Transverse vibration of an axially compressed bar with dry friction at its ends

Transverse vibration of an axially compressed bar with dry friction at its ends Francisco J. Nieves a,* , Ana Bay´ on b , F´ elix Salazar b , Francisco Gasc´ on c a Departamento Física Aplicada II, Instituto Univ. Arquitectura y Ciencias de la Construcci´ on, ETS Arquitectura, Universidad de Sevilla, Seville, Spain b E.T.S. de Ingenieros de Minas y Energía, Universidad Polit´ ecnica de Madrid, Madrid, Spain c Departamento Física Aplicada II, ETS Arquitectura, Universidad de Sevilla, Seville, Spain ARTICLE INFO Keywords: Bernoulli-Euler beam theory Bar with sliding ends Transverse vibration Axially compressed bar Dry friction ABSTRACT The transverse vibration of a bar is studied by applying the Bernoulli-Euler beam theory. The bar is placed between the platens of a hydraulic press that applies compressive stress. When the bar vibrates, its ends slide over the platens with dry friction. Boundary conditions appropriate to the existence of friction are proposed. Once the homogeneous equation of motion is solved analytically, a particular solution is obtained through elementary trigonometric series. The sum of these solutions provides the general solution that shows the movement of all the bar points. The movement is divided into successive stages. The displacement of the bar points as a function of time is calculated numerically. It is demonstrated that there is a sudden change in the shape of vibrating when a specific number of semi-oscillations is reached, going from a behaviour of sliding ends to another of fixed ends. Criteria are proposed to estimate the circumstances in which the partial stop of the vibration occurs, as well as a change in the vibration mode and its frequency. 1. Introduction The study of the vibration of beams subjected to axial loads is a classic topic discussed in textbooks [1] and in many technical publications. One reason for this is that the natural frequencies of axially loaded beams, widely used in many macro and microstructures, are of practical interest in a large number of applications. The vibration frequencies of a uniform beam under constant axial compressive loads were obtained by Bokaian [2]. This work was extended in [3] to a beam subjected to tensile axial loads for ten combinations of classical boundary conditions. A previously unreported fundamental frequency was found by Liu and Ertekin [4] in the study of vibrations of a free-free beam under tensile axial loads. Transverse vibration of a Timoshenko beam subjected to axial loads with non-classical boundary conditions was studied in reference [5] to obtain the variation of the natural frequencies with axial tensile loads in a circular thick beam. When the axial force acting on a uniform Bernoulli-Euler beam is not constant but varies linearly, free vibration transverse frequencies were calculated for different combinations of classical boundary conditions [6]. A simple closed-form equation for natural frequencies of beams, modeled as a spring-mass system, was presented by Valle et al [7]., which provides natural frequencies of beams, with various end conditions, as a function of the axial load, temperature variation or residual stress. The consideration of friction in the study of elastic vibrations complicates the solution of the problem. The system under study is nonconservative due to friction, the mechanical energy being partly transformed into internal thermal energy and partly dissipated as heat. In a classic example [8,9], a freely vibrating block slides on a rough surface and is attached to a spring fixed at the other end. With Coulomb damping, the block performs half periods of harmonic motion with decreasing maximum displacements until it stops. This is a very simplified case compared to the study presented in this work, although it has a certain similarity. Different studies can be found in literature regarding the vibration of discrete systems, in presence of dry friction forces, which are subjected to harmonic excitation, as in references [10,11] for one or two degrees of freedom, respectively. The study of stick-slip oscillations of discrete systems, induced by friction between the oscillator and a moving body, is also of interest in many technical applications, as those described in references [12–14]. The effect of dry friction on the vibrations of beams has been evaluated in different ways, such as attaching a dry friction damper to a simply supported beam [15], or modeling a rotating beam with dry * Corresponding author. E-mail address: [email protected] (F.J. Nieves). Contents lists available at ScienceDirect Forces in Mechanics journal homepage: www.sciencedirect.com/journal/forces-in-mechanics https://doi.org/10.1016/j.finmec.2025.100320 Received 11 April 2025; Received in revised form 31 May 2025; Accepted 13 June 2025 Forces in Mechanics 20 (2025) 100320 Available online 15 June 2025 2666-3597/© 2025 The Authors. Published by Elsevier Ltd. This is an open access article under the CC BY-NC-ND license ( http://creativecommons.org/licenses/bync-nd/4.0/ ). friction support boundary conditions [16]. The dynamic behavior of stick-slip vibration of beams, induced by friction, has also been addressed by many researchers using different models. In reference [17] the work is focused on the study of friction-induced vibrations of a system composed of two beams in contact, whereas in [18] the analysis of the stick-slip vibration deals with a transversely moving beam in contact with a frictional wall. Cui and Hu [19] considered a frictional sliding end within a clearance, in the study of the natural vibration of a simply supported beam subjected to a uniformly heating. In a previous work by the authors [20], the Ritz method was applied to investigate the effect of axial stress on the natural flexural frequencies of a cylinder. Nonlinear strain components were included. Sliding-sliding and clamped-clamped boundary conditions were both supposed. Laboratory experiments were carried out. A disagreement was found between the experimental values of the lowest frequency and those calculated. As the compressive force increased, the boundary conditions seemed to change from sliding-sliding conditions to clamped-clamped ends. Thus, it seems appropriate to look for a model with changing boundary conditions. To the best of our