Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_4

⚡ Mecademy AIENG정역학 · ch3  Problem Statement The 450-kg uniform I-beam supports the load shown. Determine the reactions at the supports. Problem 3/4 (a) Reaction at Support B 1. Formula: Sum of moments about point A to eliminate the reaction at A: 2. Substitution: Beam mass , Load mass , Acceleration due to gravity Beam weight Load weight Total length Load distance from A 3. Calculation: Moment from beam weight: Moment from external load: Total Moment: 4. Result: ● Final Conclusion: The vertical reaction force at support B is acting upwards. (b) Reaction at Support A M = ∑ A 0⟹R ⋅ B L−W ⋅ beam − 2 L W ⋅ load d = L 0 m = B 450kgm = L 220kg g= 9.81m/s 2 W = beam 450⋅9.81=4414.5N W = load 220⋅9.81=2158.2N L=5.6m+2.4m=8.0m d = L 5.6m R ⋅ B 8.0=(4414.5⋅4.0)+(2158.2⋅5.6) 17658N⋅m 12085.92N⋅m 17658+12085.92=29743.92N⋅m R = B = 8.0 29743.92 3717.99N R ≈ B 3.72kN 3.72kN 1. Formula: Sum of vertical forces for static equilibrium: 2. Substitution: 3. Calculation: Total weight 4. Result: ● Final Conclusion: The vertical reaction force at support A is acting upwards. ✨ Final Answer Summary (a) (upward) (b) (upward) Mecademy AI Solution · ENGProblem 3/4 F = ∑ y 0⟹R + A R − B W − beam W = load 0 R + A 3717.99−4414.5−2158.2=0 W = total 4414.5+2158.2=6572.7N R = A 6572.7−3717.99 R = A 2854.71N R ≈ A 2.85kN 2.85kN R = B 3.72kN R = A 2.85kN