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Viscous dampers



Viscous damping occurs in a mechanical system because of viscous friction that results from the contact of a system component and a viscous liquid. The damping force produced when a rigid body is in contact with a viscous liquid is usually proportional to the velocity of the body.
F= c v
Where c is called the damping coefficient and has dimensions of mass per time.
Viscous damping can occur naturally, as when a buoyant body oscillates on the surface of a lake or a column of liquid oscillates in a U-tube manometer. Viscous damping is often added to mechanical system as a means of vibration control. Viscous damping leads to an exponential decay in amplitude of free vibration control. Viscous reduction in amplitude in forced vibrations caused by a harmonic excitation. In addition, the presence of viscous damping gives rise to a linear term in the governing differential equation, and thus does not significantly complicate the mathematical modeling of the system. A mechanical device called a dashpot is added to mechanical system to provide viscous damping. A schematic of a dashpot in a one degree of freedom system is shown in next figure.
a)

b)

Figure 1 – a) schematic of one-degree-of freedom-mass-spring-dashpot system, b) Dashpot forces cx and opposes the direction of positive x’
A simple dashpot configuration is shown in next figure. The upper plate of the dashpot is connected to a rigid body. As the body moves, the plate slides over a reservoir of viscous liquid of dynamic viscosity. The area of the plate in contact with the liquid is A. The shear stress developed between the fluid and the plate creates resultant friction force acting on the plate. Assume the reservoir is stationary and the upper plate slides over the liquid with a velocity v. the reservoir depth h is small enough that the velocity profile in the liwuid can be approximated as linear, as illustrated in fig 2 b. If y is a coordinate measured upward from the bottom of the reservoir,
u(y)=v(y/h)
The shear stress developed on the plate is determined from Newton’s viscosity law
τ=µ(du/dy)= µ(v/h)
The viscous force acting on the plate is:
F= τA=(µA/h)v
Comparison of   shows that the damping coefficient for this dashpot is;
c=µA/h
Previous equation shows that a large damping force is achieve with a very viscous fluid, a small h, and a large A. A dashpot design with these parameters is often impractical and thus the devices of fig 2a is rarely used as a dashpot.

Figure 2 a) Simple dashpot model where plate is a fixed reservoir of viscous liquid b) Since h is small, a linear profile is assumed in the liquid
The analysis assumes the plate moves with a constant velocity. During the motion of a mechanical system the dashpot is connected to a particle which has a time dependent velocity. The changing velocity of the plate leads to unsteady effects in the liquid. If the reservoir depth h is small, the unsteady effects are small and can be neglected. 

Introduction to free vibrations of one degree of freedom system

           Introduction
In this post I will give an introduction to free vibrations of one degree of freedom systems. The name already tells you that we are dealing with free vibrations which mean that the object of mechanical system oscillates about system’s equilibrium position in the absence of an external excitation. Free vibrations are a result of a kinetic energy imparted to the system or of a displacement from the equilibrium position that leads to a difference in potential energy from the system’s equilibrium position.
Ok now let’s consider a mechanical system that is shown in the next figure.
Figure 1.1 - System consisting of body and the spring

From the previous figure we can see that mechanical system consists of an object with mass m, spring with stiffness k. Also you can see that a spring is a connection between the object and the fixture. When block/object is displaced at distance x0 from it’s equilibrium position then the potential energy kx0^2/2 is developed in the spring. When the system is released from equilibrium, the spring force draws the block toward the system’s equilibrium position with the potential energy being converted to kinetic energy. When the block reaches the equilibrium position, the kinetic energy reaches a maximum and motion continues. The kinetic energy is converted to potential energy until the spring is compressed a distance x0. This process of transfer of potential energy to kinetic energy and vice versa is continual in the absence of non-conservative forces. In a physical system such perpetual motion is impossible. Dry friction, internal friction in the spring aerodynamic drag, and other non-conservative mechanisms eventually dissipate the energy.
Another example of free vibrations systems with one degree of freedom are: oscillations of the pendulum about a vertical equilibrium position, the motion of a recoil mechanism of a firearm once it has been fired, and the motion of the vehicle suspension system after the vehicle encounters a pothole.
The oscillations of these systems can be described by the second-order ordinary differential equation. In this case the dependent variable is time, while the dependent variable is chosen generalized coordinate. The chosen generalized coordinate represent displacement of a particle in the system or an angular displacement and is measured from the system’s equilibrium position.
There exist two possible methods that can be used to derive the differential equation governing motion of a one-degree-of-freedom system and they are:
1)      Free body diagram method
2)      Equivalent system method
And as we discussed earlier, if the system is nonlinear, a linearizing assumption will be made.
The generalized solution of the second. Order differential equation is a liner combination of two linearly independent solutions. The arbitrary constants, called constants of integration, are uniquely determined on application of two initial conditions. The necessary initial conditions are values of the generalized coordinate and its first time derivative at a specific time, usually t=0.

