Resolução Cap 12 Hibbeler 5ed Resistência dos Materiais

12-1. An A-36 steel strap having a thickness of 10 mm and a width of 20 mm is bent into a circular arc of radius \\rho =10 m. Determine the maximum bending stress in the strap.\n\nMoment - Curvature Relationship:\n\n\\frac{M}{EI} = \\frac{1}{\\rho}\n\n\\text{however, } M = I \\sigma\\newline\n\\frac{\\sigma}{E} = \\frac{0.005}{10} \\left[ 200(10)^{3} \\right] = 100 MPa\n\n12-2. A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole is estimated by measurement to be 4.5 m. If the pole is 40 mm in diameter and it is made of a glass-reinforced plastic for which E_{c} = 131 GPa, determine the maximum bending stress in the pole.\n\nMoment - Curvature Relationship:\n\n\\frac{M}{EI} = \\frac{1}{\\rho}\n\n\\text{however, } M = I \\sigma\n\\rho = \\frac{EI}{M} = \\frac{E_{c} I}{\\left( 4.5 \\left( 10^{-3} \\right) \\right)}= 382 MPa\nAns\n\n12-3. Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant.\n\nu_{x} = M(x)\n\n\\frac{d^2u}{dx^2} = \\frac{M(x)}{EI}\n\n(1)\n\n(x1)\n\n(2)\n\n\\frac{d^2u}{dx^{2}} = \\frac{P L^{2}}{EI}\n\nw_{max} = \\frac{PL^{3}}{48EI}\n\n(3)\n\n(4)\n\n(5)\n\n(6)\n\nThus,\n\nEI\\frac{d^{4}y}{dx^{4}} = -\\frac{P L^2}{32} \\\nI = AE\\newline 12-4. Determine the equation of the elastic curve for the beam using the x coordinate that is valid for 0 \\leq x \\leq L/2. Specify the slope at A and the beam's maximum deflection. EI is constant.\n\nSupport Reactions and Elastic Curve : As shown on FBD(a).\n\nMoment Function : As shown on FBD(b).\n\nSlope and Elastic Curve :\n\nEI\\frac{d^{2}u}{dx^{2}} = M(x)\n\n(1)\n\ndu = M(x) dx\n\ndx^{2} = 2x\n\n(2)\n\n\\text{The slope: } Substitute the value of C_{1} into Eq.(1),\n\ndu/dx = 167\\frac{(A^{2}-L^{2})}{EI}\n\na_{0} = \\frac{PL}{16}\n\nAns\n\nThe Elastic Curve : Substitute the values of C_{1} and C_{2} into Eq.(2),\n\nE_{max} = \\frac{PL^{3}}{48EI}\n\nAns\n\nThe negative sign indicates downward displacement.\n\nw_{max} = 12-5. Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify the slope at A and the maximum deflection. EI is constant.\n\n\\frac{d^{2}u}{dx^{2}} = M(x)\n\\frac{d^{4}u}{dx^{4}} = \\frac{M(x)}{EI}\n\\text{For } M(x) = P x,\n\\frac{d^{2}u}{dx^{2}} = \\frac{P x}{EI}\n\\frac{1}{EI}\\frac{dM(x)}{dx}= P \\\n\\frac{1}{EI} = C_{1} + C_{2}\n\n(1)\n\n\\text{From } M(0) = P L^{2} = 0\\newline\n\\text{Thus, } \\\nC_{2}=0\\newline\n\\text{Boundary Conditions: }\n\\frac{du}{dx}|_{x=0} = 0\\newline\n\\text{Also, } u = 0 \\text{ at } x = 0.\\newline\nFrom Eq.(1) (8)\\newline\nx_{1} = (P L^{3})/(12/22)\n\nFrom Eq. (12)\\newline\n2 + \\frac{P L^{3}}{24}\n\\newline\\newline\nThe negative sign indicates downward deflection.\nAns 12-6. Determine the equations of the elastic curve for the beam using the y1 and x1 coordinates. Specify the beam's maximum deflection. EI is constant.\nSupport Reactions and Elastic Curve: As shown on FBD(a).\nMoment Function: As shown on FBD(b) and (c).\nSlope and Elastic Curve:\nEI \\frac{d^2 u}{dx^2} = M(x)\nFor M(x1) = -\\frac{P}{2}x1^2\n\\frac{d^2 u}{dx^2} = -\\frac{P}{2EI}x1\n[1]\nEI u1 = C1 x1^3 + C2 x1^2\n[2]\nFor M(x2) = -P x2\n\\frac{d^2 u}{dx^2} = -\\frac{P}{EI}x2\n[3]\nEI u2 = \\frac{P L^2}{12} + C1 x2^3 + C2\n[4]\nBoundary Conditions:\n u1 = 0 at x1 = 0.\nFrom Eq.[2], C1 = 0\n u1 = 0 at x1 = L.\nFrom Eq.[3],\n u2 = 0 at x2 = L.\nFrom Eq.[4].