Mechanical Engineering - Mechanical Systems Dynamics

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 FabioSantoroContents 1 Wave equation2 2 Stationary solution32.1 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 figure1. It is a representation of a beam with lengthL. The space variable isxand 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, whereEis 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 figure2.Figure 2:Infinitesimal portion of the beam We can study the horizontal equilibrium. The right-way contributions are positive: −m·dx·∂ 2 u(x, t)∂ t 2− N(x) +N(x+dx) = 0 −m·dx·∂ 2 u(x, t)∂ t 2−   N(x) + (   N(x) +dN(x)) = 0 m·dx·∂ 2 u(x, t)∂ t 2= dN(x) m·∂ 2 u(x, t)∂ t 2=dN (x)dx (1) Now we can express the internal axial forceN(x): N(x) =A·σ(x) =A·E·ε=A·E·∂ u (x, t)∂ x ⇒ dN (x)dx = A·E·∂ 2 u(x, t)∂ x 2 2 By substituting this last expression into (1) we can find the wave function(2): m·∂ 2 u(x, t)∂ t 2= AE·∂ 2 u(x, t)∂ x 2 ·∂ 2 u(x, t)∂ t 2=AEm · ∂ 2 u(x, t)∂ x 2 ·∂ 2 u(x, t)∂ t 2= c2 ·∂ 2 u(x, t)∂ x 2(2) c=qAE m =qE ρ is the speed of propagation of the wave in the beam. 2.Stationary solution (2) is analog to the expression of the stationary solution of the transversal vibrations of a string (all calculations are the same done for the tensioned string, download that file), so the displacement function will become, defining againγ=ωc : u(x, t) = [A·cos(γ x) +B·sin(γ x)]·cos(ωt+φ)(3) The constantsAandBare evaluated by the boundary condition equations and the value ofφis obtained by the initial condition equations 2.1.Case: Clamped-Clamped Bar Let’s analyze the case of the clamped-clamped beam1and write the two geometrical boundary con- dition equations:1.The horizontal displacement forx= 0is null:u(x, t)| x=0= 0 2.The horizontal displacement forx=Lis null:u(x, t)| x=L= 0 We can write the full expressions neglecting the time part:       u (x, t)| x=0= A= 0 u(x, t)| x=L=   : 0 A·cos(γ L) +B·sin(γ L) = 0 The second expression is null forB= 0, but it represents the trivial solution with no oscillations, and γ L=kπwithk= 1,2, . . . ,+∞, so we will have: γk=kπL ω k=kπL ·sE ρ k = 1,2, . . . ,+∞ We can evaluate the space expression for somek: •k= 1γ 1=πL ω 1= πL  rE ρ α 1( x) =B 1· sin πL x •k= 2γ 2=2 πL ω 2= 2πL  rE ρ α 2( x) =B 2· sin 2πL x 3 (a) k= 1(b) k= 2 Figure 3:Shape of axial vibration modes of a clamped-clamped beam The general expression of the vibration function of the clamped-clamped beam is obtained by the superposition of all the solutions: u(x, t) =+ ∞ X k=1B k· sin kπL x ·cos(ω kt +φ k) ω k= kπL  2sE ρ (4) 2.2.Case: Free-Free Bar Now we can consider the case of a free-free beam (beam with no constraints) it is possible to write twoFigure 4:Free-Free axial beam model equilibrium boundary condition equations (neglecting the time expression):1.The axial force in the sectionx= 0is null:N| x=0= 0 2.The axial force in the sectionx=Lis null:N| x=L= 0 We can remember the relationN=E A∂ u∂ x , so it is necessary to evaluate the first derivative of (3): ∂ u(x, t)∂ x = [ −Aγ·sin(γ x) +Bγ cos(γ x)]·cos(ωt+φ) So, we can full express the boundary conditions:    N | x=0= E A·Bγ= 0⇒B= 0 N| x=L= E A· −A·sin(γ x) = 0 The second expression is null forA= 0, but it represents the trivial solution with no oscillations, and γ L=kπwithk= 1,2, . . . ,+∞, so we will have: γk=kπL ω k=kπL ·sE ρ k = 1,2, . . . ,+∞ We can evaluate the space expression for somek: •k= 1γ 1=πL ω 1= πL  rE ρ α 1( x) =A 1· cos πL x •k= 2γ 2=2 πL ω 2= 2πL  rE ρ α 2( x) =A 2· cos 2πL x 4 (a) k= 1(b) k= 2(c) k= 0 Figure 5:Shape of axial vibration modes of a free-free beam •k= 0γ 0=0 πL = 0 ω 0= 0πL  rE ρ = 0 α 0( x) =A 0· cos 2πL 0 =A 0 The solution fork= 0corresponds to the rigid motion of the beam. The general expression of the vibration function of the free-free beam is obtained by the superposition of all the solutions: u(x, t) =+ ∞ X k=1A k· cos kπL x ·cos(ω kt +φ k) ω k= kπL  2sE ρ (5) 5