M01 Tensioned String

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Mechanical System Dynamics - Lecture Notes MSc. Mechanical Engineering A.A. 2022-2023 Vibration analysis of one-dimensional continuous systems Tensioned String Teacher: Prof. Stefano Melzi Trainer: Eng. Binbin Liu Fabio Santoro Contents 1 W ave equation 2 2 Propagative solution 3 3 Stationary solution 4 3.1 Case: Pinned-Pinned String . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1 1. Wave equation To study the transversal vibrations of a tensioned string, let’s consider the model in figure 1. It is a representation of a tensioned string (the tension is equal toT) of length L. The space variable is x and the goal is to identify the expression of the vertical displacementw(x, t) Figure 1: Tensioned string model It is necessary to assess some hypothesis: 1. small displacements: it is possible to use asymptotic approximations of functions 2. no dumping (no dissipation) 3. no concentrated load (constraints) along the span (just on the boundaries) 4. the axial deformation is negligible 5. the string is assumed with a very small cross area, so shear and bending effects are not relevant 6. the undeformed shape of the string is straight (thank to the fact we assume a very high value of the tension T) 7. homogeneous material with a constant mass per unit lengthm With these hypothesis, we can study an infinitesimal portion of the string showing the internal forces of axial load T. See the figure 2 We can write the vertical equilibrium equation of the infinitesimal Figure 2: Infinitesimal portion of the string element keeping in mind the first hypothesis of small displacements: −T · sin(α) − m · dx · ∂2w(x, t) ∂t2 + T · sin(α + dα) = 0 −T · α − m · dx · ∂2w(x, t) ∂t2 + T · (α + dα) = 0 −m · dx · ∂2w(x, t) ∂t2 + T · dα = 0 −m · dx · ∂2w(x, t) ∂t2 + T ·…