Mechanical Engineering - Mechanical Systems Dynamics

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 FabioSantoroContents 1 Wave equation2 2 Propagative solution3 3 Stationary solution43.1 Case: Pinned-Pinned String. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 1 1.Wave equation To study the transversal vibrations of a tensioned string, let’s consider the model in figure1. It is a representation of a tensioned string (the tension is equal toT) of lengthL. The space variable isx 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 ofthe tensionT) 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 loadT. See the figure2We can write the vertical equilibrium equation of the infinitesimalFigure 2:Infinitesimal portion of the string element keeping in mind the first hypothesis of small displacements: −T·sin(α)−m·dx·∂ 2 w(x, t)∂ t 2+ T·sin(α+dα) = 0   −T·α−m·dx·∂ 2 w(x, t)∂ t 2+ T·( α +dα) = 0 −m·dx·∂ 2 w(x, t)∂ t 2+ T·dα= 0 −m·dx·∂ 2 w(x, t)∂ t 2+ T·∂ w (x, t)∂ x = 0 −m·∂ 2 w(x, t)∂ t 2+ T·∂ 2 w(x, t)∂ x 2= 0 −∂ 2 w(x, t)∂ t 2+Tm · ∂ 2 w(x, t)∂ x 2= 0 2 We can rearrange this last equation by separating the variables to obtain the mono-dimensional wave equation of a string. We can also define the speed of propagation of the wave asc=qT m , so the wave equation becomes:∂2 w(x, t)∂ t 2= c2 ·∂ 2 w(x, t)∂ x 2(1) 2.Propagative solution One solution studied in vibration analysis of a string is the propagative solution where the wave of the deformed string keeps moving front and back with the speedcalong the span like in figure This kindFigure 3:Tensioned string propagative model of solution is studied in railway sector with the propagation of the wave due to the contact between the pantograph and the power wire. The power exchanged in this contact is on the order of6M W. This problem is studied particularly on high speed trains because the speed is comparable with the speed of propagation of the wave especially when there are trains with two pantographs when a wave generate by one can affect on the other device. To avoid this issue, the new trains are designed just with one pantographFigure 4:Propagative wave generated by train The expression of the solution is:w(x, t) =f 1( x−ct)−f 2( x+ct)(2) wheref 1( x, t)andf 2( x, t)are the shape function of tho waves that travel one along and the other in the opposite direction of the space variablexwith the wave propagation speedcIt is easy to prove that (2) is the solution. We can define two new variables:     s 1= x−ct s2= x+ct so the general solution becomes:w(x, t) =f 1( s 1) −f 2( s 2) 3 (a) f 1( x−ct)(b) f 2( x+ct) Figure 5:Propagative functions We can calculate the second derivatives ofw(x, t)in space and time in order to obtain the identity of the wave function (1) ∂ w(x, t)∂ t = ∂ f 1( s 1)∂ s 1· ∂ s 1∂ t + ∂ f 2( s 2)∂ s 2· ∂ s 2∂ t = ∂ f 1( s 1)∂ s 1· (−c) +∂ f 2( s 2)∂ s 2· c ∂2 w(x, t)∂ t 2=∂ 2 f1( s 1)∂ s 2 1· ∂ s 1∂ t · (−c) +∂ 2 f2( s 2)∂ s 2 2· ∂ s 2∂ t · c=∂ 2 f1( s 1)∂ s 2 1· c2 +∂ 2 f2( s 2)∂ s 2 2· c2 ∂ w(x, t)∂ x = ∂ f 1( s 1)∂ s 1· ∂ s 1∂ x + ∂ f 2( s 2)∂ s 2· ∂ s 2∂ x = ∂ f 1( s 1)∂ s 1· 1 +∂ f 2( s 2)∂ s 2· 1 ∂2 w(x, t)∂ x 2=∂ 2 f1( s 1)∂ s 2 1· ∂ s 1∂ x + ∂ 2 f2( s 2)∂ s 2 2· ∂ s 2∂ x = ∂ 2 f1( s 1)∂ s 2 1+ ∂ 2 f2( s 2)∂ s 2 2 We can substitute them into the wave function: ∂2 w(x, t)∂ t 2= c2 ·∂ 2 w(x, t)∂ x 2 ∂2 f1( s 1)∂ s 2 1· c2 +∂ 2 f2( s 2)∂ s 2 2· c2 =c2 · ∂2 f1( s 1)∂ s 2 1+ ∂ 2 f2( s 2)∂ s 2 2 It is trivial to see the identity of the equation. 