Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_126

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Repeat Prob. 5/125, where the 1500-N load has been replaced by the 4.2-kN·m couple. (Prob. 5/125: 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 .) [Note: Referring to Problem 5/125 on the previous page, the beam is supported by a pin at () and a roller at (). It is subjected to a uniform distributed load of between and . For Prob. 5/126, the point load at is replaced by a clockwise couple of .] Problem 5/126 (a) Support Reactions and Shear/Moment Diagrams 1. Formula: 2. Substitution: Equivalent concentrated load of distributed load: at . Applied clockwise couple at : . 3. Calculation: Shear Diagram: : : ; At , . (Linear decrease) : x=6 m M max Ax=0Bx=9 m 800 N/mx=2 mx=5 m 1500 Nx=7 m4.2 kN⋅ m M = ∑ A 0, F = ∑ y 0 F = dist 800 N/m× 3 m=2400 N=2.4 kNx=2+1.5=3.5 m x=7 mM = C 4.2 kN⋅ m M =∑ A 2.4(3.5)+4.2−R (9)= B 0 F =∑ y R − A 2.4+R = B 0 9R = B 8.4+4.2=12.6⟹R = B 1.4 kN R = A 2.4−1.4=1.0 kN 0<x<2 mV=1.0 kN 2<x<5 mV=1.0−0.8(x−2)x=5 mV=−1.4 kN 5<x<9 mV=−1.4 kN Moment Diagram: : ; At , . : ; At (where ), . At , . : ; At , ; At , . At : Positive jump of due to clockwise couple . : ; At , . 4. Result: Support reactions are and . ● Final Conclusion: The shear diagram consists of a constant value of , followed by a linear decrease from to , and stays constant at to the end. The moment diagram