Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_38

⚡ Mecademy AIENG정역학 · ch4  Problem Statement The members and of the loaded truss cross but are not connected to members and . Compute the forces in members , , , and . Problem 4/38 (a) Reaction Forces and Force in Member ● Calculation Process 1. Reaction Forces: We establish a coordinate system with joint at the origin . The truss is supported by a pin at and a roller at . The external loads are: at joint , acting at to the horizontal (pointing into the joint: , ). downward at . downward at . downward at . Summing moments about : Summing vertical forces: Summing horizontal forces: 2. Force in : Analyze joint . Now analyze joint . Since is the only vertical member and there are no external CJCFBI DGBCCJCIHI BC J(0,0) JF(12,0) 6 kNA(0,4)60 ∘ F = Ax 6cos(60)= ∘ 3 kNF = Ay −6sin(60)= ∘ −5.196 kN 4 kNI(3,0) 10 kNH(6,0) 8 kNG(9,0) J M = ∑ J −F (4)− Ax 4(3)−10(6)−8(9)+R (12)= F 0 −(3)(4)−12−60−72+12R = F 0⇒−156+12R = F 0⇒R = F 1 F = ∑ y J + y R − F 5.196−4−10−8=0⇒J + y 13−27.196=0 F = ∑ x J + x F = Ax 0⇒J + x 3=0⇒J = x −3 kN (to the left) BCA F = ∑ x F + AB 3=0⇒F = AB −3 kN (Compression) BBI loads at , . Summing horizontal forces at : 3. Result: (Compression) ● Final Conclusion: The force in member is in compression. (b) Force in Member ● Calculation Process 1. Joint Analysis: Joint is connected to members , , and the diagonal . From joint : (Compression, pushes down on ). Diagonal connects to . The angle with the horizontal is . . 2. Substitution: Summing vertical forces at Joint