Ejercicio 4.116 Resuélvase el problema 4.115 suponiendo que el
Transcripción
Ejercicio 4.116 Resuélvase el problema 4.115 suponiendo que el
Ejercicio 4.116 Resuélvase el problema 4.115 suponiendo que el cable EF es reemplazado or un cable unido a los puntos E y H. EF = (−30) 2 + (12) 2 + (−20) 2 = 38in FEHx = -30 in * EH = -0.79 EH 38 in FEHy = 12 in * EH = 0.315 EH 38 in FEHz = -20 in * EH = -0,526 EH 38 in ∑ MA = 0 = M WA + M AB + M AE + M AA ∑ MA = [(15” i + 10” k) x (-75 lb j)] + [(30” i) x (RBYj + RBz k)] + [(26” i + 20” k) x (-0,79 EH i + 0.315 EH j – 0.526 EH k)] = 0 ∑ MA = 0 = (-1125 lb” k + 750 lb” i)+(30” By k – 30 Bz j) + (8,19” EH k) + (13,67” EH j – 15,8” EH j – 6.3” EH i) ∑ MAx = EH = ∑ MA i = 0 = 750 lb” – 6,3” EH 119 lbs FEHx = FEHy = FEhz = -0,79 (119 lb) → FEHx = -94 lb 0,315 (119 lb) → FEhy = 37.5 lb -0,526 (119 lb) → FEhz = -62,6 lb ∑ MAY = -30 Bk + 13,67” EH – 15,8 EH = 0 -30” Bk + 1626,73” – 1880,20” = 0 Bk = -8,45 lb ∑ MAZ = ∑ MA k = -1125 lb” + 30” By + 8.19 EH = 0 -1125 lb” + 30” By + 974.61 = 0 By = 5 lb B = (5 lb) j – (8.45 lb) k ∑ Fx = 0 = Ax – 94 lb Ax = 94 lb ∑ Fy = 0 = Ay + 37.5 lb – 76 lb + 5 lbs Ay = 32,5 lb ∑ F2 = Az – 8,45 lb – 62,6 lb = 0 Az = 71 lbs A = (94 lb) i + (32,5 lb) j + (71 lb) k A = (94 lb) i + (32,5 lb) j + (71 lb) k A = 122,20 lb B = (5 lb) i – (8,45 lb) k B = 9.82 lb