The form of the solution for the differential equation depends on system parameters. A mathematical form of the solution for an un-damped system is different from the solution of a system with viscous damping. 

Springs

1.1.1. Introduction
Springs are one of the most important components in vibration analysis because they are mechanical link between two particles in a mechanical system. In general a spring is a continuous system. As usual the spring has inertia but when we compare a vale of spring inertia to the other elements in the mechanical system we can see that this value is small so we can neglect it.

The length of a spring when it is not subject to external forces is called the un-stretched length. Since the spring is made of a flexible material, the applied force F that must be applied to the spring to change its length by x is some continuous function of x.
The appropriate form of the f(x) is determined by using the constitutive equation for the spring’s material. Since f(x) id infinitely differentiable at x = 0, it can be expanded by a Taylor series about x = 0. This procedure is called MacLaurin expansion:
Since x is the spring’s change in length from its unscratched length, when x=0, F=0. Thus k0=0. When x is positive, the spring is in tension. When x is negative, the spring is negative, the spring is in compression. Many materials have the same properties in tension and compression. That is, if a tensile force F is required to lengthen the spring by, then a compressive force of the same magnitude F is required to shorten the spring by. For these materials, f(x) must be an odd function and cannot contain even powers. 
In reality all the springs are nonlinear. However in many situations x is small enough that the nonlinear terms of previous equation are small compared to the first variable that is k1*x. A linear spring obeys a force displacement law of 
where k is called the spring stiffness or spring constant and has dimensions of force pre length. The force in a spring whose force displacement law is given by (1.3) is conservative that is the work done by the spring force between two arbitrary displacement is independent of the path taken between the two displacement. Thus a potential energy function exists and for a linear spring is determined as:

A torsional spring is a link in a mechanical system where application of a torque leads to an angular displacement between the ends of the torsional spring. A linear torsional spring has a relationship between an applied moment M and the angular displacement fi of

where a torsional stiffness kt has dimensions of force times length. The potential energy function for a torsional spring is

Example 1.1

The numerical values for the parameters in the system are:
Determine the potential energy in the spring when the system is in the equilibrium. Develop an expression for the potential energy in the spring when the pulley is rotated clockwise from the equilibrium position

Figure 1.1 – Free body diagram of static equilibrium position


Let's start solving the example by defining Δ as statitc deflection of the spring, its change in length from its unstreched length when the system is in equilibrium. The static deflection is determined by applying hte equations of equilibrium to the free body diagram.











The free body diagram show that it was assumed that the spring is compressed when the system is in equilibrium. The negative sign indiates that the spring is actually stretched. Its potential energy when the system is in equilibirim is:




When the pulley is rotated through a clockwise angle θ from the system's equilibrium position, the spring is shortened in lenght 2rθ from its length when the system is in equilibrium. Thus the total change in lenght of the spring is



and the potential energy in the spring is:












          1.1.2.Helical coil springs

The helical coil spring is used in applications such as industrial machines and vehicle suspension systems. Consider a spring manufacture from a rod of circular cross section of diameter D. The shear modulus of the rod is G. The rod is formed into coil of N turns of radius r. It is assumed that the coil radius is much larger than the radius of the rod and that the normal to the plane of one coil nearly coincides with the axis of the spring. 