\n o = -\\frac{PL^3}{48} - L C2 + C4\n[5]\nContinuity Conditions:\nAt x1 = L and x2 = L - \\frac{L}{2}, \\frac{d u1}{d x} = \\frac{d u2}{d x}.\nFrom Eq.[1] and [3],\n-\\frac{PL^2}{12} = -\\frac{P L^2}{24}\nFrom Eq.(5), C1 = -\\frac{PL^2}{8}\nC3 = -\\frac{7PL}{24}\nFrom Eq. [1],\n\\frac{d u1}{d x} = \\frac{PL}{12} (L^2 - 3x1^2)\n\\frac{d u1}{d x} = 0 at x1 = L/3\nx1 = \\frac{L}{3}\nHence, u_max = \\frac{PL^3}{48E}\nAns 12-7. The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft; and at C by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Determine the equations of the elastic curve using the coordinates x1 and x2. EI is constant.\nElastic curve and slope:\nEI \\frac{d^2 u}{dx^2} = M(x)\nFor M(x1) = -P2 x1\n\\frac{d^2 u}{dx^2} = -\\frac{P}{EI} x1\n[1]\nFor M(x2) = -P2\n\\frac{d^2 u}{dx^2} = -\\frac{P2}{EI}\n[2]\nEI u1 = C1 x1^3 + C2 x1^2\n[1]\nEI u2 = C3 x2^2 + C4 x2\n[2]\nContinuity conditions:\n\\frac{d u1}{d x} - \\frac{d u2}{d x} at x1 = a and x2 = b\n[3]\nBoundary Conditions:\nv1 = 0 at x1 = a\nFrom Eq.(2), C1 = 0\nFrom Eq.(5), C2 = -\\frac{P a^3}{6E}\nFrom Eq.(4),\n\\frac{P b^4}{12} - P2 x2 = P2 [\\frac{P a^2}{6} - P2 ]\nC3 = 3\nC4 = \\frac{P^2 b^3}{3E}\nSubstitute C2 into Eq.(5)\nv1 = -\\frac{Pb}{6EI} x1^3 - \\frac{P}{EI} x1^2\nAns\nv2 = \\frac{w}{6EI} [x^4 - 12bx + 3x^2 - 2(3^2 - 3d)]\nAns 12-8. This fence board weaves between the three smooth fixed posts. If the posts remain along the same line, determine the maximum bending stress in the board. The board has a width of 6 in. and a thickness of 0.5 in. Assume the displacement of each end of the board relative to its center is 3 in.\nSupport Reactions and Elastic Curve: As shown on FBD(a).\nMoment Function: As shown on FBD(b).\nSlope and Elastic Curve:\nEI \\frac{d^2 u}{dx^2} = M(x)\nFor M(x) = \\frac{PL^2}{16}\nFrom Eq.[1] C1 = \\frac{PL^2}{16}\nC2 = 0\nFrom Eq.[2] C3 = 0\nRequire x1 = 48 in., u = -3 in. from Eq.[1].\nMaximum Bending Stress: From the moment diagram, the maximum moment is M_max = 390.625 lb. in. Applying the flexure formula,\n\\sigma_{max} = \\frac{M c}{I}\n= \\frac{390.625(0.025)}{\\frac{1}{8}(6)(0.5^3)} = 1562.5 psi = 1.56 ksi\nAns *12-12. \n\nThe shaft is supported at A by journal bearing that exerts only vertical reactions on the shaft and at B by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Draw the bending-moment diagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft's centerline. Determine the equations of the elastic curve using the coordinates x, and EI is constant. \n\nElastic Curve : As shown. \n\nMoment Function : As shown on FBD(b) and (c). \n\nSlope and Elastic Curve : \n\nFor M(x) = 300 N·m: \n\nd^{2}u/dx^{2} = M(x)/EI \n\nd^{2}u/dx^{2} = 300 \n\nEI (d^{2}u/dx^{2}) = 150x^{2} + C_{1} \quad [1] \n\nFor M(x) = 750 N·m: \n\nd^{2}u/dx^{2} = M(x)/EI \n\nd^{2}u/dx^{2} = 750 \n\nEI (d^{2}u/dx^{2}) = 375x^{2} + C_{2} \quad [3] \n\nEI (u) = 125x^{4} + C_{1} + C_{2} \quad [4]\n\nBoundary Conditions : \n\nu_{1} = 0 at x_{1} = 0. From Eq.(4). \n\nu_{2} = 0 at x_{1} = 0.4 m. From Eq.(4). \n\n0 = 125(0.4^{4}) + C_{1}(0.4) \n\nContinuity Condition : \n\nAt x_{1} = 0.15 m and x_{2} = 0. From Eqs.[1] and [3]. \n\nEI(300)(0.15) + C_{1} = -375(0.4^{2}) - 20 \nC_{1} = -85.0 \nFrom Eq.(5). \nC_{2} = 9.375 \n\nThe Elastic Curve : Substitute the values of C_{1}, C_{2}, and C_{4} into Eq.[2] and [4], respectively. \n\nu_{1} = 1/(EI)(150x^{2} - 85.0x + 9.375) N·m \nAns \n\nu_{2} = 1/(EI)(125x^{2} - 20.0x) N·m \nAns