Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_115

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Draw the shear and moment diagrams for the beam subjected to the combination of distributed and point loads. Determine the values of the shear force and bending moment at point C, which lies 3 m to the left of B. Problem 5/115 (a) Determination of Support Reactions 1. Formula: Sum of vertical forces: Sum of moments about A: 2. Substitution: First, find the total equivalent concentrated forces for the distributed loads: For the load acting over : . Its location is from A. For the load acting over : . Its location is from A. Now, apply equilibrium equations: 3. Calculation: Total downward force: 4. Result: , ● Final Conclusion: The reaction forces at the supports are upward and upward. F =∑ y 0 M =∑ A 0 2 kN/m4 mF = 1 2×4=8 kN x = 1 2+ = 2 4 4 m 3 kN/m6 mF = 3 3×6=18 kN x = 3 6+ = 2 6 9 m M = ∑ A (8 kN⋅4 m)+(5 kN⋅6 m)+(18 kN⋅9 m)+(4 kN⋅15 m 32+30+162+60=12R B 284=12R ⇒ B R = B = 12 284 23.67 kN F = total 8+5+18+4=35 kN R = A F − total R = B 35−23.67=11.33 kN R = A 11.33 kNR = B 23.67 kN R = A 11.33 kN R = B 23.67 kN (b) Values of Shear and Bending Moment at point C 1. Formula: Cut the beam at point C ( from A). Consider the left portion. 2. Substitution: Point C is to the left of B, so its position is . The portion of the load to the left of C is from to , which has a force of acting at . 3. Calculation: 4. Result: , ● Final Conclusion: At point C, the shear force is and the bending moment is . (c) Shear and Bending Moment Diagrams