knowledge, there is no previous research to obtain the dynamic response of the system described above. Since a beam is one of the fundamental components of mechanical systems, such as turbomachinery rotors, certain spacecraft structural components, solar panel supports, and so forth, understanding the vibrational behavior of a compressed beam, with sliding-sliding boundary conditions and friction at its ends, is of practical interest. Another point to consider is machine noise. When a beam vibrates with its ends in contact with a surface, friction between the surfaces can give rise to unpleasant noises such as squealing and groaning. The results obtained in this study could be used to mitigate machinery noise. In this work, the transverse vibration of a bar placed between the platens of a hydraulic press is investigated. The physical problem studied is that of an axially loaded elastic bar vibrating transversally with small deformations, which allows us to use the equation of motion corresponding to the transverse vibration of a Bernoulli-Euler beam subjected to axial loads, i.e. differential Eq. (1) given in the next section. Some of the boundary conditions are obvious and others are intuitive and postulated. More specifically, a model with sliding-sliding boundary conditions is considered at the beginning of the vibration. The friction force is introduced as a step force, concentrated at each end, acting in the direction opposite to the velocity of the ends. The effect of such a force is evaluated by the particular solution of the differential equation, which is added to the solution of the homogeneous equation to obtain the general solution. Of all the obtainable solutions, the mathematically simplest ones have been chosen. Despite the simplicity of the approach described above, the proposed model allows us to evaluate the effect of axial compressive loads and dry friction forces at the ends on the transverse vibration of the bar. The compression force is assumed to remain constant throughout the vibration. The results obtained with the proposed model show that after a given number of semi-oscillations, the boundary conditions of the bar ends change from sliding-sliding to clamped-clamped. The general solution for a symmetric mode is obtained by numerically calculating the resulting vibration in consecutive half-periods. In each half period, the bar is assumed to vibrate in only a mode. That is, the frequency does not change during vibration under s-s conditions, but the amplitude progressively decreases until reaching c-c conditions, which produces a change in the vibration frequency. It is then assumed that the bar vibrates at the lowest c-c frequency for the conditions considered, the amplitude remaining constant since there is no energy dissipation. The calculations in this paper refer to a commercial cylindrical aluminum specimen. The properties of the bar, measured or calculated from measurements made in the laboratory, are: length L =0.150 m, diameter D =0.0150 m, density ρ =2829 kg/m3, Young’s modulus E = 70.764 GPa, and H =175.85 Nm 2 (Product of Young’s modulus and the moment of inertia). 2. Flexural vibration of a slender bar between the platens of a press A cylindrical aluminum bar is held by axial compression using a hydraulic press, as shown schematically in Fig. 1. The bar is deformed transversely and elastically and then allowed to vibrate freely. Let us now study the motion to which the bar will be subjected. 2.1. Equation of motion of a bar under compressive loads The model presented in this work is an idealization and simplification of reality. The so-called Bernoulli-Euler beam theory is suitable for slender bars, assuming that a set of basic restrictions are met. The most important assumption is that the plane cross-sections of the beam remain plane and perpendicular to the neutral axis during bending. This theory will be applied in this paper. Fig. 2 represents the curved bar transversely vibrating at time t. Two large steel platens of a hydraulic press apply compression forces, both with magnitude T but with opposite sense. The displacement y(t) is exaggerated in this figure, since in the present study y(t) is much smaller than L. If the movement (velocity v) of the ends of the bar is in the negative direction of the OY axis, that is, downwards, the platens exert friction forces of magnitude F r and upward at both ends. The resultant of the external forces acting on the bar will be upward and, therefore, the centre of mass of the bar will move with an upward acceleration. In general, if in addition to the axial compression forces T there is a transverse distributed force on the bar, with linear density q, the equation of motion for the transverse vibration of the bar under prestress [1] is given by H ∂ 4y ∂ x4+T ∂ 2y ∂ x2+ ρ S ∂ 2y ∂ t2=q(x,t),(1) where y(t) is the transverse displacement of a point located at the position x, t is the time, ρ is the density, and S is the cross-sectional area. The OX axis is located along the neutral axis of the bar with its origin at the left end. 2.2. Solution of the homogeneous equation Let us solve the homogeneous differential Eq. (1) with q =0, by applying the method of separation of variables. The solution proposed is yh=Y(x) τ (t), and substituting it in Eq. (1), it leads to a harmonic movement in time of frequency ω , τ (t) = C1cos( ω t) + C2sin( ω t), whereas the transverse displacement Y(x) of the bar is governed by the differential equation, Fig. 1. The figure represents, by means of a rectangle, a very massive hydraulic press, which axially compresses a slender aluminum bar. The X and Y coordinate axes are used to quantify the movement of the bar points. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 2 d4Y dx4+T H d2Y dx2− ρ S H ω 2Y=0,(2) whose solution is Y=Acos ( α x) + Bsin ( α x) + Ccosh (βx) + Dsinh (βx),(3) A, B, C and D being constant coefficients, and α , and β are given, respectively, by: α ={T/(2H) + [(T/(2H))2+ ω 2 ρ S/H]1/2}1/2 ,β ={−T/(2H) + [(T/(2H))2+ ω 2 ρ S/H]1/2}1/2 .