Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_32

⚡ Mecademy AIENG정역학 · ch7  Problem Statement The uniform bar of mass and length is supported in the vertical plane by two identical springs each of stiffness and compressed a distance in the vertical position . Determine the minimum stiffness which will ensure a stable equilibrium position with . The springs may be assumed to act in the horizontal direction during small angular motion of the bar. Problem 7/32 (a) Determination of Minimum Spring Stiffness for Stability 1. Formula: The stability of a system at an equilibrium position is determined by the second derivative of its total potential energy with respect to the degree of freedom . For stable equilibrium at , we must have: The total potential energy is the sum of gravitational potential energy and elastic potential energy : 2. Substitution: Taking the pivot point at the bottom as the datum (), the height of the center of mass for a small angle is . In the vertical position (), both springs are compressed by a distance . For a small angular displacement , the top of the bar moves horizontally by . One spring's compression increases to , while the other decreases to . mL kδθ=0 kθ= 0 Vθ θ=0 > dθ 2 dV 2 θ=0 0 VV g V e V=V + g V e y=0 θh = G cosθ 2 L V = g mgh = G cos θ 2 mgL θ=0δ θx= Lsinθδ+Lsinθ δ−Lsinθ V = e k ( δ + 2 1 Lsinθ)+ 2 k ( δ − 2 1 Lsinθ) 2 Substituting these into the total potential energy expression: 3. Calculation: Simplify the elastic potential energy term: Total potential energy: First derivative (equilib