Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_191

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Plot the shear and moment diagrams for the beam subjected to the two concentrated forces and combination of distributed loads. State the largest positive and negative values of the bending moment and their locations along the beam. Problem 5/191 (a) Support Reactions 1. Formula: 2. Substitution: Consider the left end of the beam as the origin (). The beam's geometry and loads are: Pin support A at . Roller support B at . Downward triangular load: total force at centroid . Upward uniform load: total force at centroid . Downward concentrated force: at . Downward concentrated force: at the right end . Moment equation about support A: 3. Calculation: M = ∑ A 0, F = ∑ y 0 x=0 x=3 m x=3+18=21 m 12 kNx=4 m 0.75 kN/m×18 m=13.5 kN x=3+9=12 m 6 kNx=12+4=16 m 3 kNx=3+18+3+ 4=28 m ↺M = ∑ A 0:−12(4−3)+13.5(12−3)−6(16−3)+R (18)− B 3 −12(1)+121.5−78+18R − B 75=0 18R = B 12−121.5+78+75=43.5 R = B = 18 43.5 2.4167 kN (upward) Vertical equilibrium: 4. Result: , ● Final Conclusion: The reaction at support A is upward and the reaction at support B is upward. (b) Shear and Bending Moment Equations 1. Formula: 2. Substitution: From to : : ; . At . : ; . At . : ; . At . Shear is zero at . : ; . At . Shear is zero at . : ; . At . 3. Calculation: Find critical points where shear is zero to determine local extrema for the moment: In , at : . ↑F = ∑ y 0:R + A R + B 13.5−12−6−3=0 R + A 2.4167+13.5−21=0 R = A 21−15.9167=5.0833 kN (upward) R = A 5.083 kNR = B