(4) The applicable boundary conditions depend on the experiment performed. In our case, the ends of the bar are in contact with the platens of the press. If the ends of the bar can slide on the platens, the applicable displacement boundary conditions would be of the sliding-sliding type. However, if the bar were glued to the platens, the simple model to apply would be the clamped-clamped model. Since our bar slides on the platens, we adopt here as the boundary conditions model that corresponding to the sliding-sliding one. To calculate the four constants appearing in (3), the following two pairs of boundary conditions are taken, according to the previously mentioned model: 1) If all points at both ends of the bar are in continuous contact with the respective platens, the deformed axis of the bar will remain, all the time, perpendicular to the platen surface at both ends, i.e. there is no rotation, then the slope of the deformed axis must be zero at both ends: dY dx =0,∀x=0,x=L.(5) 2) The ends of the bar can slide on the platens, at least in the first moments, then the sliding-sliding conditions are applicable at both ends: dY3 dx3=0,∀x=0,x=L.(6) The friction force applied on each end has a component along the OY axis with a value F r = μ T, μ being the friction coefficient. Then, the shear stress at each end is equal to the friction force μ T. On the other hand, the shear stress is proportional to the third derivative of the displacement, therefore this derivative should not be zero. Furthermore, the inclusion of the friction force in Eq. (6) leads to an incompatible system of equations, when the constants in Eq. (3) are determined. In the proposal made in this article we use an alternative model, namely the friction force will be replaced by an equivalent force, distributed along the bar in a non-uniform way, as we will explain in obtaining the particular solution of the nonhomogeneous differential equation (Eq. (1)). Applying the boundary conditions (5) and (6) to (3), a homogeneous system of four equations with the four unknowns is obtained: A, B, C, and D. Solving such an equation system, the characteristic equation is determined: sin( α L) =0 ⇒ α =n π /L, n =1, 2, 3… and the mode shapes coefficients [2]: A =arbitrary constant and B =C =D =0. With these values, Eq. (3) reduces to Y=Acos( α x), α =n π L,∀n=1,2,3,…(7) This result corresponds to the mode shapes and shows that the barshape is harmonic in x, with the spatial frequency α given by (7). For odd values of n, the figure is antisymmetric with respect to the centre of the bar. For even n values the figure is symmetrical. The arbitrary constant A indicates the maximum displacement during vibration. Fig. 2 depicts Eq. (7) with n =2. The natural frequency is obtained from the first of Eq. (4) and the value of α obtained from Eq. (7), resulting ω n= π n L(H ρ S)1/2(n2 π 2 L2−T H)1/2 ∀n=1,2,3,…(8) In order to apply Eq. (8) to our specimen, it has been considered that the applied compression, T =18 kN, is less than the buckling compression, and n =2 (first symmetric mode), then Eq. (8) gives the frequency f = ω / (2 π ) =5082.4 Hz. Finally, taking into account the aforementioned procedure, the solution of the homogeneous differential equation is obtained: yh=Acos(n π x/L)[C1cos( ω t) + C2sin( ω t)].(9) From the equation of motion (9), it can be deduced that the displacement obtained with the homogeneous equation is purely harmonic, both in the x coordinate of the bar points and in time t. In general, the solution of the homogeneous equation will be a linear combination of normal modes, since (2) is linear. 2.3. Complete equation and friction force In the proposed model, the friction force is distributed along the bar of length L in a discontinuous manner: it is concentrated at the two ends. This dependence can be expressed by forces equal to μ T and Dirac´s deltas of the x coordinate. We will assume hereafter, for reasons of simplicity, that the bar begins the movement preferably according to the modal form of the symmetric mode number n, whose frequency ω is given by Eq. (8) for the symmetric mode considered, and its oscillation period is 2 π / ω . If the ends of the bar move downward, their velocities are downward and the friction forces are upward. For a given frequency ω , half a period is the time elapsed between t and t + π / ω . We call this semi-period “a stage” because it is the time it takes for the speed (and forces) to change its direction. Let us suppose that the ends of the bar can stop after a certain time and begin an upward movement; if so, the friction forces change their direction. This phenomenon of change of direction may be repeated periodically. In this case, Fig. 3 would represent the variations in the direction of the friction force as a function of time, through the transverse component of the force divided by its magnitude. A more precise analytical expression of this change in direction of the force at the origin of the bar is − − [( ∂ y/ ∂ t)/| ∂ y/ ∂ t|](x=0). As a result, it turns out for the transverse force applied per unit length: Fig. 2. The two steel platens of a press apply axial compression forces T at the ends of the bar, which in its natural state is straight and of length L. The bar vibrates in bending at the instant t, whose ends move with speed v. The bar deflected shape is represented in the figure. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 3 q(x,t) = μ T[δ(x) + δ(x−L)]⎡ ⎢ ⎣(− ∂ y ∂ t)  ∂ y ∂ t ⎤ ⎥ ⎦(x=0) .(10) Eq. (10) shows a distribution of friction force symmetrical with respect to the plane x =L/2, since that force is of the same value and identical direction at both ends of the bar. This density of force is, likewise, an odd function of time. 3. Vibration at the beginning of the motion 3.1. Frictional forces and particular solution To simplify the mathematical calculation, in this Section we will limit ourselves to the analysis of the displacement of the bar points during the time interval in which the friction force applied to the bar at each end is constantly upward and with modulus μ T. The part of Eq. (10) that depends on time is equal to one in this time interval, so it is reduced to: q= μ T[δ(x) + δ(x−L)].(11) An appropriate expression for the pair of deltas is the one obtained by the Fourier series expansion of the sum of both Dirac´s deltas [21]. We start from a hypothetical distribution of the external force, applied on the bar per unit length, so that, in the intervals from x =0 to x =Δx and from x =L - Δx to x =L, the linear density of force is µT/Δx, and is zero in the rest of the bar. When Δx tends to zero, the expansion for the delta located at the origin results: q0=2 μ T L[1 2+∑ ∞ i=1 cos ( π ix L)].