3.Stationary solution The other solution typically studied in engineering is the so called stationary solution. It is obtained by the product of two functions, one space-dependentα(x)and the other time-dependentβ(t) w(x, t) =α(x)·β(t)(3) 4 Figure 6:Tensioned string stationary model α(x)is the shape function and it is scaled on the time functionβ(t). Let’s evaluate the second partial derivative in time and the second partial derivative in space1 of (3) ∂2 w(x, t)∂ t 2= α(x)·d 2 β(t)dt 2= α(x)·¨ β(t) ∂2 w(x, t)∂ x 2=d 2 α(x)dx 2· β(t) =αI I (x)·β(t) and substitute them into (1) and separate the variables: α(x)·¨ β(t) =c2 ·αI I (x)·β(t) ¨ β(t)β (t)= c2 ·α I I (x)α (x)= ω2 ω∈R>0(4) we can separate the two functions:¨ β(t) +ω2 ·β(t) = 0αI I (x) +ω 2c 2· α(x) = 0(5) the first one is the equation of motion of free undamped single linear system. The general solution has the form:β(t) =C·cos(ωt+φ)(6) We can define the parameterγ2 =ω 2c 2 and solve the space second order ODE by imposing the general exponential solutionα(x) =α 0eλt . The equation becomes: αI I (x) +γ2 ·α(x) = 0 λ2 +γ2
·α 0· eλt = 0(7) The trivial solution of this equation isα 0= 0 that corresponds to a straight beam (no oscillations). The non-trivial solution corresponds to putting equal to zero the determinant: λ2 +γ2 =⇒λ=±iγ(8) The general space solution is: α(x) =α 0,1· eiγ t +α 0,2· e− iγ t =A·cos(γ x) +B·sin(γ x)(9) The complete stationary solution of the transverse vibrations of a tensioned string is obtained by combining (3), (6) and (9). We can incorporate the constantCin the other constantsAandB. w(x, t) = [A·cos(γ x) +B·sin(γ x)]·cos(ωt+φ)(10) The constantsA,Bare evaluated by imposing the boundary conditions (in this case are required 2 equations as BCs) and the phase delayφis obtained by the initial conditions.1 For a total derivative operation it is possible to use a simplified notation:d ( i) f(x)dx i= f( i) (x). Usually in the second notation thei−thderivative is indicated by roman numbers 5 3.1.Case: Pinned-Pinned String Let’s analyze the case of an unloaded pinned-pinned string, in particular the stationary solution.Figure 7:Pinned-Pinned string case The boundary conditions (two geometrical) due to the two pins are:1.Vertical displacement forx= 0is null:w(x, t)| x=0= 0 2.Vertical displacement forx=Lis null:w(x, t)| x=L= 0 Here are the mathematical expressions of the BCs sorted by the previous list. We can neglect the time function from the BCs in order to lighten up the notation:     w (x, t)| x=0= A= 0(11a) w(x, t)| x=L= B·sin(γ L) = 0(11b) The solution of the second BC can beB= 0, but it represents the trivial solution (straight string with no oscillations), andγ L=kπwithk= 1,2, . . . ,+∞, so we will have γk=kπL ω k= γ k·rT m = kπL ·rT m k = 1,2, . . . ,+∞ We can evaluate the space expression for somek: •k= 1γ 1=πL ω 1= πL  rT m α 1( x) =B 1· sin πL x •k= 2γ 2=2 πL ω 2= 2πL  rT m α 2( x) =B 2· sin 2πL x •k= 3γ 3=3 πL ω 3= 3πL  rT m α 3( x) =B 3· sin 3πL x(a) k= 1(b) k= 2(c) k= 3 Figure 8:Shape of vibration modes of a pinned-pinned string The general expression of the vibration function of the pinned-pinned string is obtained by the super- position of all the solutions: w(x, t) =+ ∞ X k=1B k· sin kπL x ·cos(ω kt +φ k) ω k=kπL ·rT m (12) 6