Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_116

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Plot the shear and moment diagrams for the beam loaded as shown. State the maximum magnitudes of the shear force and bending moment for the beam. Problem 5/116 (a) Support Reactions ● Calculation Process 1. Present Final Formula: We apply equilibrium equations to the entire beam. Let be at the left end. Support is at and support is at the right end, . Resultant of uniform load: at . Concentrated load: at . Resultant of triangular load: at . The equilibrium equations are: 2. Substitute Values: x=0Ax=6mB x=6+3+3+12=24m F = 1 w L = 11 4kN/m⋅6m=24kN x=3m P=24kNx=9m F = 2 w L = 2 1 22 ⋅ 2 1 6kN/m⋅12m= 36kNx=12+ (12)= 3 2 20m M = ∑ A 0, F = ∑ y 0 M = ∑ A −24(6−3)+24(9−6)+36(20−6)−R (24− B 6)=0 −24(3)+24(3)+36(14)−18R = B 0 3. Partial Operations: From : 4. Final Calculation: ● Final Conclusion: The support reactions are and . (b) Maximum Magnitude of Shear Force ● Calculation Process 1. Present Final Formula: Define the shear force for each segment. : . At , . : jumps by . At , . : . : jumps by . At , . : . : . At , . 2. Substitute Values: Compare magnitudes across all segments. Max in segment 1: Max in segment 2: Max in segment 3: Max in segment 4: (at ) 3. Final Calculation: The maximum absolute value of shear occurs in the segment between support and the concentrated load. 4. Result: −72+72+504−18R = B 0 18R = B 504 R = B 28kN F =∑ y 0 R + A R = B F + 1 P+F = 2 24+24+36=84kN R = A 84−28=56kN R = A 56kNR = B 28kN V(x) 0≤x<6V(x)=−4xx=6