Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_127

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Plot the shear and moment diagrams for the beam subjected to the concentrated force and distributed load. State the values of the largest positive and largest negative bending moments and give the location in the beam where each occurs. Problem 5/127 (a) Support Reactions 1. Formula: First, we determine the load function . Based on the diagram, the load is defined from to as . We calculate the resultant force of the distributed load and its moment about support at : Then, apply equilibrium equations for the whole beam (length ): 2. Substitution: From the diagram, and . At : . At : . Now calculate and : 3. Calculation: Substitute into equilibrium equations: w(x) x=0x=10 mw=w − 0 kx 3/2 RM R Ax= 0 R= w (x)dx, M = ∫ 0 10 R x w (x)dx ∫ 0 10 L=13 m M = ∑ A 0⟹R (10)− B M − R 7(13)=0 F = ∑ y 0⟹R + A R − B R−7=0 w(0)=2.5 kN/mw(10)=1.5 kN/m x=02.5=w − 0 k(0)⟹w = 0 2.5 kN/mx=101.5= 2.5−k (10)⟹ 3/2 k= ≈ 10 10 1 0.03162RM R R= (2.5− ∫ 0 10 x )dx= 10 10 1 3/2 2.5x− ⋅ = [ 10 10 1 2.5 x 2.5 ] 0 10 25− 2. M = R x(2.5− ∫ 0 10 x )dx= 10 10 1 3/2 1.25x− ⋅ = [ 2 10 10 1 3.5 x 3.5 ] 0 10 12 10R − B 96.43−91=0⟹10R = B 187.43⟹R = B 18.743 kN R + A 18.743−21−7=0⟹R = A 28−18.743=9.257 kN 4. Result: (upward) (upward) ● Final Conclusion: The support reactions are at the pin support and at the roller support. (b) Shear and Moment Equations 1. Formula: Segment 1 (): Segment 2 (): By cutting from the right end (): 2. Substitution: For : 3. Calculation: Check b