Statistics > SOLUTIONS MANUAL > Assignment 6 solutions - University of Ontario Institute of Technology MECE 2230U (All)

Assignment 6 solutions - University of Ontario Institute of Technology MECE 2230U

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The boom DF of the jib crane and the column DE have a uniform weight of 50 If the hoist and load weigh 300 lb, determine the normal force, shear force, and moment in the crane at sections passing t ... hrough points A, B, and C. lb>ft. SOLUTION Ans. Ans. a Ans. Ans. Ans. a Ans. Ans. Ans. a+ ©MC = 0; -MC - 650(6.5) - 300(13) = 0; MC = -8125 lb Ans. # ft + c ©Fy = 0; NC - 650 - 300 - 250 = 0; NC = 1200 lb :+ ©Fx = 0; VC = 0 + ©MB = 0; -MB - 550(5.5) - 300(11) = 0; MB = -6325 lb # ft + c ©Fy = 0; VB - 550 - 300 = 0; VB = 850 lb :+ ©Fx = 0; NB = 0 + ©MA = 0; -MA - 150(1.5) - 300(3) = 0; MA = -1125 lb # ft + c ©Fy = 0; VA - 450 = 0; VA = 450 lb :+ ©Fx = 0; NA = 0 5 ft 7 ft C D F E B A 300 lb 2 ft 8 ft 3 ft *7–12. Determine the internal normal force, shear force, and the moment at points C and D. SOLUTION Support Reactions: FBD (a). a Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have Ans. Ans. a Ans. Applying the equations of equilibrium to segment BD [FBD (c)], we have Ans. Ans. a MD = 16.5 kN Ans. # m + ©MD = 0; 8.485132 - 611.52 - MD = 0 + c ©Fy = 0; VD + 8.485 - 6.00 = 0 VD = -2.49 kN : ND = 0 + ©Fx = 0; MC = 4.97 kN # m + ©MC = 0; MC - 3.515 cos 45°122 = 0 a+©Fy¿ = 0; 3.515 sin 45° - NC = 0 NC = 2.49 kN Q+ ©Fx¿ = 0; 3.515 cos 45° - VC = 0 VC = 2.49 kN :+ ©Fx = 0 Ax = 0 + c ©Fy = 0; Ay + 8.485 - 12.0 = 0 Ay = 3.515 kN By = 8.485 kN + ©MA = 0; By 16 + [Show More]

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