Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_21

⚡ Mecademy AIENG정역학 · ch4  Problem Statement The loading of Prob. 4/20 is shown applied to a Howe roof truss. Neglect any horizontal reactions at the supports and solve for the forces in all members. Compare with the results of Prob. 4/20. Problem 4/21 (a) Determination of support reactions and truss geometry 1. Formula: From equilibrium of the entire truss: 2. Substitution: Total vertical load = . Due to symmetry, the reactions at and are equal: 3. Calculation: The truss has a total span of () and a central height at of . The angle of the top chord members () with the horizontal is: For the diagonals and , they connect to . The horizontal distance is and the vertical is . Thus, they have the same angle with the horizontal. 4. Result: Support reactions: , . ● Final Conclusion: The external reactions are determined as at each end support. The slope angle for chords and diagonals is . (b) Force in each member of the loaded Howe truss F = ∑ y 0, M = ∑ A 0 1+1+1+1+1=5 kN AE R = A R = E = 2 5 kN 2.5 kN 8 m4×2 mC 2 mθAB,BC,CD,DE tanθ= = 4 m 2 m 0.5⟹θ=26.57 ∘ sinθ= ,cos θ = 5 1 5 2 BGDGB(2,1)G(4,0) 2 m1 mθ R = A 2.5 kN↑R = E 2.5 kN↑ 2.5 kN tanθ=0.5 1. Formula: Apply the method of joints at each connection: , . 2. Substitution: Joint A: Joint H: No vertical diagonal or load at , so . Thus . Joint B: 3. Calculation: At A: (Compression); (Tension). At H: ; (Tension). At B: Let (pushed into joint). Solving the system: (C); (C). At G: By symmetry, (T), (C). Vertical equilibrium: 4. Resu