Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_10

⚡ Mecademy AIENG정역학 · ch4  Problem Statement If the maximum tensile force in any of the truss members must be limited to 24 kN, and the maximum compressive force must be limited to 35 kN, determine the largest permissible mass which may be supported by the truss. Problem 4/10 (a) Largest permissible mass 1. Formula: First, we determine the geometry and then the member forces in terms of the load . From the diagram, the horizontal width of each bay is . At joint , the angle between member and the horizontal member is . The height of the truss is: Using the method of joints starting from joint : Joint : Joint : Joint : (Member angle ) m m W=mgw=4 m CBCCD30 ∘ h h=wtan30= ∘ 4tan30= ∘ ≈ 3 4 2.309 m C C ↑F = ∑ y 0⟹F sin30− BC ∘ W=0⟹F = BC = sin30 ∘ W 2W →F = ∑ x 0⟹−F − CD F cos30= BC ∘ 0⟹F = CD −2Wco B ↑F = ∑ y 0⟹−F − BD F sin30= BC ∘ 0⟹F = BD −2W(0.5 →F = ∑ x 0⟹−F + AB F cos30= BC ∘ 0⟹F = AB 2 W( 2 DADφ=30 ∘ ↑F = ∑ y 0⟹F + BD F sin30= AD ∘ 0⟹−W+F (0.5)= AD →F = ∑ x 0⟹F − CD F − ED F cos30= AD ∘ 0⟹ − W − 3 2. Substitution: Identify the maximum tensile and compressive forces: Apply the given limits: Tensile limit: Compressive limit: 3. Calculation: The governing maximum weight is the smaller of the two limits: Calculate mass using : 4. Result: ● Final Conclusion: The largest permissible mass is (or ). ✨ Final Answer Summary (a) (or ) Mecademy AI Solution · ENGProblem 4/10 T = max 2W C = max 2 W ≈ 33.464W 2W≤24 kN⟹W≤12 kN 3.464W≤35 kN⟹W≤ ≈ 2 3 35 10.104 kN W = max 10.104 kN=10104 N