(12) Since the expression for the expansion of the delta at x =L is identical to (12), but centred at x =L, we obtain that the expansion of the two deltas together is given by q=2 μ T L+4 μ T L∑ ∞ i=1 cos(2 π ix L).(13) The first addend of (13) represents a homogeneous force density, applied on the bar, which would cause a uniform and upward acceleration at each point on the bar. That is, the first addend of (13) indicates a constant braking force acting on all the particles of the bar, when their ends move downwards. In this work, we are not interested in the consequences of its existence, because it does not imply vibration dependent on x, unlike what happens to the other addend of (13). Moreover, if other methods are used to obtain the solution of the non-homogeneous differential equation, as modal analysis [22], the results obtained do not include the rigid motion of the bar. Therefore, we will limit ourselves to studying the effect of linear force density: q=4 μ T L∑ ∞ i=1 cos(2 π ix L).(14) In our actual numerical calculations, the index i is finite, but large. Fig. 4 represents q for a number of addends I =100. The friction coefficient considered is µ =0.2 and the compression T =18 kN. From this figure it can be seen that q is 10 7 N/m at both ends, and 10 5 N/m at a distance as short as approximately 10 −3 m from the ends. That is, Eq. (14) is an expression that approximates two deltas, each one at one end of the interval [0,L]. Starting from this distribution of forces (14), we try a particular solution of the complete differential Eq. (1) in the following way: yp=L[∑ ∞ i=1 C3icos(2 π ix L)],(15) where C 3i are constants to be determined. The substitution of this y p in (1) and equating term-by-term on each side of the equation, for each value of i, gives the sufficient solution: C3i=4 μ TL2 (2 π i)2[H(2 π i)2−TL2].(16) Taking into account Eq. (15) and Eq. (16), the particular solution results: yp=∑ ∞ i=1 4 μ TL3 (2 π i)2[H(2 π i)2−TL2]cos(2 π ix L).(17) y p given by (17) makes sense if you consider the sum, not each addend. This particular solution is adequate, as it fulfils both, Eq. (1) with q given by (14), and the four boundary conditions. Note that y p does not depend on time during the time interval studied. Observe also in (17), that the basic spatial frequency is 2 π /L. Fig. 5 represents the particular solution y p corresponding to the vibration of the bar under study and has been calculated by applying (17) with the following data: bar-platen friction coefficient µ =0.2, compression T =18 kN and number of addends I =100. The figure of y p represents the result of a series of addends, as can be seen in (17), all of the cosine type, and looks like a cosine. In Fig. 5, it can be seen how y p is positive in the first part of the bar, while it is negative in the second and Fig. 3. Value of the unitary transverse component of the friction force on the ends of the bar as a function of time. Every half-period π / ω the friction force changes its direction. Fig. 4. The function q approaches zero for all values of x, except at the ends where it tends to infinity. It represents approximately two Dirac deltas at the ends of the bar. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 4 third regions, and becomes positive again in the fourth portion. 3.2. General solution for the first moments The general solution of Eq. (1) is the sum of the solution of the homogeneous Eq. (9) and a particular solution (17). Therefore, the general solution for a single frequency ω , in the time domain, and symmetric oscillation of order n, is y=Acos( α x) [C1cos( ω t) + C2sin( ω t)] + yp∀ α =n π /L;n=2,4,6,… (18) The first addend of (18), solution of the homogeneous, represents the shape of the bar at any instant of time, as if there were no friction. The second addend, particular solution y p , modifies the shape of the bar if there is friction. Now we impose the initial conditions for the displacement and velocity: y =y 0 and v= ∂ y/ ∂ t=0∀t=0. From the initial velocity condition and Eq. (18) we deduce: y=C0cos( α x)cos( ω t) + yp∀ α =n π /L;n=2,4,6,…,(19) where C 0 =AC 1 is an arbitrary constant. In Eq. (19), as well as in everything that follows, we will consider a single mode of vibration of frequency ω . Eq. (19) gives the displacement of any point of the bar at any of the first time instants. From this equation it is deduced that each point of the bar (constant x) experiences a simple harmonic motion, parallel to the OY axis, centred at a point located at distance y p and with vibration amplitude C 0 cos( α x). More specifically, during the first instants, the oscillation of a point located at distance x from the origin is around the point with coordinates, (x,(y0+y π / ω )/2)=(x,yp),(20) where it has been taken into account that from Eq. (19) it can be deduced that y0=C0cos( α x) + yp∀t=0.(21) In the same way, we obtain y at t = π / ω (y π / ω ). Equality (21) relates the maximum initial displacement y 0 to the vibration amplitude C0cos( α x)and the displacement y p relative to the OX axis, due to friction. It should not be forgotten that y p is a function of x, as it is the result of the sum of periodic addends in x, which have different spatial frequencies. Furthermore, y p is not a function of time in the interval considered, so it can be stated that the movement in the first instants is vertical with the centre of oscillation at the points given by y =y p . From Eq. (21) it also follows that the ends, cos( α x) =1, have a vibration amplitude C 0 , and then C 0 represents the vibration amplitude of the ends. Also note in Eq. (21) that at the initial instant, and at one end of the bar, it is verified that y 0 =C 0 +y p , ∀x =0, t =0. In an ideal experiment, the bar would be placed between the platens of the press and perpendicular to them. Applying the compressive forces of value T, and knowing the value of μ , C 3i can be calculated using Eq. (16) and y p using Eq. (17). If the amplitude of the vibration of the end of the bar, C 0 , is taken as input data, the initial displacement or maximum displacement y 0 can be calculated. This initial displacement, suitable to produce a vibration according to mode n, must be in accordance with Eq. (21), whose absolute value is the initial distance of the point x of the deformed bar to the OX axis. By using Eq. (21), Eq. (19) takes the following form: y=(y0−yp)cos( ω t) + yp.