Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_30

⚡ Mecademy AIENG정역학 · ch7  Problem Statement The torsional spring at has a stiffness and is undeformed when bars and are in the vertical position and overlap. Each uniform bar has mass . Determine the equilibrium configurations of the system over the range and the stability of the system at each equilibrium position for , , and . Problem 7/30 (a) Potential Energy Formulation 1. Formula: The total potential energy of the system is the sum of the gravitational potential energy and the elastic potential energy of the torsional spring . The height of the center of mass for each bar of length is . Since there are two identical bars, the total gravitational potential energy (taking the base line as datum) is: The torsional spring at joint is undeformed when , at which point the interior angle between the bars is zero. For a general angle , the interior angle is . The angular deformation is thus . The elastic energy is: 2. Substitution: Substitute the given values: , , , and . Ak T OAAB m 0≤θ≤90 ∘ m=1.25 kgb=750 mmk = T 1.8 N⋅m/rad V V g V e V=V + g V e bh = c bsinθ 2 1 OB V = g 2⋅ m g sinθ = ( 2 b )mgbsinθ Aθ=90 ∘ θφ= π−2θδ=φ−0=π−2θ V = e kδ = 2 1 T 2 k ( π − 2 1 T 2θ) 2 m=1.25 kgb=0.750 mg= 9.81 m/s 2 k = T 1.8 N⋅m/rad mgb=(1.25)(9.81)(0.750)=9.196875 J k = T 1.8 J/rad (as energy coefficient) 2 3. Calculation: 4. Result: ● Final Conclusion: The potential energy of the system as a function of is established for equilibrium analysis. (b) Equilibrium Configurations 1. Formula: Equil