and Given Data:\nElectric Motor\nn = 1780 rpm\n55% rated power\nFluid Coupling -- performance curves -- Fig. 19.11\nDriven Machine\n\nAssumptions:\n1. The fluid coupling is the type shown in Fig. 19.10.\n2. The motor and coupling performance curves are given in Fig. 19.11.\n3. The rated motor output corresponds to 1750 rpm, 100% rated torque.\n\nAnalysis:\n(a) The 1780 motor rpm corresponds to approximately 2% slip. Machine input rpm is 1780 (0.98) = 1744 rpm.\n(b) As assumed, 100% of motor torque reaches the machine, hence 2% of the motor power is converted to heat.\n(c) From Fig. 19.11, at 100% rated torque, slip = 3.5%.\n(d) From Fig. 19.11, at maximum coupling torque, the stall speed is approximately 775 rpm.\nAt this point, the motor delivers 168% of the rated torque at 775/1750 = 44% rated speed. Hence power is (1.68)(0.44) = (0.744) times rated power, or approximately 74% of rated power is converted into heat.\n\nSOLUTION (19.16)\nKnown: A fluid coupling connects an electric motor to a driven machine. The motor and coupling curves are given.\n\nFind: Determine the factor by which the fluid coupling diameter could be reduced if the (i) power required is reduced by half (ii) the rotational speed is doubled.\n19-15 Schematic and Given Data:\n\nElectric Motor\nn = 1750 rpm\nPower = 10 kW\nFluid Coupling -- 10 kW size\nDriven Machine\n\nElectric Motor\nn = 1750 rpm\nPower = 5 kW\nFluid Coupling -- 5 kW size\nDriven Machine\n\nElectric Motor\nn = 3500 rpm\nPower = 5 kW\nFluid Coupling\nDriven Machine\n\nAssumption: The fluid couplings are geometrically similar (given).\n\nAnalysis:\n1. From Eq. (19.7), T ∝ ω2D5. Hence, power ∝ ω3D5.\n2. If power is halved, D5 can be halved, giving Dnew = 0.5 D5old.\n3. Dnew = √0.5 Dold = 0.87 Dold\n4. If power is halved and ω is doubled, then with D ∝ √(power / ω^3):\nDnew = Dold √(5 / (0.5)^2) = D and Dnew = 0.574 Dold\n\n19-16 SOLUTION (19.17)\n\nKnown: A hydrodynamic torque converter is used to provide a known torque multiplication of the known torque of a motor.\n\nFind: Estimate the torque applied to the one-way clutch.\n\nSchematic and Given Data:\n\nHydrodynamic Torque Converter - provides a 2.4 torque multiplication\nDriving Motor\nTorque = 100 N m\nDriven Machine\n\nAssumption: The hydrodynamic torque converter is like the one represented in Fig. 19.13.\n\nAnalysis:\n1. From Eq. (19.9), Ti + To + Tr = 0; where Tr is the reaction torque applied by the one-way clutch.\n2. With To = -2.4 Ti, we have Ti + (-2.4 Ti) + Tr = 0.\n3. Solving for Tr gives Tr = 1.4 Ti.\n\n19-17