(22) To cause vibration of the bar in mode n, appropriate transverse static forces can be applied to the sample until an initial deformation y 0 is achieved according to the form given in (21). Having achieved this objective, at t =0, leaving the elastic bar free, it tries to recover its original shape, which was straight, and the movement begins according to Eq. (22). The friction force tries to stop the movement of the bar. This theoretical picture for starting the movement is too complicated. For this reason, in practical applications the procedure can be replaced by applying a sudden blow perpendicularly to the bar, which is simple, but it will cause the bar to vibrate in a wide range of vibration modes. At the end of the first half-period, t = π / ω , the displacement is, according to Eq. (19), y π / ω = − C0cos( α x) + yp,(23) whose absolute value is the distance of the x-coordinate point to the OX axis. Considering Eq. (22) at t = π / ω , it follows that y π / ω = − y0+2yp(24) where the dependence on x is included in y 0 and y p . It means that at t = π / ω , in which the bar reaches zero speed (see Eq. (19)), the maximum displacement of the points of the bar has decreased by the amount 2y p . Note that, at the end of this first half-period, the points of the bar are closer to their rectilinear position before starting the experiment, OX axis, then the deformation of the bar has decreased and, therefore, its elastic potential, that is, its elastic energy has become smaller. Therefore, taking the above into account, we can say that from the dynamic point of view, friction moves the centre of vibration of the points of the bar by the amount y p . Fig. 6 shows the movement of a point located at the x coordinate of a hypothetical bar. The point starts at time t =0 with an initial displacement y 0 =C 0 cos( α x)cos( ω ⋅0) +y p =1 +0.2 =1.2 (arbitrary unities), performs half oscillation around y =y p =0.2 and ends at y =-0.8. 4. General solution for later times The bar, whose ends were descending, stops at the instant t = π / ω with a displacement y π / ω =- y 0 +2y p , as we have just seen in (24). From that moment on, the second time interval begins, with a different time scale t’, related to the previous one by t =t’ + π / ω . At t’ =0, the displacement is yʹ 0= − y0+2yp,(25) and zero initial velocity. In this second time interval, everything hapFig. 5. Particular solution y p for the displacement calculated when the friction forces at the ends of the bar are upward, which is obtained by adding cosinetype terms. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 5 pens in a similar way to the first, in particular the frequency remains the same, given by Eq. (8). But the friction force changes its direction, which is reversed, then the sign of q(x) is opposite, leading to the particular solution y p ’=- y p . That is, if the ends rise, the sign of y p changes. Eq. (19) is still applicable, but with the initial conditions found, it gives: yʹ=Cʹ 0cos( α x)cos( ω tʹ) − yp.(26) That is, each point of the bar describes simple harmonic motion, with the same frequency as in the first stage, but around a point located under the OX axis at a distance y p . Since at t ′ =0, the initial displacement y 0 ′ , given by Eq. (26), is equal to that given by Eq. (25), we can obtain the vibration amplitude in the second half-period, C 0 ′ cos( α x), as a function of y 0 and y p . By substituting it in Eq. (26), it follows that the transverse displacement y ′ of the points of the bar is given by: yʹ=(−y0+3yp)cos( ω tʹ) − yp,(27) where the dependence on x is included in y 0 and y p . The second half-period ends when ω t ′ = π , with the displacement yʹ π / ω =y0−4yp,(28) and with zero velocity. As mentioned previously, the movement of the bar during the second time interval analysed is of the same temporal frequency ω as the first, but the oscillation of each point of the bar occurs with a lower amplitude and around a point, on the vertical line that passes through the initial xcoordinate of the point and with the y-coordinate equal to - y p . Thus, the vibration of the bar is not pure periodic. Despite all this, we could call the initial time interval during which the friction force is upwards the first half-period and the following interval, with the force downwards, the second half-period, or first and second stage, respectively. The third half-period begins at t ″ =0, with the particular solution y p (17), the same as in the first half-period, and with zero velocity, so the expression for the displacement in the third half-period is yʹʹ =Cʹʹ 0cos( α x)cos( ω tʹʹ) + yp.(29) Equating (29), particularized for t ″ =0, with the displacement given by (28), at the end of the second half-period, the vibration amplitude C 0 ″ cos( α x) is obtained, which substituted in (29) gives for the third halfperiod: yʹʹ =(y0−5yp)cos( ω tʹʹ) + yp,(30) the displacement at the end of the third half-period being yʹʹ π / ω = − y0+6yp.(31) In the same way, for the fourth stage, it is found that yʹʹʹ 0=yʹʹ π / ω = − y0+6yp.(32) From Eqs. (22), (25), (28), and (32), it can be inferred that the maximum displacements of the successive half-periods are: y 0 , -y 0 +2y p , y 0 - 4y p , - y 0 +6y p , which are also the initial displacements of each halfperiod. These values follow a precise and simple pattern, so the maximum displacement of half-period number j can be written as: yj0= (− 1)j−1[y0−2(j−1)yp]∀j=1,2,3,…J,(33) where j =1 refers to the first half-period and y p is determined by Eq. (17). Note that the rate of decrease of the maximum displacement depends only on y p , whose magnitude is directly proportional to the friction force, μ T, and also depends on the properties of the bar, L and H. In the same way, Eq. (22) can be generalized for the half-period number j, resulting: yj=[C0cos( α x)− (2j−2)yp]cos[ ω tj+(j−1) π ]+(− 1)(j+1)yp∀j=1,2,3,…J, (34) where t j is the time measured during the half-period j; that is, t j is included in the interval [0, π / ω ]. Introducing Eq. (34) the successive values j =1, 2, 3, …, J, it is verified that the numerical coefficients that multiply y p in the first addend of Eq. (34) are: 0, 2, -4, 6, …. This decrease in amplitude is reminiscent of those published in classical literature for the movement of a block with friction, held by a spring [9]. As we see, this classic study is a particularly simple case of what we developed here. Therefore, these numerical results show that the movement of each of the points of the bar can be described as a succession of half-periods of simple harmonic movements, each of which has a vibration amplitude smaller than the previous one by the amount 2y p . At each point of the bar, this harmonic motion is centred at a point on the perpendicular to the axis of the bar, at a distance y p from the axis, above or below the axis, alternately, in each half-period. As stated before, since at each stage the vibration amplitude at each point becomes progressively smaller, oscillating around a different point, an oscillation pattern is produced that is not purely periodic. 