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

SOLUTION (7.1)\nKnown: A 950-lb gin pole is supported by a stranded steel cable with known cross section and modulus of elasticity. As the gin pole is descending at a constant velocity, an accident causes the top of the cable, 70 ft above the gin pole, to suddenly stop.\nFind: Estimate the maximum elongation and maximum tensile stress developed in the cable.\nSchematic and Given Data:\nSteel cable\nA = 0.110 in^2\nE = 12 x 10^6 psi\nv = 30 fpm\nAssumptions:\n1. The mass of the cable is negligible.\n2. Neglect any stress concentrations.\n3. Ignore damping due to internal friction within the cable.\n4. The cable responds elastically to the impact.\n5. The cable is attached at the top of the gin pole (L = 70 feet).\nAnalysis:\nSolution 1 - Using Eqs. (7.1a)\n1. δ = δt [1 + √(1 + v^2/gδt)]\nwhere δt = PL/AE = (950 lb/(840 in-l))/(0.11 in^2)(12 x 10^6 psi) = 0.60 in.\n2. δmax = (0.60 in) [1 + √(1 + (6 in/3)^2/(386 in/s^2)(0.60 in))] = (0.60 in)(2.07) = 1.24 in.\n3. Impact factor = δmax/δt = 1.24/0.60 = 2.07\n4. σmax = Fmax/A = (950 lb)(2.07)(0.11 in^2) = 17.9 ksi\n\n7-1 Solution 2 - Using Eq. (7.3a)\n1. δ = δt√(v^2/gδt) = (0.60 in)√(6/738(60.60) = 0.237 in.\nBut this is due to the accident only. Adding the pre-existing deflection of 0.60 in. gives a total δmax = 0.837 in.\n2. Using Eq. (7.4a), Fε = W√(v^2/g0) = (950)(√(6 in/s)/(386 in/s^2)(0.60 in) = 375 lb\n3. σmax = W + P/A = (950 + 375)/0.11 = 12.0 ksi\n\nSolution 3 - Using Basic Energy Relationships\n1. The rope stretch due to impact involves converting the kinetic energy of the elevator mast to elastic energy in the cable:\n\nlmv^2 = 1/kx^2\n2. Using k = P/δ = AE/L = (0.11)(12 x 10^6)/(840) = 1,571 lb/in.\nUsing Eq. (7.3b), δ = √(2U/k) where U = 1/2 mv^2\nδ = (950/386)(6) = 0.237 in.\n3. Pre-accident equilibrium deflection of 0.60 in. is added to give δmax = 0.837 in. Hence, σmax = 12.0 ksi (as in Solution 2)\n\nSolution 4 - Using Differential Equation of Mass Motion\n1. ΣFvent = 0: kx + mx'' = 0 (neglecting damping, and also the gravity force which is in equilibrium with the pre-accident cable force; and where x is measured from the equilibrium point.)\n2. The solution is x = x0 sin(√(k/m)t where xmax = x0√(k/m)t. This is known to be 6 in/s.\n3. Therefore, 6 in = x0√(1,571 lb/in./950 lb/386 in/s^2); x0 = 0.237 in.\n4. Adding 0.60 gives δmax = 0.837 in., and σmax = 12.0 ksi as in Solution 2.\n\n7-2 SOLUTION (7.2)\nKnown: A 5-ton elevator is supported by a stranded steel cable with known cross section and modulus of elasticity. As the elevator is descending at a constant velocity, an accident causes the top of the cable, 70 ft above the elevator, to suddenly stop.\nFind: Estimate the maximum elongation and maximum tensile stress developed in the cable.\nSchematic and Given Data:\nSteel cable\nA = 2.5 in^2\nE = 12 x 10^6 psi\nv = 400 fpm\nAssumptions:\n1. The mass of the cable is negligible.\n2. Neglect any stress concentrations.\n3. Ignore damping due to internal friction within the cable.\n4. The cable responds elastically to the impact.\nAnalysis:\nSolution 1 - Using Eqs. (7.1a)\n1. δ = δt [1 + √(1 + v^2/gδt)]\nwhere δt = PL/AE = (10,000 lb/(840 in.))(2.5 in.^2)(12 x 10^6 psi) = 0.28 in.\n2. δmax = (0.28 in.) [1 + √(1 + (80 in/s)^2/(386 in/s^2)(0.28 in.))] = (0.28 in.)(8.76) = 2.45 in.\n3. Impact factor = δmax/δt = 2.45/0.28 = 8.75\n4. σmax = Fmax/A = (5 tons)(2000 lb/ton)(8.75)/2.5 in.