Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_15

⚡ Mecademy AIENG정역학 · ch4  Problem Statement Each member of the truss is a uniform 20-ft bar weighing 400 lb. Calculate the average tension or compression in each member due to the weights of the members. Problem 4/15 (a) Analysis of Support Reactions and Member Forces 1. Formula: To analyze a truss including member weights, we distribute half of each member's weight to its end joints. Let be the weight of each member. The load at each joint is: Equilibrium equations for the whole structure (due to symmetry, ): Member forces are then found using the method of joints: 2. Substitution: Joint Loads: Joint (members ): Joint (members ): Joint (members ): Joint (members ): Joint (members ): Total Weight: Support Reactions (by symmetry): W=400 lb L = j m at j ∑ 2 W A = x 0 F = ∑ y 0⟹R + A R = C L ∑ j F = ∑ x 0, F = ∑ y 0 AAE,ABL = A + 2 400 = 2 400 400 lb BAB,BE,BD,BCL = B 4× = 2 400 800 lb CBC,CDL = C + 2 400 = 2 400 400 lb EAE,BE,EDL = E 3× = 2 400 600 lb DED,BD,CDL = D 3× = 2 400 600 lb W = tot 400+800+400+600+600=2800 lb R = A R = C = 2 2800 1400 lb 3. Calculation: Joint : Joint : Using Symmetry: 4. Result: ● Final Conclusion: The member forces due to their own weight are: 1155 lb (Compression) for and ; 577 lb (Tension) for and ; 462 lb (Tension) for and ; and 808 lb (Compression) for . ✨ Final Answer Summary (a) , , , A F =∑ y 0⟹R − A L + A F sin60= AE ∘ 0 1400−400+F = AE 2 3 0⟹F = AE − ≈ 3 2000 −1154.7 lb F =∑ x 0⟹F + AB F cos60= AE ∘ 0 F + AB (−1154.7)(0.5)=0⟹F = AB 577.35