5. Results of the displacement by numerical calculation When performing the numerical calculation of the displacement, the data to take into account are: the values L, H, ρ and S, of the bar, the applied compression T, the friction coefficient μ , the considered mode n (even), the constant C 0 , and the number of addends I in the sum of y p Eq. (17). With these data and Eq. (8), the vibration frequencies are calculated. Applying Eq. (17) we obtain y p , which is a function of x. Once ω and y p are known, the displacement of each of the points of the bar, in each of the half-periods, can be obtained by Eq. (34). The results y j , corresponding to each value of j, form an array of two-dimensional matrices, whose dimension depends on the number of half-periods considered. In each half-period, the elements of each two-dimensional matrix correspond to the values of the displacement of the points of the bar at each instant. For each value of x in the bar, the results obtained for each j are concatenated, obtaining a single vector with the displacements as a function of time. Care must be taken to avoid duplicates. Fig. 6. Theoretical displacement of a point of a bar in arbitrary length units, as a function of time, for ω =1 rad/s, C 0 cos( α x) =1 and y p =0.2. The point starts from y 0 =1.2 and reaches y π / ω =- 0.8. The displacement is around the axis y = y p =0.2. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 6 Fig. 7 represents the transverse displacement of one end of the studied bar, calculated as a function of time. The compressive force in the bar is equal to that mentioned in Section 2.2, T =18 kN, and the vibration mode n =2. The calculations have been carried out with the following values of coefficient C 0 and coefficient of friction μ : C 0 =3 mm and μ =0.2, respectively. In each half-period the motion is harmonic, the vibration occurs around a position that changes from y p to - y p every half-period. In each successive half-period, the frequency f remains unaltered. The displacement at the end of a half-period is identical to the initial displacement of the next half-period. It can be noted that the maximum displacement in each period as a function of time is completely rectilinear, decreasing linearly with time. The reason for such a decrease is due to the fact that in each half-period the maximum displacement is reduced by the same amount, 2y p , a constant value. Let us remember that the decrease in the maximum displacement is exponential in the case of viscous friction, quite different from that obtained in the case studied. It should be noted in the Fig. 7 that said linear decrease in the maximum displacement is observed until the seventh half-period. However, this figure shows also the theoretical displacement that would be obtained with Eq. (34) for times greater than those of the seventh half-period, in order to visualize the limit of application of the proposed theory. The origin of previously used times at each stage, t, t ′ , t ″ , …, correspond to the abscissa values of the successive maxima and minima, but in this figure a single time t is used. 6. Estimation of the stop half-period of the ends of the bar The ends of the bar will stop forever when the force of static friction is predominant. We propose two criteria for permanent detection of the ends of the bar. The results obtained with these criteria allow us to estimate the number of half-periods carried out by the ends of the bar, prior to stopping. For numerical application, the following data have been taken: μ =0.2, T =18 kN, I =100, C 0 =3 mm, n =2, then f =5082 Hz, y pmax =0.203 mm at x =0. Criterion A1.- The bar will stop when the absolute value of the maximum displacement at the ends, y j0 , is less than or equal to twice the displacement y p of the ends, obtained from the particular solution: abs(yj0)≤abs(2yp)(∀ x=0).(35) The reason for this criterion is that discussed previously in Section 5: Starting from a maximum or minimum displacement in a certain halfperiod, the next minimum or maximum displacement decreases by 2y p . Therefore, if Eq. (35) is verified, the ends will not cross the OX axis. This criterion allows us to put a limit on the number of calculable halfperiods without producing mathematical contradictions. In the case of Fig. 7, it can be seen, through simple visual extrapolation, that the ends would stop, according to this criterion, after performing approximately 7 half-oscillations. By introducing Eq. (33) into Eq. (35), considering the value of y 0 , at x =0 (given by Eq. (21)), it results: (− 1)j−1[C0+yp−2(j−1)yp]≤2yp(∀ x=0).(36) Therefore, the estimated number of half-periods completed by the end before stopping will be: j≥(C0+yp)/(2yp)→jmax =(C0+yp)/(2yp)(∀ x=0).(37) For the proposed numerical data, it turns out that the number of halfperiods, j ess , with s-s boundary conditions, described by each end before its stop occurs, is: jess =7.89→7stages.(38) Criterion A2.- From the comment made at the end of Section 5, the decrease in the maximum displacement (absolute value), per stage, is equal to 2y p , it can be deduced that the straight lines that join the points of maximum and minimum displacements of the ends of the bar are, respectively, y1= − 4ypf t +y0and y2=4ypf t −y0,(39) so the crossing occurs at the instant tc=(C0+yp)/(4ypf).