M01 Axial vibrations of a bar

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Mechanical System Dynamics - Lecture Notes MSc. Mechanical Engineering A.A. 2022-2023 Vibration analysis of one-dimensional continuous systems Axial Vibrations of a bar Teacher: Prof. Stefano Melzi Trainer: Eng. Binbin Liu Fabio Santoro Contents 1 W ave equation 2 2 Stationary solution 3 2.1 Case: Clamped-Clamped Bar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2 Case: Free-Free Bar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1 1. Wave equation To study the axial vibrations of a beam, let’s consider the model represented in figure 1. It is a representation of a beam with length L. The space variable is x and the goal is to identify the expression of the horizontal displacementu(x, t) It is necessary to assess some hypothesis: Figure 1: Clamped-Clamped axial beam model 1. the beam moves (deforms) just along the axial direction (no transversal motion) 2. linear elastic behaviour: σ = E · ε, where E is the Young’s modulus; isotropic relationship between stress and strain in the beam material (typical behaviour for metallic materials) 3. no concentrated loads (or constraints) along the span 4. no damping (no dissipation) 5. homogeneous material: • constant transversal areaA • constant mass per unit lengthm • constant Young’s modulusE With these hypothesis, we can study an infinitesimal portion of the beam showing the internal axial forces. See the figure 2. Figure 2: Infinitesimal portion of the beam We can study the horizontal equilibrium. The right-way contributions are positive: −m · dx · ∂2u(x, t) ∂t2 − N (x) + N (x + dx) = 0 −m · dx · ∂2u(x, t) ∂t2 −N (x) + (N (x) + dN (x)) = 0 m · dx · ∂2u(x, t) ∂t2 = dN (x) m · ∂2u(x, t) ∂t2 = dN (x) dx (1) Now we can express the internal axial forceN (x): N (x) = A · σ(x)…