^2 = 35.0 ksi\n\n7-3 Solution 2 - Using Eq. (7.3a)\n1. \\u03b4 = \\u03b4_s (\\u221av^2/g)\\u2206_\\textit{it} = (0.28 in)\\u221a(80)/(386)(0.28) = 2.155 in.\n\n2. Using Eq. (7.4a), F_e = W\\u221a(\\u00a6) = (10,000)\\u221a(80 in./s^2) = 76,950 lb\n\n3. \\u03c3_max = W + F_e/A = 10,000 + 76,950 = 34.8 ksi\n\nSolution 3 - Using Basic Energy Relationships\n1. \\u00bd mv^2 = kx^2\n where k = P = AE/L = (2.5)(12 x 106) = 35,714 lb/in.\n2. Using Eq. (7.3b), \\u03b4 = \\u221a(2U/k) where U = 1/2 mv^2\n \\u03b4 = \\u221a(10,000/386)(80)^2 = 2.155 in.\n\n3. Pre-accident equilibrium deflection of 0.28 in. is added to give \\u03b4_max = 2.435 in. hence, \\u03c3_max = 34.8 ksi (as in Solution 2).\n\nSolution 4 - Using Differential Equation of Mass Motion\n1. \\u2211F_ver = 0: kx + mx = 0 (neglecting damping, and also the gravity force which is in equilibrium with the pre-accident cable force; and where x is measured from the equilibrium point.)\n2. The solution is x = x_0 sin \\u221ak/m t where x_max = x_0 \\u221ak/m. This is known to be 80 in./s.\n3. Therefore, 80 in./s = x_0 \\u221a(35,714 lb/in./10,000 lb/in^2) = 2.155 in.\n4. Adding 0.28 gives \\u03b4_max = 2.435 in., and \\u03c3_max = 34.8 ksi as in Solution 2.\n\n7-4 SOLUTION (7.3)\nKnown: A tensile impact bar is fractured in service. A new bar is made exactly like the old one except that the middle third is enlarged to twice the diameter of the ends.\nFind: Compare the impact capacities of the new and old bars\nSchematic and Given Data:\n\nAssumptions:\n1. Neglect stress concentrations.\n2. The static load capacities of the old and the new bars are the same.\n3. The mass of the bars are negligible.\n4. Ignore damping due to internal friction.\n\nAnalysis: Neglecting stress concentration, the load capacities of the old and the new bars are the same.\n\n7-5 Solution 1.\n1. The corresponding deflections are:\n\n2. The new bar deflects only 3/4 as much as the old bar. Energy \\u2200 (load capacity) (deflection). Hence, the new bar can absorb 3/4 times the energy of the old bar. \nSolution 2.\n1. The elastic capacity for the old bar is determined directly from the Eq. (7.5a), where \\u03c3 = S_y:\n\n2. For the new bar, the energy absorbed is equal to the energy absorbed by the middle section plus the energy absorbed by the two end sections. The end sections are weakest and can be loaded to a stress of S_y. Their volume is (2V/3) where V is the volume of the old rod.\n\nHence, the new bar can absorb 3/4 times the energy of the old bar. \n\nComment: The stress concentrations at the changes in section of the stepped bar would further reduce its impact capacity and would tend to promote brittle fracture at a step.\n\n7-6 SOLUTION (7.4)\nKnown: A vertical rod is subjected to an axial impact by a 100 lb weight dropped from a height of 2 ft. The rod is made of steel, with Sy = 45 ksi and E = 30 X 10^6 psi.\n\nFind: The length of the member to avoid yielding for a diameter of (a) 1 in., (b) 1.5 in., and (c) 1 in. for half of its length, and 1.5 in for the other half.\n\nSchematic and Given Data:\n\nDrop weight\n\n24 in.\n\nAssumptions:\n1. The mass of the members are negligible.\n2. Neglect any stress concentrations.\n\nAnalysis: Neglecting stress concentration, the load capacities of the old and the new bars are the same.