(40) With the proposed data, the number of stages before the end of the s-s movement would be jess =(C0+yp)/(2yp)=7.89,(41) which is the maximum limit value and as it must be an integer, the ends of the bar must stop at the end of stage number 7. Furthermore, at that moment of maximum displacement, in the negative direction, the speed of the ends is zero and the friction force applied is maximum because the coefficient of static friction is higher than that of kinetic friction. The stop of the ends of the bar will occur at the final instant of the s-s oscillation, that is, at time t ess =j ess ( π / ω ss ), where ω ss refers to Eq. (8) when the bar is vibrating in the corresponding s-s mode (even). 7. Second vibration: bar with fixed ends At the instant t ess when the ends of the bar stop forever, the points of the bar will be displaced by the amount y(x,t ess ) according to Eq. (34). As these displacements do not generally coincide with those corresponding to the static equilibrium of all the points of the bar, since the bar is deformed, it will continue to vibrate from the position reached when the ends stop moving due to friction. The model, once the ends stop moving, will be that of a bar with both end fixed. Then, the applicable boundary conditions are clamped-clamped, Y(x) = 0∀x=0,x=L, dY(x)/dx =0∀x=0,x=L.(42) The bar will vibrate according to a sum of its own modes. The initial Fig. 7. Representation of the displacement 10 −3 y (m) of one end of the bar as a function of time 10 −3 t (s), calculated for the successive half-periods. In this figure, T =18 kN, a friction coefficient of 0.2 and C 0 =3 mm have been assumed. The number of terms in the calculation of y p is I =100. It can be seen that in the first 7 half-periods the maximum displacement decreases linearly with time. Note that a single time t is used. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 7 conditions are equal to the final conditions of the s-s mode mentioned above, that is, y(x,t ess ). The frequency of a beam fixed at both ends, ω cc , is numerically calculable from the frequency equation shown below, obtained from (3) and with the boundary conditions (42), corresponding to the c-c model [2]: By applying Eq. (43) to the bar with compression T =18 kN, the lowest frequencies are obtained: f cc1 =2882.5 Hz, f cc2 =8059 Hz, f cc3 = 15,913 Hz. When calculating the mode shapes corresponding to these frequencies [2], it turns out that the shape corresponding to f cc1 has some resemblance to the symmetric shape, n =2, during s-s vibration. Note that the ends are fixed in the c-c mode, while the initial displacement at the ends in the s-s mode is maximum, but decreases progressively over time due to friction. Moreover, the form corresponding to f cc2 is antisymmetric. The initial conditions in the second type of vibration (c-c) must be identical to those obtained at the end of the first (s-s). Furthermore, taking into account that the shape of the vibration of the bar at the end of the last s-s stage has a close resemblance to the modal shape of the first c-c mode, we assume that the bar will continue to vibrate in the c-c mode and n’ =1. This assumption that both forms of vibration are almost identical represents a simplification of the problem; since the final s-s displacement will lead to the excitation of different c-c vibration modes, the first being the one expected to be preferentially excited. Remembering the value f ss2 =5082.4 Hz, corresponding to the first s-s symmetric mode, and that corresponding to the c-c vibration, f cc1 =2882.5 Hz, a sudden decrease in frequency of approximately 43 % is obtained. It is important to note that as the ends will remain fixed after the s-s vibration, the vibration of the rest of the points of the bar will be around the line y(x =0, t ess ) that joins the ends of the bar at the moment of stop . The previous considerations allow us to propose the c-c solution: ycc(x,t) = Y0cc(x)cos( ω cct) + y(x=0,tess).(44) As aforesaid, there must be continuity of displacements when changing the type of vibration. If we consider, for example, the centre point of the bar, x =L /2, the value of Y 0cc (L/ 2) in (44) can be obtained by Y0cc(x=L/2) = yess(x=L/2,tess) − y(x=0,tess).(45) The platens exert a force on each end of the bar vibrating in this c-c mode, but the work done is zero since both ends are fixed and, consequently, the energy lost due to friction is zero. 8. General case: Vibration with s-s and c-c conditions consecutively Fig. 8 shows the vibration of the centre of the bar during the first approximately 4.2 ms, vibrating in the aforesaid s-s and c-c symmetric modes, successively. This figure has been obtained in accordance with the theory developed in this work and calculated with the same data as in Fig. 7. In this new figure the representation of the displacement is that of the central point of the bar, instead of the end, because the ends remain at rest from the instant t ess , that is, from the last s-s oscillation. The vibration represented in Fig. 8 has two clearly distinguishable time intervals: (a) One formed by 7 semi-oscillations of decreasing maximum displacement, similar to those in Fig. 7. (b) Another, theoretically with an infinite number of harmonic oscillations, of which only the first are represented. As previously indicated, the frequency of the first interval is 5082.4 Hz while that of the second is 2882.5 Hz. To obtain Fig. 8, with x =L/2, the data taken is: μ =0.2, T =18 kN; with them Eq. (17) gives y p,L/2 =-1.7869×10 −4 m. With the added data n =2 and C 0 =3×10 −3 m, we have applied Eq. (34) to the centre of the bar in the first moments. The initial displacement of the centre is y 0ss (x =L/2, t =0) =-C 0 +y p,L/2 , clearly negative. Eq. (38) gives j ess =7 semioscillations. To obtain the displacement of the centre at the end of the s-s vibration, it is enough to consider in Eq. (34): j =7 semi-oscillations and t j = π / ω ss , which corresponds to a time t =t ess =7 π / ω ss . With this value of time, Eq. (34) gives the displacement of the centre at the instant at which the s-s condition ends, y ess (x =L / 2, t ess ) =6.770×10 −4 m. In the same way, at the end we obtain y(x =0, t ess ) =- 3.598×10 −4 m With these values, Y 0cc (x =L / 2)=0.0010 m, given by (45). With the value of the displacement of the centre, the frequency ω cc , and (44) for L /2, the second part of Fig. 8 is drawn. It is observed in this figure that the c-c vibration occurs around y(x =0, t ess ) (dotted line); an apparent increase in amplitude is also observed when going from s-s to c-c vibration. Fig. 8. Vibration of the centre of the bar, x =L / 2, in its two successive symmetrical modes and with lower frequencies. In the first 7 semi-oscillations the bar vibrates with its sliding ends, but with friction, and the rest of the time with both ends fixed. Note the changes in amplitude and frequency from s-s to c-c vibration. cosh〈{−T+[T2+16 π 2H ρ Sf2]1/2}1/2/(2H)1/2〉cos〈{T+[T2+16 π 2H ρ Sf2]1/2}1/2/(2H)1/2〉 −1+sinh〈{−T+[T2+16 π 2H ρ Sf2]1/2}1/2L/(2H)1/2〉sin〈{T+[T2+16 π 2H ρ Sf2]1/2}1/2L/(2H)1/2〉 T 4 π f(H ρ S)1/2=0. (43) F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 8 9. Discussion and some ideas for future research The results previously obtained show that to each number of order n, associated to the harmonics of the s-s vibration, corresponds a value of the spatial frequency α , according to Eq. (7). In turn, each α corresponds to a value of the frequency ω , according to Eq. (4). Therefore, each value of n corresponds to a frequency ω , given by Eq. (8). The study carried out so far has referred to a single s-s vibration mode, obtained with specific initial conditions, and has made possible to obtain the instant t ess at which the ends stop. If an experiment were carried out by applying a percussion, a spectrum with a wide range of frequencies would be obtained, each component with its amplitude and phase. In this case, we would have to know the positions, velocities and accelerations of the ends in order to be able to add them properly and calculate the possible stop time of the s-s vibration. In the study of the vibration of a beam simply supported at both ends and subjected to central percussion, a harmonic series results whose amplitude is inversely proportional to the square of the number n [1]. If we suppose this result is applicable to our case, we can assume, as an approximation, that only the three lowest frequency harmonics are of interest. With this hypothesis, we could calculate with more precision when the ends of the bar stop and expand the study carried out in this work. The second part of the vibration depends on the final shape of the first and is expected to also have a complicated spectrum. In the preparation of this article, two simplifications have been introduced, among others: The Bernoulli-Euler beam theory and the hypothesis that the friction coefficient is small. The first hypothesis is widely used, but the second could be eliminated in future works in order to expand the range of friction values. To study the influence of the friction coefficient on the vibration pattern, Table 1 is presented with the results obtained for some magnitudes, which describe the main characteristics of the vibration pattern. Note that other values of the friction coefficient μ are used, rather than only use the value 0.2. In the table, the second column represents the number of half-periods or stages of s-s vibration (j ess ), while the third shows the quotient of the initial amplitude at the ends (C 0 ) and the stop time for the s-s vibration (t ess ), which in turn corresponds to the slope of the envelope of the damped curve for s-s vibration in Fig. 7. The values in the fourth and fifth columns correspond, respectively, to the vibration amplitude at the center of the bar for the c-c vibration (Y 0cc ) and the displacement at the ends when the s-s vibration ends (y(x=0,t ess )), that is, the position around which the c-c vibration occurs, dashed line in Fig. 8. Finally, the sixth column shows the instant at which the ends of the bar stop (t ess ). Fig. 9 shows the results given in the table. Representative values for the friction coefficient have been chosen in the first column of the table. The second column shows how the number of half-periods of the s-s vibration, j ess , decreases with μ , this decrease being very noticeable for values of friction coefficient in the range between 0 and 0.1, as also observed in Fig. 9. The third column, the slope of the envelope of the s-s vibration (C 0 /t ess ) increases with μ , due to the decrease in the value of t ess with the friction coefficient, t ess being the time of the s-s vibration, i.e., until the ends stop. The results in the fourth column, values of Y 0cc for x =L/2, given by Eq. (45), seem strange and exhibit irregular behavior. In the fifth column, relative to y(x=0,t ess ), the positions around which the c-c vibration occurs are given, which correspond to the ”Centered c-c” curve in Fig. 9. As in the fourth column, the trend is not clear. It should be noted that depending on whether the half-period number, j ess , corresponding to the last half-period of the s-s Table 1 Parametrical study of the influence of friction coefficient µ on the vibration behavior of the bar. Results for µ =0.2 are shown in Figs. 7 and 8. Note that frequencies do not depend upon µ. CoF μ No. of halfperiods of initial vibration Slope of envelope (m/s) Amplitude of latter vibration c-c (m) Amplitude around that vibration occurs (m) Time t ess of pattern change (s) 0.01 148 0.20604188 0.00036727 4.42E-06 0.01456015 0.02 74 0.41208376 0.00038636 1.46E-05 0.00728007 0.03 49 0.62233057 0.00044362 4.50E-05 0.00482059 0.04 37 0.82416751 0.00042454 3.49E-05 0.00364004 0.05 30 1.01647326 0.00036727 4.42E-06 0.00295138 0.10 15 2.03294653 0.00046271 5.52E-05 0.00147569 0.15 10 3.04941979 0.00055816 0.00010597 0.00098379 0.20 7 4.35631398 0.00103538 0.00035983 0.00068866 0.25 6 5.08236632 0.00074905 0.00020751 0.00059028 0.30 5 6.09883958 0.00084449 0.00025828 0.00049190 0.35 4 7.62354947 0.00132171 0.00051215 0.00039352 0.40 4 7.62354947 0.00065360 0.00015674 0.00039352 0.45 3 10.1647326 0.00170349 0.00071524 0.00029514 0.50 3 10.1647326 0.00122627 0.00046137 0.00029514 Fig. 9. The figure represents the magnitudes presented in Table 1 for the values of μ collected in that summary table and for up to 50 other μ values ranging from 0.01 to 0.50. For simple grouping, the amplitudes describing c-cvibration and time scaled 1:10 are represented along the left vertical axis, while the number of stages and the quotient or slope, scaled by a factor of 10, are given along the right vertical axis. Fig. 10. Vibration patterns or shapes of the bar for the first 7 s-s semioscillations or stages, when maximum displacements are reached, obtained from Eq. (34). Note that the vibration patterns are obtained at the beginning of each stage, the time being a multiple of π / ω when a single time is considered. F.J. Nieves et al. Forces in Mechanics 20 (2025) 100320 9