Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_37

⚡ Mecademy AIENG정역학 · ch4  Problem Statement Determine the force in member for the loaded truss. The horizontal distances between vertical members are each. The bottom chord is inclined at below the horizontal, and the segment of the top chord is inclined at above the horizontal. Assume the top chord segments and are collinear. Problem 4/37 (a) Force in member BF 1. Formula: The equilibrium of joints is governed by the equations: Geometric relations based on the given horizontal spacing and angles (top chord) and (bottom chord) are used to find joint coordinates. 2. Substitution: Coordinates with origin at : Angle of member (): 3. Calculation: Joint A: BF 5 ′ AGFE15 ∘ AB30 ∘ AB,BC,CD F = ∑ x 0, F = ∑ y 0 d=5 ′ θ = 1 30 ∘ θ = 2 15 ∘ A(0,0) G=(5,−5tan15)= ∘ (5,−1.340) ′ B=(5,5tan30)= ∘ (5,2.887) ′ F=(10,−10tan15)= ∘ (10,−2.679) ′ BFθ BF θ = BF tan = −1 ( x −x FB y −y FB ) tan = −1 ( 10−5 −2.679−2.887 )−48.06 ∘ F =∑ x F cos30+ AB ∘ F cos15= AG ∘ 0⟹F = AG −0.8966F AB F =∑ y F sin30− AB ∘ F sin15− AG ∘ 9=0 F (0.5+ AB 0.8966sin15)= ∘ 9⟹0.7321F = AB 9⟹ F = AB 12.294 kips (T) Joint G: Joint B: -- (1) -- (2) From (2), . Substituting into (1): 4. Result: ● Final Conclusion: The force in member is acting in compression. ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 4/37 F = AG −11.023 kips (C) F =∑ x 0⟹F = GF F = AG −11.023 kips (C) F =∑ y 0⟹F + GB (11.023−11.023)sin15− ∘ 3= 0⟹F = GB 3 kips (T) F =∑ x F cos30+ BC ∘ F cos(−48.06)− BF ∘ F cos30= AB ∘ 0 0.8660F + BC 0.6684F