Projetos e Elementos de Maquinas - Juvinall -exercicios Resolvidos - Load Analysis

SOLUTION (2.1D)\nKnown: Definitions are needed for the terms: free-body diagram, equilibrium analysis, and bioengineering.\n\nFind: Write definitions of above terms.\n\nHint: Definitions of the terms are given in Section 2.1.\n\nAnalysis:\n1. A free-body diagram is a drawing or a sketch of a body (or part of a body) that shows all the forces from the surroundings acting on that body. The forces could be caused by gravitational attraction, centrifugal acceleration, magnetic repulsion or attraction, or another body.\n\n2. An equilibrium analysis is an analytical method that employs the basic equations of equilibrium to determine unknown loads (forces and/or moments).\n\n3. The term bioengineering defines a relatively new field of engineering that applies the scientific principles of biology to benefit humanity.\n\nSOLUTION (2.2D)\nKnown: A vertical wall channel C holds cylindrical rods A and B of known density. The width of channel C is not given.\n\nFind: Select a metal for rods A and B. Draw free body diagrams for rod A, rod B, and channel C. Determine the magnitude of the forces acting on A, B, and C.\n\nSchematic and Given Data:\n\n w = 4 in.\n rod length = 2.0 in.\n d = diameter = 2.5 in. Decisions:\n1. Select steel which has a known density of p = 0.28 lbm/in.3.\n2. Select w = 4.0 in. for analysis.\n\nAssumptions:\n1. The channel is open upward and supported on the bottom by two knife edges at G and H.\n2. The friction forces between the contacting bodies are negligible.\n3. The rods A and B and channel C are in static equilibrium.\n4. The force of gravity is the only body force.\n5. The weight of the channel C is negligible.\n\nAnalysis:\n1. From the free body diagram for A: D = (W/(sin θ)), C = (W/(tan θ)).\n\n2. From the free body diagram for B: E = (W/(tan θ)), F = 2W. For w = 4 in., d = 2.5 in.; and rod length L = 2.0 in.; the rod mass = ρV = (0.28 lbm/in.3)(2.0 in.)(π(1.25 in.)²/4 = 0.687 lbm.\n4. The weight of each rod is W = F = ma = (0.687 lbm)(32.2 ft/s²)/g_c = 0.687 lb.\n5. θ = cos⁻¹((d-d)/d) = 53.13°.\n6. D = 0.88 lb, C = 0.515 lb, F = 1.374 lb.\n7. From the free body diagram for the channel, the forces G and H can be obtained from force equilibrium.\n\nSOLUTION (2.3)\nKnown: An automobile of weight W and wheel base L slides while braking on pavement with given coefficient of friction. The location of the center of gravity is specified.\n\nFind: Draw a free-body diagram of the automobile.\n\nSchematic and Given Data:\n\nAssumptions:\n1. The friction force is constant during braking.\n2. The vehicle deceleration is uniform.\n3. The motor exerts negligible torque on the wheels (the motor is disconnected).\n\nAnalysis:\n1. From summation of moments at point R,\n ΣM_R = -F(L) - Ah + Wc = 0\n\n2. From summation of forces in the vertical direction,\n ΣF_y = 0 = R - W + F = 0 3. From summation of forces in the horizontal direction,\n∑ F_x = 0 = A + μR + μF = 0\n4. Solving by eliminating A and R gives\n-FL + μhW + W_c = 0\n5. Solving for F gives, F = W(c + μh)\nL\n\nComment: For a 4000 lb vehicle, with L = 120 in., h = 26 in., c = 70 in., and\nμ = 0.7:\n1. F = 4000 lb[70 in + (0.7)(26 in)] / 120 = 2940 lb\n2. With the vehicle stationary, the static force on the two front tires is 2330 lb. SOLUTION (2.4D)\nKnown: Two spheres A and B are in a container C.\nFind: Draw the free body diagrams of A, B, and C. Also determine the forces\nacting on these bodies.\n\nSchematic and Given Data:\nThe diagram includes the forces acting on the spheres and the container. Assumptions:\n1. The container is cylindrical in shape and supported on two knife edges, each at a\ndistance of (1/3)L (where L is the diameter of the container cross section) from\nthe base circle center or (2/3)L apart.