\n\nArea = 0.785 in² Area = 1.77 in² Area = 1.77 in²\nd = 1.0 in\nd = 1.5 in\nd = 1.0 in; d = 1.5 in\n\n7-7 (a) Diameter is 1.0 inches\n1. For a rod with uniform axial stress, σ = √(2UE/V), or\n\nV = 2UE/σ² = (2)(100)(24)(30 x 10^6)/45,000² = 71 in.³\n\n2. L = V/A = 71/π(1)² = 90.5 in.\n\n(b) Diameter is 1.5 inches\n1. V = 71 in.³\n\n2. L = (71)/(π)(1.5)² = 40 in.\n\n(c) Diameter of 1.0 inches and 1.5 inches\n1. Let VS = small end volume. (Large end volume = (1.5)²VS)\n2. Small end stress = Sy = 45,000 psi\n\n3. Large end stress = 45,000\n\n(1.5)²\n\n5. Large end energy = UL = V1σ1L/2E = (2.25√(45,000)²/(2)(2)(30 x 10²)\n\n6. U = US + UL = 2400 in. lb. = VS(45,000)²/(2)(30 x 10²)[(1)+(1/2.25)]\n\n7. Therefore, VS = 49.1 in.³\n\n8. LS = 49.1/π = 62.6 in.\n\n9. Therefore, L = 125.2 in.\n\nComment: Note physically why, if the \"excess diameter\" were machined off, the rod could be shortened from 125 inches to 90 inches.\n\n* This assumes a deflection which is negligible in comparison with the 24 in. drop.\n\n7-8 SOLUTION (7.5)\nKnown: A rescue car, of 1400 kg mass, attempted to jerk a stuck vehicle back onto a road using a 5-m steel tow cable of stiffness k = 5000 N/mm. The rescue car reached a speed of 4 km/hr at the instant the cable became taut.\n\nFind: Estimate the maximum impact force and resulting elongation that develops in the cable.\n\nSchematic and Given Data:\n\nRope\n\nk = 5000 N/mm\nv = 4 km/hr\nL = 5 m\nm = 1400 kg\n\nAssumptions:\n1. The cable is attached rigidly to the masses of the cars.\n2. Ignore the mass of the rope.\n3. Neglect any stress concentrations.\n4. Ignore damping due to internal friction within the rope.\n5. The rope responds to the impact elastically.\n\nAnalysis:\n1. From Eq. (7.4a), Fe = W/√(v²/gₒ), where δu = W\nW = gravity force = mg = (1400 kg)(9.81 m/s²) = 13,734 N\n\nTherefore, δu = 13,734 N/(5000 N/mm) = 2.75 mm; v = 4 km/hr = 1.11 m/s\n\n2. Fe = (13,734 N)/(9.81 m/s²)(0.00275 m) = (1.11 m/s²)(13,734)(6.76) N\n\n3. Maximum impact force = 6.76 times the gravity weight of the car,\n or 92.9 kN.\n\n4. Cable elongation = (92.9 kN)/(5 kN/mm) = 18.6 mm.\n\n7-9 SOLUTION (7.6) Known: A rescue car attempted to jerk a stuck vehicle back onto a road using a 12 m elastic cable of stiffness 2.4 N/mm. The rescue car was able to reach 12 km/hr at the point of becoming taut.\n\nFind: Estimate the impact force developed and the resulting cable elongation. Determine the energy stored in the cable. State the warning that you suggest be provided with elastic cables sold for this purpose.\n\nSchematic and Given Data:\nRope\nk = 2.4 N/mm\nv = 12 km/hr\nm = 1400 kg\nL = 12 m\n\nAssumptions:\n1. The cable is attached rigidly to the masses of the cars.\n2. Ignore the mass of the rope.\n3. Neglect any stress concentrations.\n4. Ignore damping due to internal friction within the rope.\n5. The rope responds to the impact elastically.\n6. The stuck vehicle does not move significantly until the rescue car has just come to a stop.\n\nAnalysis:\n1. From Eq. (7.4a), Fe = W/v^2/2 δt, where δt = W/k\nW = gravity force = mg = (1400)(9.81) = 13,734 N\nδt = 13734 N = 5723 mm 2.4 N/mm\nv = 12 km = 3.33 m/s\n\nFe = (13,734 N)√((3.33)/(9.81)(5.723)) = (13,734)(0.44) = 6043 N\n\nMaximum impact force is 6043 N\n\nCable elongation = 6043 N / 2.4 N/mm = 2518 mm ≈ 2.5 m\n\n2. Energy = (average force)(deflection) = (6043 N)(2.5 m) = 7554 N·m\nGravity force on a 100 kg mass is (100 kg)(9.81 m/s²) = 981 N\n7-10