Hibbler 5 Edicao Questoes de 1 a 60 Resistencia dos Materiais

Problem 1-11\nThe beam supports the distributed load shown. Determine the resultant internal loadings on the cross sections through points D and E. Assume the reactions at the supports A and B are vertical.\nGiven: \nw1 = 4.5 kN/m\nw2 = 6.0 kN/m\na = 1.8m\nb = 1.8m\nc = 2.4m\nd = 1.35m\ne = 1.35m\nSolution:\nL1 = a + b + c\nL2 = d + e\nSupport Reactions:\n∑M_D = 0; B_y L1 - (w1 L1)(0.5 L1) - (0.5 w2 L2) = 0\nB_y = (w1 L1)(0.5) + (0.5 w2 L2) = 1 + L2 / 3 L1\nBy = 22.82 kN\nAy = 12.29 kN\nFor point D:\n∑F_y = 0; N_D = 0\nAns\n∑F_y = 0; [A_y - w1(a)] - V_D = 0\nV_D = A_y - w1(a)\nV_D = 4.18 kN\nAns\nM_D + [w1 L1](0.5 a) - A_y(a) = 0\nM_D = -[w1 L1](0.5 a) + A_y(a)\nM_D = 14.823 kN.m\nAns\nFor point E:\n∑F_y = 0; N_E = 0\nAns\n∑F_y = 0; V_E - 0.5 w2(0.5 e) = 0\nV_E = 0.5 w2(0.5 e)\nV_E = 2.03 kN\nAns\n∑M_E = 0; -M_E - [0.5 w2(0.5 e)] = 0\nM_E = -[0.5 w2(0.5 e)]\nM_E = -0.911 kN.m\nAns\nNote: Negative sign indicates that M_E acts in the opposite direction to that shown on FBD. Problem 1-2\nDetermine the resultant internal torque acting on the cross sections through points C and D of the shaft. The shaft is fixed at B.\nGiven: \nT_A = 250N.m\nT_CD = 400N.m\nT_DB = 300N.m\nSolution:\nEquations of equilibrium:\nTA - TC = 0\nTC = TA\nTC = 250N.m\nAns\nTA - TC - TD = 0\nTD = T_CD - TA\nTD = 150N.m\nAns Problem 1-3\nDetermine the resultant internal torque acting on the cross sections through points B and C.\nGiven: \nT_D = 500N.m\nT_BC = 350N.m\nT_AB = 600N.m\nSolution:\nEquations of equilibrium:\n∑M_x = 0; T_B + T_BC - T_D = 0\nT_B = -T_BC + T_D\nT_B = 150N.m\nAns\n∑M_x = 0; T_C - T_D = 0\nT_C = T_D\nT_C = 500N.m\nAns Problem 1-4\nA force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A.\nGiven: P = 80N\ng = 30deg \u03b8 = 45deg a = 0.3m b = 0.1m\n\nSolution:\nEquations of equilibrium:\n\u21d0 \\sum F_{x}=0;\nN_{A} = P \\cdot \\cos(\\theta) - 0 = 0\nN_{A} = P \\cdot \\cos(\\theta) = P \\cdot \\cos(45) = 77.27 N\nAns\n\n\u21d0 \\sum F_{y}=0;\nV_{A} - P \\cdot \\sin(\\theta) = 0\nV_{A} = P \\cdot \\sin(\\theta) = 20.71 N\nAns\n\n\u21d0 \\sum M_{A}=0;\nM_{A} = P \\cdot \\cos(\\theta) \\cdot a \\cdots P \\cdot \\sin(\\theta) \\cdot (b + a \\sin(\\theta)) = 0\nM_{A} = -P \\cdot \\cos(\\phi) \\cdot a \\cdots P \\cdot \\sin(\\phi) \\cdots (b + a \\cdots)\nM_{A} = -0.555 N-m\nAns\nNote: Negative sign indicates that M_{A} acts in the opposite direction to that shown on FBD. Problem 1-5\nDetermine the resultant internal loadings acting on the cross section through point D of member AB.\nGiven: M_{E} = 70N.m\na = 0.05m b = 0.3m\nSolution:\nSegment AB: Support Reactions\n\u21d0 \\sum M_{A}=0;\n-M_{E} - B_{y}(2a + b) = 0\nB_{y} = -\\frac{M_{E}}{2a + b} = -175 N\nAt B:\nB_{x} = -\\frac{150}{200} = -131.25 N\n\nSegment DB:\nN_{B} = -B_{x}\nV_{B} = -B_{y}\n\n\u21d0 \\sum F_{x}=0;\nN_{D} + N_{B} = 0\nN_{D} = -N_{B}\n\n\u21d0 \\sum F_{y}=0;\n-V_{D} + V_{B} = 0\nV_{D} = -V_{B} = 175N\nAns\n\n\u21d0 \\sum \\Sigma M_{D}=0;\n-M_{D} - M_{E} - B_{y}(a + b) = 0\nM_{D} = -M_{E} - B_{y}(a + b)\nM_{D} = -8.75 N-m\nAns Problem 1-6\nThe beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal loadings acting on the cross section at point D.\nGiven: P = 5000N\na = 0.8m b = 1.2m c = 0.6m d = 1.6m\ne = 0.6m.\nSolution:\u03b8 = atan(b/d) = 36.87 deg\n\u03c6 = atan(a/b) - \u03b8 = 14.47 deg\nMember AB:\n\n\u21d0 \\sum M_{A}=0;\nF_{BC} \\sin(\u03phi)(a + b) - P \\cdot (b) = 0\nF_{BC} = \\frac{P (b)}{\\sin(\u03phi)(a + b)}\nF_{BC} = 12.01 kN\n\nSegment BD:\n\u21d0 \\sum F_{y}=0;\n-N_{D} \\cdot \\cos(\u03phi) - P \\cdot \\cos(\u03theta) = 0\nN_{D} = F_{BC} \\cdot \\cos(\u03phi) - P \\cdot \\cos(\u03theta) = 0\n\n\u21d0 \\sum F_{x}=0;\nV_{D} + F_{BC} \\sin(\u03phi) - P \\cdot \\sin(\u03theta) = 0\nV_{D} = -F_{BC} \\sin(\u03phi) + P \\cdot \\sin(\u03theta)\nV_{D} = 0 kN\nAns\n\n\u21d0 \\sum \\Sigma M_{D}=0;\n(F_{BC} \\sin(\u03phi) - P \\cdot \\sin(\u03theta)) \\cdot \\frac{d - c}{\\sin(\u03theta)} - M_{D} = 0\nM_{D} = 0 kN-m\nAns\nNote: Member AB is the two-force member. Therefore the shear force and moment are zero. Problem 1-7\nSolve Prob. 1-6 for the resultant internal loadings acting at point E.\nGiven: P = 5000N\na = 0.8m b = 1.2m c = 0.6m d = 1.6m e = 0.6m\nSolution:\n\\u03b8 = atan( \\frac{b}{d}) = 36.87 deg\n\\u03d5 = atan( \\frac{a + b}{d}) - \\u03b8 = 14.47 deg\n\nMember AB:\n\\u27b1 %u03a3M_E = 0:\nF_{BC} \\sin(\\u03b8)(a + b) - P(b) = 0\nF_{BC} = \\frac{P(b)}{\\sin(\\u03b8)(a + b)}\nF_{BC} = 12.01 kN\n\nSegment BE:\n\\u27b1 %u03a3F_x = 0:\n-N_E - F_{BC} \\cos(\\u03b8) - P \\cos(\\u03b8) = 0\nN_E = -F_{BC} \\cos(\\u03b8) - P \\cos(\\u03b8) = -15.63 Ans\n\\u27b1 %u03a3F_y = 0:\nV_E + F_{BC} \\sin(\\u03b8) - P \\sin(\\u03b8) = 0\nV_E = -F_{BC} \\sin(\\u03b8) + P \\sin(\\u03b8) = 0 kN\nAns\n\\u27b1 %u03a3M_E = 0:\n(F_{BC} \\sin(\\u03b8) - P \\sin(\\u03b8)) e - M_E = 0\nM_E = (F_{BC} \\sin(\\u03b8) - P \\sin(\\u03b8)) e = 0 kN.m\nAns\nNote: Member AB is the two-force member. Therefore the shear force and moment are zero. Problem 1-8\nThe boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and load weight 1500 N, determine the resultant internal loadings in the crane on cross sections through points A, B, and C.\nGiven: P = 1500N w = 750 N/m\na = 2.1m b = 1.5m c = 0.6m d = 2.4m e = 0.9m\nSolution:\nEquations of Equilibrium: For point A:\n\\u27b1 %u03a3F_x = 0:\nN_A = 0 Ans\n\\u27b1 %u03a3F_y = 0:\nV_A - w - P = 0\nV_A = w e + P = 2.17 kN Ans\n\\u27b1 %u03a3M_A = 0:\n-M_A - (w-e)(0.5 e) - P(e) = 0\nM_A = - (w-e)(0.5 e) - P(e) = -1.654kN.m Ans\nNote: Negative sign indicates that M_A acts in the opposite direction to that shown on FBD.\nEquations of Equilibrium: For point B:\n\\u27b1 %u03a3F_x = 0:\nN_B = 0 Ans\n\\u27b1 %u03a3F_y = 0:\nV_B = w(d + e) - P = 0\nV_B = w(d + e) + P = 3.98 kN Ans\n\\u27b1 %u03a3M_B = 0:\n-M_B - [w(d + e)][0.5(d + e)] - P(d + e) = 0\nM_B = -w(d + e)[0.5(d + e)] - P(d + e) = -9.034kN.m Ans\nNote: Negative sign indicates that M_B acts in the opposite direction to that shown on FBD.\nEquations of Equilibrium: For point C:\n\\u27b1 %u03a3F_x = 0:\nV_C = 0 Ans\n\\u27b1 %u03a3F_y = 0:\n-N_C - w(b + c + d + e) - P = 0\nN_C = -w(b + c + d + e) - P = -5.55 kN Ans\n\\u27b1 %u03a3M_C = 0:\n-M_C - [w(c + d + e)][0.5(c + d + e)] - P(c + d + e) = 0\nM_C = - w(c + d + e)[0.5(c + d + e)] - P(c + d + e) = -11.554kN.m Ans\nNote: Negative sign indicates that N_C and M_C act in the opposite direction to that shown on FBD. Problem 1-9\nThe force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a-a.\nGiven: P = 400N \\u03b8 = 30deg \\u03c6 = 45deg\na = 4mm b = 5.75mm\nSolution:\n\\u03b1 = \\u03c6 - \\u03b8 = \\u03c6 = 45deg\nEquations of equilibrium: For section a-a:\n\\u27b1 %u03a3F_x = 0:\nV_A - P \\cos(\\u03b1) = 0\nV_A = P \\cos(\\u03b1) = 386.37N Ans\n\\u27b1 %u03a3F_y = 0:\nN_A - P \\sin(\\u03b1) = 0\nN_A = P \\sin(\\u03b1) = 103.53N Ans\n\\u27b1 %u03a3M_A = 0:\n-M_A - P \\sin(\\u03b1)a + P \\cos(\\u03b1)b = 0\nM_A = -P \\sin(\\u03b1)a + P \\cos(\\u03b1)b = 1.808 N.m Ans