\n2. The friction forces between the contacting bodies are negligible.\n3. The sphere and containers are in static equilibrium.\n4. The force of gravity is the only body force.\n\nAnalysis:\n1. From the free body diagram for A, D = [1000] N, C = [1000] N.\n2. From the free body diagram for B, E = [1000] N, F = 1125 N.\n3. From the above diagram and the free body diagram for C, the forces G and H can\nbe obtained.\n\nComments:\n1. The geometry of the container must be clearly determined before proceeding to\nsolve this problem. For example, if the container is assumed to be rectangular in\nshape, then the spheres would not be in stable equilibrium. Also, until the\ncontainer shape is defined it is impossible to draw the free body diagram.\n2. The container should be supported such that it is in stable equilibrium. To\nconsider the container as a free body, all the forces from the surroundings acting\non the body must be shown. SOLUTION (2.5)\nKnown: The geometry and the loads acting on a pinned assembly are given.\nFind: Draw a free-body diagram for the assembly and determine the magnitude of the forces acting on each member of the assembly.\n\nSchematic and Given Data:\n1500 N\n 1500 N\n B C\n Link 5\n 45°\n Link 3\n 45°\n Link 4\n 1000 mm\n \nA D\n 45° \n Link 1\n\nAssumptions:\n1. The links are rigid.\n2. The pin joints are frictionless.\n3. The weight of the links are negligible.\n4. The links are two force members and are either in tension or compression.\n\nAnalysis:\n1. We first draw a free-body diagram of the entire structure.\n\n1500 N 1500 N\n B 1\n 45° 45°\n √2 1\n A 45°\n Ax Ay\n Dy\n\n2. Taking moments about point A and assuming clockwise moments to be positive,\n Σ MA = 0 = 1500(2) + 1500(1) - Dy(1)\n3. Solving for Dy gives Dy = 4500 N.\n4. Summation of forces in the y-direction and assuming vertical forces positive,\n Σ Fy = 0 = Ay + Dy - 1500 - 1500 = Ay + Dy - 3000.\n 2-7 5. Since Dy = 4500 N, Ay = 3000 - Dy = -1500 N.\n6. Σ Fx = 0 gives Ax = 0.\n7. We now draw a free-body diagram for a section at C.\n\n1500 N\n 45°\n CB\n DC\n\n8. Σ Fx = 0 = - CB + DC sin 45° and\n9. Σ Fy = 0 = DC sin 45° - 1500.\n10. Solving simultaneous equations gives DC = 2121 N, CB = 1500 N.\n11. We now draw a free-body diagram for a section at A.\n\n AB\n 45°\n DA\n 1500 N\n\n12. Σ Fy = 0 = AB sin 45° * 1500.\n Σ Fx = 0 = AB cos 45° - DA.\n13. Solving simultaneous equations gives AB = 2121 N, DA = 1500 N.\n14. We now draw a free-body diagram for a section at D. 14. Σ Fy = 0 = 4500 - BD - 2121(sin 45°)\n Hence, BD = 3000 N.\n15. The free-body diagrams for links DC, BC, AB, AD, and BD are:\n\n1500 N 3000\n B C\n 1500 N 1500\n B C\n 3000 D\n D\n 2121 N \n16. We can now draw a free-body diagram of pin B:\n\n1500 N\n B\n 1500 N\n 45°\n 2121 N\n 3000 N\n\n17. Checking for static equilibrium at pin B gives:\n Σ Fx = 2121 cos 45° - 1500 = 0\n Σ Fy = 1500 + 2121 sin 45° - 3000 = 0\n18. We can also draw a free-body diagram for pin C:\n\n1500 N\n 2121 N\n 45°\n C 19. Checking for static equilibrium at pin C gives:\nΣ F_x = 1500 - 2121 cos 45° = 0\nΣ F_y = 1500 - 2121 sin 45° = 0\nComment: From force flow visualization, we determine that links 1, 3, and 4 are in compression and that links 2 and 5 are in tension.\nSOLUTION (2.6)\nKnown: A 1800 rpm motor is rotating a blower at 6000 rpm through a gear box having a known weight.\nFind: Determine all loads acting on the gear box when the motor output is 1 hp, and sketch the gear box as a free-body in equilibrium.\nSchematic and Given Data:\n\nAssumption: The friction losses in the gear box are negligible.\nAnalysis: