Model and Simulations of Contact Between a Vibrating Beam and an Obstacle Using the Damped Normal Compliance Condition

Abstract

This work constructs a new mathematical model for the vibrations of a Bernoulli beam that can come in contact with a reactive obstacle situated below its right end. The obstacle reaction is described by the Damped Normal Compliance (DNC) contact condition. This condition, unlike the usual Normal Compliance (NC) contact condition, takes into account the energy dissipation during the contact process. The steady states of the model are described and the model is studied computationally for different values of obstacle stiffness and damping. The computational scheme is shown numerically to converge with a rate higher than 1. The numerical simulations illustrate how the beam’s end penetration and vibrations differ in soft vs. stiff obstacle environments, and how damping modifies the dynamic behavior. The results may be useful for vibration control and material interaction in settings when collisions or repetitive contacts occur. By providing computational and analytical insights, the study is an addition to the currently maturing Mathematical Theory of Contact Mechanics (MTCM).

1. Introduction

The interaction of vibrating structures via contact with other system parts and components is an important topic in mechanical and structural engineering, with applications ranging from robotics, transportation, and automotive, to aerospace systems and MEMS. When a beam repeatedly strikes an obstacle, its dynamic response is governed in part by the condition used to describe the contact process. Classical contact or impact models, such as the Hertz condition, often assume rigid, instantaneous contact with a restitution coefficient, but such models fail to capture the finite stiffness and energy dissipation observed in real materials; see, e.g., [1,2,3,4,5,6], for the mathematical aspects. For this reason, the normal compliance condition was introduced in [5]. The normal compliance condition introduces a finite stiffness that allows small penetrations; however, without damping, they still underestimate energy losses during collisions [4]. To address these shortcomings, the damped normal compliance (DNC) law has recently been proposed that augments the normal compliance contact law with a damping term [7]. This modification yields a more realistic contact boundary condition because it permits controlled penetration while explicitly accounting for energy loss or dissipation during the process and thus is likely to improve the agreement with experiments. Various aspects of the vibration of beams in contact can be found in [8,9,10] and the references therein. We note that some information on the use of layered beams for vibration control can be found in [11].

The novelty in this study is the use of the DNC condition to model the contact between a vibrating Bernoulli beam and a reactive obstacle located below its free end. Such settings can be found in many MEMS systems, in actuators and grippers. This is in contrast to the usual Normal Compliance which does not address the energy dissipation during contact. We construct a new partial differential equation model that incorporates beam bending and the contact boundary conditions of elastic compliance of the obstacle and damping. The main interest here is to study the effects of the beam’s flexural stiffness, the obstacle’s stiffness and its damping on the vibrations amplitude and frequencies of the beam. This may shed more light on the various noise characteristics of such systems. We note that the obstacle problems for the Bernoulli and Gao beams have been studied in the literature using the normal compliance contact condition. This work constitutes a contribution to the currently maturing discipline of Mathematical Theory of Contact Mechanics (MTCM).

Once the DNC condition is used in the design of various engineering applications, the results are likely to be more accurate. We note that processes in which bodies come into contact are ubiquitous in industrial settings, in transportation, and many other settings. Thus, our work is also a contribution to engineering design.

The remainder of the article is structured as follows. Section 2 describes in detail the setting and constructs the classical mathematical model, with the usual fourth order beam equation. In particular, it concentrates on the new DNC contact condition at the right end of the beam where the obstacle is situated. Energy considerations can be found in Section 3, where the energy balance is derived. Section 4 studies the steady states of the system. The condition on the force f, the position of the obstacle $y_{-}$, and the beam’s stiffness ${\alpha }^2$, for the existence of contact, is obtained. Section 5 states briefly, in Theorem 1, the existence of weak solutions to the model, the proof of which can be found in [12].

To explore the system dynamics numerically, we construct an implicit finite difference algorithm in Section 6. The numerical simulations are reported in Section 7. They illustrate how different combinations of obstacle stiffness and damping affect the beam’s penetration, vibrations, and approach to equilibrium. For very stiff obstacles with small damping, the behavior approaches the rigid Signorini contact with noticeable rebounds, whereas moderate stiffness and larger damping lead to deeper penetration and faster decay of oscillations. These computational experiments underscore the wide range of behaviors captured by the DNC law and highlight its ability to bridge between purely elastic and highly dissipative contacts [7]. The numerical convergence of our numerical scheme is described in Section 8, and it is seen that for small space and time steps, the convergence is better than linear. This provides us with confidence that the algorithm produces reliable simulations. The paper ends with Section 9, which presents a short summary, some conclusions, and some unresolved problems, which we consider worth further study.

2. The Setting and the Model

This section presents the setting and constructs a classical model for the contact process between a vibrating Bernoulli beam and a reactive object or obstacle, which is located below the right end of the beam. The setting is illustrated in Figure 1.

The system consists of a Bernoulli beam that in its reference configuration occupies the (scaled with L—the original length of the beam) interval $[0,1]$. The beam’s displacement from this reference configuration is denoted by $u(x,t)$, scaled by L, chosen as positive when above the x-axis. The beam is acted upon by a distributed force $f=f(x,t)$. The right end of the beam, at $x=1$, can come in contact with a reactive obstacle, marked in Figure 1 as ‘obstacle.’ The position of the obstacle is at $y=y_{-}$, and it is where contact can take place. Our main interest lies in what happens during contact between the beam’s end and the obstacle. We denote partial derivatives with subscripts, so $u_x=\partial u/\partial x$, etc. The force applied to the beam is such that the beam can press against the obstacle, possibly leading to repeated contact.

We turn to the classical formulation of the model for the process

Model 1.

Given the force $f=f(x,t)$, and the coefficients $\alpha ,\kappa ,\beta$, all positive constants, find the displacement field $u:[0,1]\times [0,T]\to \mathbb{R}$, for $0<T$, such that

$$\begin{array}{cc}\hfill & u_{tt}+{\alpha }^2{u}_{xxxx}=fin{\Omega }_T=\{0<x<1,0<t<T\},\hfill \end{array}$$

(1a)

$$\begin{array}{cc}\hfill & u(0,t)=u_x(0,t)=0for0\le t\le T,\hfill \end{array}$$

(1b)

$$\begin{array}{cc}\hfill & u_{xx}(1,t)=0,for0\le t\le T,\hfill \end{array}$$

(1c)

$$\begin{array}{cc}\hfill & -{\alpha }^2{u}_{xxx}(1,t)=\left\{\begin{array}{cc}0,\hfill & u(1,t)>y_{-},\hfill \\ \kappa {(u(1,t)-y_{-})}_{-}-\beta u_t(1,t),\hfill & u(1,t)\le y_{-}.\hfill \end{array}\right.\hfill \end{array}$$

(1d)

$$\begin{array}{cc}\hfill & u(x,0)=u_0\left(x\right),u_t(x,0)=v_0\left(x\right),0\le x\le 1.\hfill \end{array}$$

(1e)

The initial displacement $u(x,0)=u_0\left(x\right)$ and the initial velocity $v(x,0)=v_0\left(x\right)$ are given functions, since the beam is clamped at $x=0$ and free ate $x=1$. These satisfy the compatibility conditions $u_0\left(0\right)=0$, $v_0\left(0\right)=0$, $u_{0xx}\left(1\right)=0$ and also $u_0\left(1\right)>y_{-}$ and then $u_{0xxx}\left(1\right)=0$.

Here, u and x are scaled with L; ${\alpha }^2=E_{Y}I/\rho A{L}^4$, $\rho$ is the density of the beam material, A is the cross-sectional area, $E_Y$ is Young’s modulus (elasticity modulus), and I is the second moment of the beam’s cross-sectional area. The external force f is scaled with $LA$, so it is force per unit volume of the beam. In addition, ${(·)}_{-}$ denotes the negative part function, that is ${\left(r\right)}_{-}=|r|$ when $r<0$ and ${\left(r\right)}_{-}=0$ when $r>0$.

Remark 1.

A note on the DNC contact condition (1d). When there is no contact, $u(1,t)>y_{-}$, the traction or stress at $x=1$ is zero since the end is free. When there is contact, $u(1,t)\le y_{-}$, the obstacle reaction is given by

$$R=\kappa {(u(1,t)-y_{-})}_{-}-\beta u_t(1,t),$$

where κ is the obstacle stiffness and β is its damping coefficient. Thus, the DNC condition explicitly takes into account the energy loss during contact. It is also noted that R may be positive or negative, although the stiffness term is always positive so it points up. However, when the obstacle is soft (small κ) and the damping is large (large β) and the velocity is large and, in the upward direction, R may be negative. As we show below, the energy dissipation is always negative when the end is moving. Furthermore, $|u(1,t)-y_{-}|$ is the depth of penetration of the beam into the obstacle, when in contact.

For simplicity, we introduce the cutoff function, which is the characteristic function of the interval $(-\infty ,y_{-}]$, by

$${\chi }_{*}\left(u\right)=\left\{\begin{array}{cc}0\hfill & u(1,t)>y_{-}\hfill \\ 1\hfill & u(1,t)\le y_{-}.\hfill \end{array}\right.$$

(2)

It vanishes when there is no contact and, therefore, the obstacle resistance can be written concisely as

$$R=\left(\kappa (y_{-}-u(1,t))-\beta u_t(1,t)\right){\chi }_{*}\left(u\right).$$

(3)

We next consider the system energy balance and show that the obstacle dissipates energy.

3. Energy Balance and Dissipation

We derive the energy balance in the system, assuming that the solutions are sufficiently smooth for the manipulations to make sense. We multiply (1a) with $u_t$ and integrate over $[0,1]\times [0,t]$, for $0<t\le T$. We deal with the different terms separately. The first term, after integration by parts, yields

$${\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^2{|}_0^{t}dx={\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^2(x,t)dx-{\displaystyle \frac{1}{2}}{\int }_0^1{v}_0^2\left(x\right)dx=E_k\left(t\right)-E_k\left(0\right).$$

Here, the kinetic energy, at time t, is given by

$$E_k\left(t\right)={\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^2(x,t)dx.$$

(4)

Next, we note that

$${\alpha }^2{\int }_0^t{\int }_0^1{u}_{xxxx}{u}_{t}dxd\tau ={\alpha }^2{\int }_0^t{u}_{xxx}{u}_t{|}_0^{1}d\tau -{\alpha }^2{\int }_0^t{\int }_0^1{u}_{xxx}{u}_{xt}dxd\tau .$$

We have, using (3),

$${\alpha }^2{\int }_0^t{u}_{xxx}{u}_t{|}_0^{1}d\tau =-{\int }_0^t\left(\kappa (y_{-}-u(1,\tau ))u_t(1,\tau )-\beta u_t^2(1,\tau )\right){\chi }_{*}\left(u\right)d\tau .$$

Moreover, using the initial condition $u_0\left(1\right)>y_{-}$, we find

$${\int }_0^t\left(\kappa (u(1,\tau )-y_{-})u_t(1,\tau )\right){\chi }_{*}\left(u\right)d\tau ={\displaystyle \frac{\kappa }{2}}{(u(1,t)-y_{-})}^2{\chi }_{*}\left(u\right).$$

Furthermore, using integration by parts, the boundary conditions and the facts that $u_x(0,t)=0$ and $u_{xx}(1,t)=0$, we obtain

$${\alpha }^2{\int }_0^t{\int }_0^1{u}_{xxx}{u}_{xt}dxd\tau ={\alpha }^2{\int }_0^t{u}_{xx}{u}_{xt}{|}_0^{1}d\tau -{\displaystyle \frac{{\alpha }^2}{2}}{\int }_0^1{u}_{xx}^2{|}_0^{t}dx$$

$$=-{\displaystyle \frac{{\alpha }^2}{2}}{\int }_0^1{u}_{xx}^2(x,t)dx+{\displaystyle \frac{{\alpha }^2}{2}}{\int }_0^1{u}_{0xx}^2\left(x\right)dx.$$

We now define the potential energy of the system, at time t, as

$$E_p\left(t\right)={\displaystyle \frac{{\alpha }^2}{2}}{\int }_0^1{u}_{xx}^2(x,t)dx+{\displaystyle \frac{\kappa }{2}}{(u(1,t)-y_{-})}^2{\chi }_{*}\left(u\right).$$

(5)

Then, using (4) and (5) and the above expressions, we obtain the following balance of the system energy.

Result 1.

The system energy balance is given by

$$E\left(t\right)=E_k\left(t\right)+E_p\left(t\right)$$

$$={\displaystyle \frac{1}{2}}{\int }_0^1{u}_t^2(x,t)dx+{\displaystyle \frac{{\alpha }^2}{2}}{\int }_0^1{u}_{xx}^2(x,t)dx+{\displaystyle \frac{\kappa }{2}}{(u(1,t)-y_{-})}^2{\chi }_{*}\left(u\right)$$

$$=E_k\left(0\right)+E_p\left(0\right)+{\int }_0^t{\int }_0^{1}f(x,t)u_t(x,t)dxd\tau -\beta {\int }_0^t{u}_t^2(1,\tau ){\chi }_{*}\left(u\right)d\tau .$$

(6)

The first term on the right describes the initial kinetic energy and the second term the initial elastic potential energy. The third term describes the energy supplied or removed by the applied force, and the last term is the energy dissipated by the obstacle. It follows, since the last term is negative when the end is in motion and in contact, that the obstacle’s resistance dissipates energy from the system.

We next study the steady states of the system and find the threshold force needed to have contact.

4. Steady States

This section studies the steady states of the system. We assume that the applied force f is a negative constant, since if $f>0$ the end does not come into contact with the obstacle. The steady solutions, denoted by $\overline{u}\left(x\right)$, are found by setting the time derivatives in (1b)–(1d) to zero. Thus, we obtain the following ODE model:

Model 2.

Given the force $f=const.$, $\alpha ,\kappa ,$ positive constants, find the displacement field $\overline{u}$: $[0,1]\to \mathbb{R}$, such that:

$$\begin{array}{cc}\hfill & {\alpha }^2{\overline{u}}_{xxxx}=f,0<x<1,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & \overline{u}\left(0\right)={\overline{u}}_x\left(0\right)=0,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & {\overline{u}}_{xx}\left(1\right)=0,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & -{\alpha }^2{\overline{u}}_{xxx}\left(1\right)=\kappa (\overline{u}\left(1\right)-y_{-}){\chi }_{*}\left(\overline{u}\left(1\right)\right).\hfill \end{array}$$

(10)

Integrating four times with respect to x gives the general polynomial solution for (7) and applying the boundary conditions in (8), we obtain

$$\overline{u}\left(x\right)={\displaystyle \frac{1}{2}}B{x}^2+{\displaystyle \frac{1}{6}}A{x}^3+{\displaystyle \frac{f}{24{\alpha }^2}}{x}^4.$$

Applying the boundary conditions at $x=1$, we have

$${\overline{u}}^{″}\left(1\right)=B+A+{\displaystyle \frac{f}{2{\alpha }^2}}=0,$$

hence

$$\overline{u}\left(x\right)=-\left({\displaystyle \frac{f}{4{\alpha }^2}}+{\displaystyle \frac{A}{2}}\right)x^2+{\displaystyle \frac{A}{6}}{x}^3+{\displaystyle \frac{f}{24{\alpha }^2}}{x}^4.$$

(11)

and

$$-{\alpha }^2{\overline{u}}^{‴}\left(1\right)=-{\alpha }^{2}A-f=\kappa \left(\overline{u}\left(1\right)-y_{-}\right){\chi }_{*}\left(\overline{u}\left(1\right)\right).$$

(12)

This implies that there are two cases to consider:

(i)

No contact: When $\overline{u}\left(1\right)-y_{-}>0$ then ${\chi }_{*}\left(\overline{u}\left(1\right)\right)=0$, and we find $A=-{\displaystyle \frac{f}{{\alpha }^2}}$. Therefore, in this case, the solution is

$$\overline{u}\left(x\right)={\displaystyle \frac{f}{24{\alpha }^2}}\left(x^2-4x+6\right)x^2.$$

(13)

Evaluating this at $x=1$, we find $\overline{u}\left(1\right)={\displaystyle \frac{f}{8{\alpha }^2}}$; therefore,

Result 2

(No contact). The condition for no contact is

$${\displaystyle \frac{f}{8{\alpha }^2}}>y_{-}.$$

(14)

Then ${\chi }_{*}\left(\overline{u}\left(1\right)\right)=0$ and the solution is just the usual one of a linear beam with a free right end and is given in (13).

Physically, this means that the (scaled) force is insufficient to bend the beam enough to contact the obstacle, since the beam stiffness opposes it.

(ii)

Contact: When $\overline{u}\left(1\right)-y_{-}<0$, then $\left(\overline{u}\left(1\right)-y_{-}\right){\chi }_{*}\left(\overline{u}\left(1\right)\right)=y_{-}-\overline{u}\left(1\right)$, ${\chi }_{*}\left(\overline{u}\left(1\right)\right)=1$, and from (11) and (12) we find that

$$-{\displaystyle \frac{{\alpha }^2}{\kappa }}A-{\displaystyle \frac{f}{\kappa }}=y_{-}-\overline{u}\left(1\right)=y_{-}-\left(-{\displaystyle \frac{5f}{4{\alpha }^2}}-{\displaystyle \frac{A}{3}}\right),$$

$$=y_{-}+{\displaystyle \frac{5f}{24{\alpha }^2}}+{\displaystyle \frac{A}{3}}.$$

Then, after rearranging and some manipulations, we obtain

$$A={\displaystyle \frac{-3\kappa }{(\kappa +3{\alpha }^2)}}\left(y_{-}+f\left({\displaystyle \frac{5}{24{\alpha }^2}}+{\displaystyle \frac{1}{\kappa }}\right)\right),$$

(15)

which depends on ${\alpha }^2,f$ and $\kappa$. We note that $f<0,y_{-}<0$, hence $A>0$. Thus, with A given in (15), the solution of the steady problem when contact takes place is

$$\overline{u}\left(x\right)=-\left({\displaystyle \frac{f}{4{\alpha }^2}}+{\displaystyle \frac{A}{2}}\right)x^2+{\displaystyle \frac{A}{6}}{x}^3+{\displaystyle \frac{f}{24{\alpha }^2}}{x}^4.$$

(16)

In particular,

$$\overline{u}\left(1\right)={\displaystyle \frac{1}{(\kappa +3{\alpha }^2)}}\left({\displaystyle \frac{3}{8}}f+\kappa y_{-}\right).$$

(17)

We conclude that in the case with contact, we have

Result 3

(Contact). Assume that

$${\displaystyle \frac{f}{8{\alpha }^2}}\le y_{-}.$$

(18)

Then, there is contact between the beam and the obstacle and the steady solution $\overline{u}$ is given in (16).

Physically, this means that the (scaled) force to the beam’s stiffness ratio is sufficient to bend the beam so it contacts the obstacle.

Figure 2 shows the computed solutions of the steady states, using formulas (13) or (16), depending on inequality (14). In the simulations ${\alpha }^2=1$ and $y_{-}=-0.2$ and constant force f values are used. The figure shows the steady bending of the beam for various values of the force, and contact is established when the force is close to $f=-1.60$.

It is found that under a constant distributed load $f<0$ and the DNC contact condition (or the NC contact condition) with stiffness $\kappa >0$ and obstacle position $y_{-}$, the beam deflection naturally divides into two cases characterized by

$$\overline{r}={\displaystyle \frac{f}{8{\alpha }^2{y}_{-}}}.$$

When $\overline{r}<1$, there is no-contact, and the free end deflection is above the obstacle. When $\overline{r}\ge 1$ there is contact. The value $\overline{r}=1$ is the transition when the end of the beam just touches the obstacle, without penetrating it.

In the example, Figure 2, the transition case $\overline{r}=1$ is $f=-1.60$. This can be clearly seen in the figure. The positions of the beam’s end are given by (17) for the different values of f. The decrease in the penetration as the stiffness of the obstacle increases is clearly seen.

We next discuss an abstract approach to the problem.

5. Weak Formulation and Existence

This section is rather abstract and may be skipped upon first reading. It presents a weak formulation of the problem and also a regularized problem with viscosity both of which involve a substantial functional analysis background. The detailed proof of the existence of a weak solution to the problem will be found in [12]. Here, we just provide a short description and state the main existence theorem. The proof is based on a sequence of regularized problems and passing to the regularization coefficients limits. The main issue is that the boundary conditions for $v=u_t$ are not well defined when we only know that $u_t\left(t,·\right)$ has values in $L^2(0,1)$. Also, the discontinuity of ${\chi }_{*}$ causes some difficulty. To regularize the problem, we add viscosity to the beam equation in the form of $\epsilon u_{xxxxt}$, for $\epsilon >0$, which allows us to obtain better estimates. We note the problem with viscosity has interest in its own right, since many materials, such as metals, exhibit viscosity.

We first consider a problem in which ${\chi }_{\delta }$ is regularized by replacing it with a piecewise linear Lipschitz approximation given by

$${\chi }_{\eta }\left(r\right)=\left\{\begin{array}{cc}0\hfill & r>y_{-}\hfill \\ r\hfill & \left(y_{-}-\eta \right)\le r\le y_{-}\hfill \\ 1\hfill & r\le \left(y_{-}-\eta \right),\hfill \end{array}\right.$$

for a small $0<\eta$. Eventually, we pass to the limits $\epsilon ,\eta \to 0$.

We set the problem in terms of $v=u_t$, and then

$$u\left(x,t\right)=u_0\left(x\right)+{\int }_0^{t}v\left(x,s\right)ds.$$

We use the notation ${\Omega }_T=\{0<x<1,0<t<T\}$,  and let $H=L^2(0,1)$ and $V=\{w\in H^2(0,1);w\left(0\right)=w_x\left(0\right)=0\}$, with the usual inner products and norms.

The classical formulation of the regularized problem is:

Model 3.

Find $v:[0,T]\to V$ such that

$$v_t+\epsilon v_{xxxx}+{\alpha }^2{u}_{xxxx}=fin{\Omega }_T,$$

(19)

$$u(0,t)=0,u_x(0,t)=0for0\le t\le T,$$

(20)

$$u_{xx}(1,t)=0,for0\le t\le T,$$

(21)

$$-{\alpha }^2{u}_{xxx}(1,t)={\chi }_{\eta }\left(u\left(1,t\right)\right)\left(\kappa (y_{-}-u(1,t))-\beta v(1,t)\right),$$

(22)

$$v(x,0)=v_0\left(x\right),v_0\in L^2\left(0,1\right),$$

(23)

$$u\left(x,t\right)=u_0\left(x\right)+{\int }_0^{t}v\left(x,s\right)ds.$$

(24)

Next, we set the problem in a variational form. The function space in which we seek weak solutions is the Sobolev space

$$V=\left\{u(·,t)\in H^2(0,1)|u(0,t)=0,u_x(0,t)=0\right\},$$

and we let

$$\mathcal{V}=L^2(0,T,V).$$

For the time dependence on $t\in [0,T]$, we require $u\in \mathcal{V}$ and $v=u_t\in L^2(0,T)\times H^2(0,1)$. We denote this space by $\mathcal{W}$,

$$\mathcal{W}=\left\{u:[0,T]\to \mathcal{V}|u_t\in L^2(0,T)\times \left(H^2(0,1)\right)\right\}.$$

We choose $\varphi (x,t)\in L^2(0,T;V)$ as a test function. Multiplying Equation (19) by $\varphi$ and integrating on $[0,1]$, using various manipulations, Green’s identities, and the boundary conditions, we obtain the following:

$$\begin{array}{cc}\hfill & {\int }_0^T{\int }_0^1{u}_{tt}\varphi dxdt+{\alpha }^2{\int }_0^T{\int }_0^1{u}_{xx}{\varphi }_{xx}dxdt+\epsilon {\int }_0^T{\int }_0^1{u}_{xxt}{\varphi }_{xx}dxdt\hfill \\ \hfill & -{\int }_0^T\left[\kappa (u(1,t)-y_{-})-\beta u_t(1,t)\right]{\chi }_{\eta }\left(u(1,t)\right)\varphi (1,t)dt={\int }_0^T{\int }_0^{1}f\varphi dxdt.\hfill \end{array}$$

(25)

The weak or variational formulation of the problem with viscosity, Model 3, (19)–(24), is the following:

Model 4.

Given $f\in L^2\left(0,T\right)\times L^2\left(0,1\right)$, and positive constants $\alpha ,\kappa ,\beta ,\epsilon ,\eta$, find $u\in \mathcal{V}$, such that $u_t\in \mathcal{V}$, and for every test function $\varphi \in \mathcal{V}$ that satisfies $\varphi (x,T)=0$, (25) holds, together with

$$u(0,t)=0,u_x(0,t)=0,0\le t\le T,$$

and

$$u(x,0)=u_0\left(x\right),u_t(x,0)=v_0\left(x\right),0\le x\le 1.$$

As noted in [12], we need to control the boundary term, and hence, we introduce the truncation function.

$$m\left(r\right)=\left\{\begin{array}{c}0,0\le r,\hfill \\ r,-M\le r\le 0,\hfill \\ -M,r\le -M.\hfill \end{array}\right.$$

Then, we replace the term $\kappa (u(1,t)-y_{-})$ with

$$m\left(\kappa (u(1,t)-y_{-})\right),$$

which is bounded and Lipschitz continuous.

We have the following existence theorem, established in [12].

Theorem 1

([12]). There exists a weak solution u to Model 4, in which the term $\kappa (u(1,t)-y_{-})$ is truncated as $m\left(\kappa (u(1,t)-y_{-})\right)$, and ${\chi }_{\eta }$ is replaced with ${\chi }_{*}$. The solution with $\eta =0,\epsilon =0$ is weaker.

The proof uses a fixed-point argument and the regularization. Then, sufficient estimates are obtained that allow us to consider the case without regularization and viscosity, that is, when $\eta \to 0$ and $\epsilon \to 0$.

To depict graphically what the solutions of the model look like, we consider numerical simulations of the solutions. First we describe the algorithm for the simulations.

6. Finite Difference Algorithms

We use an implicit second-order finite difference algorithm for the simulations of the problem. The beam is divided into M uniform segments $\Delta x=L/M$. The simulations are performed over the time interval $[0,T]$, and the time step is given by $\Delta t=T/N$, where N is the number of time steps. Let $u_i^j$ be the approximation of the solution at the node $x_i=i\Delta x$ at time $t_j=j\Delta t$, that is, $u_i^j\approx u(x_i,t_j)$, for $i=0,1,\dots ,M$ and $j=0,1,\dots ,N$. First, we note that, thanks to the initial conditions in Equation (1e), we have

$$u_i^0=u_0\left(x_i\right),u_i^1=\Delta t{v}_0\left(x_i\right)+u_0\left(x_i\right),0\le i\le M.$$

(26)

Moreover, thanks to the Dirichlet boundary condition in Equation (1b), we have

$$u_0^j=0,\forall 1\le j\le N.$$

(27)

For $i=1,2,\dots ,M$ and $j=1,2,\dots ,N-1$, the implicit finite difference scheme is given by

$${\displaystyle \frac{u_i^{j+1}-2u_i^j+u_i^{j-1}}{\Delta t^2}}+{\alpha }^2{\displaystyle \frac{u_{i+2}^{j+1}-4u_{i+1}^{j+1}+6u_i^{j+1}-4u_{i-1}^{j+1}+u_{i-2}^{j+1}}{\Delta x^4}}=f_i^{j+1}.$$

(28)

Rearranging terms and defining $\gamma ={\alpha }^2{\displaystyle \frac{\Delta t^2}{\Delta x^4}}$, we have

$$\gamma u_{i-2}^{j+1}-4\gamma u_{i-1}^{j+1}+(1+6\gamma )u_i^{j+1}-4\gamma u_{i+1}^{j+1}+\gamma u_{i+2}^{j+1}=\Delta t^2{f}_i^{j+1}+2u_i^j-u_i^{j-1}.$$

(29)

Equation (29) is used to construct a linear system at every time level which is solved to obtain the discrete solution at that level. We note that when $i=1$, we have the ghost node $u_{-1}^{j+1}$, for which we use the boundary condition in Equation (1b). In particular, using a second-order approximation of the x-derivative, this boundary condition gives us

$$u_x(0,t_{j+1})\approx {\displaystyle \frac{u_1^{j+1}-u_{-1}^{j+1}}{2\Delta x}}=0\Rightarrow u_{-1}^{j+1}=u_1^{j+1}.$$

Therefore, the finite difference scheme Equation (29) for $i=1$ becomes

$$(1+7\gamma )u_1^{j+1}-4\gamma u_2^{j+1}+\gamma u_3^{j+1}=\Delta t^2{f}_1^{j+1}+2u_1^j-u_1^{j-1}.$$

(30)

We also note that when $i=M-1$ and when $i=M$, there are two ghost nodes: $u_{M+1}^{j+1}$ and $u_{M+2}^{j+1}$. Using the boundary condition in Equation (1c), we have

$$u_{xx}(1,t_{j+1})\approx {\displaystyle \frac{u_{M+1}^{j+1}-2u_M^{j+1}+u_{M-1}^{j+1}}{\Delta x^2}}=0⟹u_{M+1}^{j+1}=2u_M^{j+1}-u_{M-1}^{j+1}.$$

Thus, the finite difference scheme for $i=M-1$ becomes

$$\begin{array}{cc}\hfill \gamma u_{M-3}^{j+1}-4\gamma u_{M-2}^{j+1}& +(1+5\gamma )u_{M-1}^{j+1}\hfill \\ \hfill & -2\gamma u_M^{j+1}=\Delta t^2{f}_{M-1}^{j+1}+2u_{M-1}^j-u_{M-1}^{j-1}.\hfill \end{array}$$

(31)

The contact boundary condition Equation (1d) is taken into account depending on whether the beam had contact with the obstacle at the previous time or not. We have the following two cases:

• No contact: $u_M^j>y_{-}$

  $$u_{xxx}(1,t_{j+1})\approx {\displaystyle \frac{-\frac{1}{2}{u}_{M-2}^{j+1}+u_{M-1}^{j+1}-u_{M+1}^{j+1}+\frac{1}{2}{u}_{M+2}^{j+1}}{\Delta x^3}}=0$$

  $$\Rightarrow u_{M+2}^{j+1}=u_{M-2}^{j+1}-2u_{M-1}^{j+1}+2u_{M+1}^{j+1}=u_{M-2}^{j+1}-4u_{M-1}^{j+1}+4u_M^{j+1}.$$
  Thus, in the case of no contact, the finite difference scheme is

  $$2\gamma u_{M-2}^{j+1}-4\gamma u_{M-1}^{j+1}+(1+2\gamma )u_M^{j+1}=\Delta t^2{f}_M^{j+1}+2u_M^j-u_M^{j-1}.$$

  (32)
• Contact: $u_M^j\le y_{-}$

  $$\begin{array}{cc}\hfill -{\alpha }^2{u}_{xxx}(1,t_{j+1})& \approx -{\alpha }^2{\displaystyle \frac{-\frac{1}{2}{u}_{M-2}^{j+1}+u_{M-1}^{j+1}-u_{M+1}^{j+1}+\frac{1}{2}{u}_{M+2}^{j+1}}{\Delta x^3}}\hfill \\ \hfill & =\kappa (y_{-}-u_M^{j+1})-\beta {\displaystyle \frac{u_M^{j+1}-u_M^j}{\Delta t}}\hfill \\ \hfill \Rightarrow u_{M+2}^{j+1}& =-{\displaystyle \frac{2\kappa }{{\alpha }^2}}\Delta x^3(y_{-}-u_M^{j+1})+{\displaystyle \frac{2\beta }{{\alpha }^2}}\Delta x^3{\displaystyle \frac{u_M^{j+1}-u_M^j}{\Delta t}}\hfill \\ \hfill & +u_{M-2}^{j+1}-4u_{M-1}^{j+1}+4u_M^{j+1}\hfill \end{array}$$
  Thus, in the case of contact, the finite difference scheme is

  $$\begin{array}{c}\hfill 2\gamma u_{M-2}^{j+1}-4\gamma u_{M-1}^{j+1}+\left(1+2\gamma \left({\displaystyle \frac{\Delta x^3}{{\alpha }^2}}\left(\kappa +{\displaystyle \frac{\beta }{\Delta t}}\right)+1\right)\right)u_M^{j+1}\\ \hfill =\Delta t^2{f}_M^{j+1}+2\left(1+\gamma {\displaystyle \frac{\beta \Delta x^3}{{\alpha }^2\Delta t}}\right)u_M^j-u_M^{j-1}+2\gamma {\displaystyle \frac{\Delta x^3}{{\alpha }^2}}\kappa y_{-}.\end{array}$$

  (33)

7. Simulations

The implicit finite difference Algorithm 1 was implemented in MATLAB, (version R2025b) and the backslash function was used to solve the linear system. In all simulations $v_0\left(x\right)\equiv 0$, the beam length was $L=1$, the final time was $T=5$, and $\Delta x=\Delta t=0.01$. Moreover, the force was taken as $f=-0.25$ and ${\alpha }^2=1.296$. The obstacle was located at $y_{-}=-0.02$. In this section, we report some of the more interesting computer experiments. Typically, each simulation took about 10–15 s, running in MATLAB. The values of the various parameters are shown in

Table 1. Symbols and values of the parameters in the simulations.

Parameter	Value	Name
$\Delta t$	0.01	time step
$\Delta x$	0.01	space step
T	5	final time
f	−0.25	applied force
${\alpha }^2$	1.296	elastic beam modulus
$y_{-}$	−0.02	position of the obstacle
$\kappa$	0.1, 1, 10, 1000	obstacle stiffness
$\beta$	0.001, 0.1, 1, 5	obstacle damping

. The computer algorithm follows.

Algorithm 1 Implicit finite difference scheme for the vibrating beam.

Initialization:   1: for $i=0$ to M do   2:       Set $u_i^0$ and $u_i^1$ using Equation (26).   3: end for   3: Time Marching:   4: for $j=1$ to $N-1$ do   4:     (1) Set Dirichlet boundary condition on the left using Equation (27).   4:     (2) Construct the $M\times M$ matrix A and $M\times 1$ vector b   5:       Assign the first row of A and first entry of b according to Equation (30)   6:       for $i=2$ to $M-2$ do   7:             Assign i-th row of A and i-th entry of b according to Equation (29)   8:       end for   9:       Assign $M-1$-th row of A and $M-1$-th entry of b according to Equation (31) 10:       Assign M-th row of A and M-th entry of b 11:       if $u_M^j>y_{-}$ then                                                                   ▹ No-contact scenario 12:             Use Equation (32) 13:       else                                                                                            ▹ Contact scenario 14:             Use Equation (33) 15:       end if 15:     (3) Solve the linear system with matrix A and right hand side b 16: end for

The following are example simulations for various obstacle stiffnesses, $\kappa$, and damping coefficients $\beta$. Typically, in Figure 3, we show the position of the right end of the beam in time for different values of $\kappa$ and $\beta$. It is seen that, progressively, the penetration decreases, and substantially changes between the values $\kappa =10$, in which the obstacle is relatively soft and $\kappa =1000$, in which the obstacle is rigid. In the latter case, there is virtually no penetration. In the figure, the dashed black line is the location of the obstacle. The colored lines represent the motion of the tip for various values of the damping $\beta$, for each fixed value of the stiffness $\kappa$. The red line corresponds to $\beta =0.001$, the blue line to $\beta =0.1$, the green line to $\beta =1$ and yellow is $\beta =5$. Progressively, as the stiffness increases, the penetration decreases, and substantially changes between the values of $\kappa =10$, mildly soft and $\kappa =1000$, which is essentially rigid.

Result 4.

We summarize our results with these observations: 

1. 

For large κ (≥$100$) and small β (≈$0.1$), the contact process behaves similarly to that with a Signorini (rigid) contact condition: very little penetration, high contact force and repeated bounces when the initial energy is sufficient.

2. 

For moderate stiffness κ (≈$1-10$), one obtains a normal compliance-like penetration, with partial rebounds damped out, depending on the values of β.

3. 

For small κ (≈$0.1$) and large β (≈$10$), the obstacle effectively behaves like a “soft dissipative” material, such as tissue, allowing for deeper penetration and higher energy dissipation.

These results confirm that the DNC contact condition can model a whole spectrum of contact behaviors, bridging the gap between purely elastic and purely dissipative extremes.

Next, we provide a numerical indication that our algorithm is reliable. Therefore, we have confidence that so are the numerical solutions.

8. Numerical Convergence

In this section, we numerically demonstrate the convergence of the algorithm. We consider a reference solution $\tilde{u}$ which is obtained using Algorithm 1 with $\Delta x=\Delta t=2^{-12}$. The parameters are $\kappa =1$, $\beta =0.001$, $f=-0.25$, and ${\alpha }^2=1.296$. We calculate the numerical solution $u_i^j$ with $\Delta x=2^{-k}$, for $k=4,\dots ,10$, and calculate the ${\ell }^{\infty }$ norm of the error at the final time $T=2$. The rate of convergence is calculated as

$$\mathrm{rate}={\displaystyle \frac{1}{log\left(0.5\right)}}log\left({\displaystyle \frac{\parallel \tilde{u}-u_{\Delta x/2}{\parallel }_{{\ell }^{\infty }}}{\parallel \tilde{u}-u_{\Delta x}{\parallel }_{{\ell }^{\infty }}}}\right),$$

where $u_{\Delta x}$ and $u_{\Delta x/2}$ are the numerical solutions obtained with $\Delta x$ and $\Delta x/2$, respectively. The results are presented in

Table 2. Rates of convergence at final time $T=2$ with $\Delta t=\Delta x$, for case in Section 8.

$\Delta \mathit{x}$	$\parallel \tilde{\mathit{u}}{-\mathit{u}\parallel }_{{\mathit{\ell }}^{\mathit{\infty }}}$	Rate	$\Delta \mathit{x}$	$\parallel \tilde{\mathit{u}}{-\mathit{u}\parallel }_{{\mathit{\ell }}^{\mathit{\infty }}}$	Rate
$2^{-4}$	1.10 $\times {10}^{-2}$	-	$2^{-8}$	1.06 $\times {10}^{-3}$	0.99
$2^{-5}$	7.07 $\times {10}^{-3}$	0.64	$2^{-9}$	4.51 $\times {10}^{-4}$	1.23
$2^{-6}$	4.01 $\times {10}^{-3}$	0.81	$2^{-10}$	9.05 $\times {10}^{-5}$	2.31
$2^{-7}$	2.12 $\times {10}^{-3}$	0.91

, where we observe that the subsequent numerical solutions converge to the reference solution, and the error decays at a rate of at least 1.

This result provides substantial confidence in the numerical method and, therefore, in the numerical solutions.

9. Conclusions and Unresolved Questions

This work studies the vibrations of a Bernoulli beam that can come into contact with an obstacle that is situated below its right end. The contact process is described by the Damped Normal Compliance boundary condition. The model consists of the usual fourth-order dynamic beam equation, together with the initial and boundary conditions. The interest lies in the resistance of the obstacle R, (3), which includes the stiffness and damping terms, and is active when the right end is in contact.

The energy balance of the system is derived in Section 3, showing that once the beam is in contact, the system loses energy due to damping. The steady states of the system are found in Section 4. The ratio $\overline{r}$,

$$\overline{r}={\displaystyle \frac{f}{8{\alpha }^2{y}_{-}}},$$

controls the contact. When $\overline{r}<1$, there is no contact in the steady state, and when $\overline{r}\ge 1$, the beam is in contact, and when $\overline{r}=1$, the beam’s end just touches the obstacle. The variational formulation of the problem is provided in Section 5, which includes the functional setting. It is noted that the term $v=u_t(1,t)$ needs special treatment, so we introduce the space $\mathcal{W}$ and moreover, we consider regularized problems with viscosity, the solutions of which are sufficiently regular to allow the trace to exist. The existence of variational solutions to the regularized problems is stated in Theorem 1. Moreover, the solutions of the problem without viscosity are obtained as limits when $\epsilon \to 0$ and can be found in the space $\mathcal{W}$. The proofs will be found in [12].

A finite difference scheme and an algorithm for the simulations of the problem can be found in Section 6. The algorithm is implicit and special attention is paid to the discretized contact conditions. Section 7 provides the simulation results. It is found that for a force above a threshold value, the beam penetrates the obstacle. Then, depending on stiffness and damping, the simulations indicate that as stiffness decreases and damping increases, the penetration increases and the oscillations decay faster. These are summarized in Section 4. We conclude that by applying the DNC condition instead of the NC condition, the results are likely to better represent the process.

These results confirm that the DNC condition can model a whole spectrum of contact behaviors, bridging between purely elastic and purely dissipative extremes. Moreover, in Section 8, we show that the algorithm and the resulting simulations are reliable. Indeed, we show that as the space and time steps decrease, the accuracy of the simulations substantially increases.

Once the DNC condition is applied in the design and computer software of various engineering applications, the results are likely to be more accurate. We note that processes in which bodies come into contact are ubiquitous in industrial settings, in transportation, and many other settings. Thus, this work is also a contribution to engineering design.

We turn to some unresolved problems that are of interest. More simulations and numerical studies of the dependence of vibration frequencies on the beam’s elastic modulus and the obstacle’s stiffness may be of practical interest, since these interact in basic ways. Thus, it may provide a better way to control unwanted vibrations. The case when there are two obstacles, one above and the other below, is of interest, especially numerically, since the vibration frequencies of the beam depend on the beam’s rigidity and the stiffness of the obstacles.

Further mathematical analysis of the problem is of interest. However, this will be done elsewhere, since the needed mathematical tools are from advanced functional analysis, and are outside of the scope of this work.

Another aspect that may be of interest is the possible addition of friction to the contact, so that when the beam’s end, when in contact, also undergoes friction and the associated frictional energy dissipation. This will add another layer of complexity to the problem, since friction is inherently connected with non-smooth models. However, it may be of interest to compare the frictional energy loss with that caused by the obstacle damping.