Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_125

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Plot the shear and moment diagrams for the beam loaded with both distributed and point loads. What are the values of the shear force and bending moment at ? Determine the maximum bending moment . Problem 5/125 (a) Support Reactions and Shear/Moment Diagrams ● Calculation Process 1. Present Final Formula: To find the support reactions at () and ( ), apply the equations of equilibrium: The total load from the distributed section is , acting at . 2. Substitute Values: 3. Calculation: Shear Diagram (): : : . At , . : : Moment Diagram (): x=6 m M max Ax=0Bx= 9 m M = ∑ A 0, F = ∑ y 0 F = w w⋅L = w 800 N/m⋅3 m= 2400 Nx=2+1.5=3.5 m M = ∑ A (2400)(3.5)+(1500)(7)−R (9)= B 0 F = ∑ y R + A R − B 2400−1500=0 9R = B 8400+10500=18900⟹R = B 2100 N R = A 3900−2100=1800 N V 0≤x<2 mV=1800 N 2≤x<5 mV(x)=1800−800(x−2)=3400−800x x=5V=−600 N 5≤x<7 mV=−600 N 7≤x<9 mV=−600−1500=−2100 N M (parabolic curve) 4. Result: Reactions are and . ● Final Conclusion: The shear diagram starts at , decreases linearly across the distributed load to , stays constant until the point load, then drops to . The moment diagram consists of linear segments and a parabolic arch reaching its peak within the distributed load region. (b) Shear Force and Bending Moment at ● Calculation Process 1. Present Final Formula: At , which is in the region : 2. Substitute Values: 3. Calculation: 4. Result: , . ● Final Conclusion: At , the shear force is and the bending moment is . (c) Maximum