Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_110

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the shear force and bending moment at a section of the loaded beam 200 mm to the right of . Problem 5/110 (a) Support Reactions 1. Formula: To find the reactions, we first determine the resultant force of the distributed load and its position. Then we apply the equilibrium equations for the entire beam: 2. Substitution: Total length of the beam: . Distributed load magnitude: . Length of distributed load: . Resultant of distributed load: . Distance of from : . Distance of from : . For moment about : 3. Calculation: Using vertical equilibrium: 4. Result: Support reactions are and . VM A M = ∑ B 0, F = ∑ y 0 L=300 mm+300 mm=600 mm= 0.6 m w=6 kN/m L = w 300 mm=0.3 m W=w⋅L = w 6 kN/m⋅0.3 m=1.8 kN WAd = A = 2 0.3 m 0.15 m WBd = B 0.6 m−0.15 m=0.45 m B A (0.6 m)− y (1.8 kN)(0.45 m)=0 A = y = 0.6 1.8⋅0.45 1.35 kN A + y B − y W=0⇒1.35+B − y 1.8=0 B = y 0.45 kN A = y 1.35 kNB = y 0.45 kN ● Final Conclusion: The reaction at support is upwards. (b) Internal Shear Force 1. Formula: We consider the free-body diagram of the left portion of the beam from to the section at . 2. Substitution: Resultant of the distributed load on this 200 mm segment: . 3. Calculation: 4. Result: ● Final Conclusion: The internal shear force at the section 200 mm to the right of is . (c) Internal Bending Moment 1. Formula: Applying the moment equilibrium about the cut section for the left portion: 2. Substitution: 3. Calculation: 4. Result: A1.35 kN V A x=20