Sinh viên: Nguyễn Thanh Phong MSSV: G1002398 Lớp: GT10HK KHÍ ĐÀN HỒI Giảng viên: TS Trần Tiến Anh Ngày nộp: 06/03/2014 BÀI TẬP 1-4 Introduction to Structural Dynamics and Aeroelasticity Problem 1: Show that the equation of motion for longitudinal vibration pf a uniform beam is the same ass that for a string, viz EA Solution: ∂ 2u ∂ 2u = m ∂x ∂t y x ds T+ u v u+ ∂u dx ∂x ∂T dx ∂x dx We have: With β = So : ∂ ∂ 2u T =m ∂x ∂t & 0=m ∂ 2v ∂t Such as the force of the rod vertically then u turn stretches over time, and v is constant (the derivative with time 0) Equation (2.19) T = EAε Let us presuppose the existence of a static-equilibrium solution of the string deflection so that whereT0 and ε o are constants and δ=l−l0 is the change in the length of the string between its stretched and unstretched states If the steady-state tension T0 is sufficiently high, the perturbation deflections about the staticequilibrium solution are very small Thus, we can assume: we find that the equations of motion can be reduced to two linear partial differential equations: ∂ 2uˆ ∂ 2uˆ EA = m ∂x ∂t Problem 2: Show that the equation of motion for longitudinal vibration pf a uniform beam is the same ass that for a string, viz l  ∂u  P = ∫ EA  ÷ dx  ∂x  Solution: To solve problems involving the forced response of strings using Lagrange’s equation, we need an expression for the strain energy, which is caused by extension of the string, viz where, as before and the original length is0 To pick up all of the linear terms in Lagrange’s equations, we must include all terms in the energy up through the second power of the unknowns Taking the pertinent unknowns to be perturbations relative to the stretched but undeflected string, we can again write ForEAequal to a constant, the strain energy is: Because : Strain energy simplifies to the strain energy becomes We assuming that v is const and ε o