Figure 1.2. Free body diagram of cut coil spring exposes resultant shear force and resultant torque.
Consider a helical coil spring when subject to an axis load F. Imagine cutting the rod with a knife at an arbitrary location in a coil, slicing the spring in two sections. The cut exposes an internal shear force F and an internal resisting torque Fr, as illustrated in previous figure. Assuming elastic behavior, the shear stress due to the resisting torque varies linearly with distance from the center of the rod to a maximum of 
Where J = Pi*D^4/32 is the polar moment of inertia of the rod. The shear stress due to the shear force varies nonlinearly with distance from the neutral axis. For r/D>>1 the maximum shear stress due to the resisting torque, and its effect is neglected.
Principles of mechanics of materials can be used to show that the total change in length of the spring due to an applied force F is:
 Comparing Eq.(1.9) with Eq.(1.4) leads to the conclusion that the under the assumptions stated a helical coil spring can be modeled as a linear spring of stiffness.

Example 1.2
A tightly wound spring is made from a 20 mm diameter bar of 0.2% C hardened steel (G=80e9 N/m^2). The coil diameter is 20 cm. the spring has 30 coils. What is the largest force that can be applied such that the elastic strength in shear of 220 e6 N/m^2 is not exceeded? What is the change in length of the spring when this force is applied?
Assuming the shear force is negligible, the maximum shear stress in the spring when a force F is applied is: 




Thus the maximum allowable force is:



The stiffness of this spring is calculated by:




The total change in length of the spring due to application of the maximum allowable force is:



          1.1.3.Elastic elements as spring
Application of a force F to the block of mass m of figure 1.3, results in a displacement x. The block is attached to a uniform thin rod of elastic modulus E, un-stretched length L, and cross sectional area A. Application of the force results in uniform normal strain in the rod of: 

The total strain energy developed due to the work of the force in stretching the rod is:


If the force is suddenly removed, the block will oscillate about its equilibrium position. The initial strain energy is converted to kinetic energy and vice versa, a process which continues indefinitely. If the mass of the rod is small compared to the mass of the block, then inertia of the rod is negligible and the rod behaves as a discrete spring. From strength of materials, the force F required to change the length of the rod by x is:

A comparison of Eq. (1.13) and Eq. (1.4) implies that the stiffness of the rod is:

Figure1.3 – Longitudinal vibrations of mass attached to end of uniform thin rod can be modeled as a linear mass-spring system with k=AE/L
The motion of a particle attached to an elastic element can be modeled as a particle attached to a linear spring, provided the mass of the element is small compared to the mass of the particle and a linear relationship between force and displacement exists for the element. In figure 1.4 A particle of mass m is attached to the midspan of a simply supported beam of length L, elastic modulus E, and cross-sectional moment of inertia I. The transverse displacement of the midspan of the beam due to applied static load F is

Thus a linear relationship exists between transverse displacement and static load. Hence if the mass of the beam is small, the vibrations of the particle can be modeled as the vertical motion of a particle attached to a spring of stiffness.

In general the transverse vibrations of a particle attached to a beam can be modeled as those of a particle attached to a linear spring. Let w(z) represent the displacement function of the beam due to  a concentrated unit load applied at z = a. then the displacement at z=a due to a load F applied at z =a is:

Then the spring stiffness for a particle placed at z = a is

Figure 1.4 – a) The transverse vibrations of a block attached to a simply supported beam, b) are modeled by the mas spring system, provided the mas of the block beam is small compared to the mass of the block
Figure 1.5 – Torsional stiffness of the shaft is JG/L
Torsional oscillations occur in the system of Fig 1.5. A thin disk of mass moment of inertia I is attached to a circular shaft of length L, shear modulus G and polar moment of inertia J. When the disk is rotated through an angle from its equilibrium position, a moment 

develops between the dis and the shaft. Thus if the polar mass moment of inertia of the shaft is small compared with I, then the shaft acts as a torsional spring of stiffness:

Example 1.3

A 200 kg machine is attached to the end of a cantilever beam of length L =2.5 m. Elastic modulus E=200e9 N/m^2, and cross sectional moment of inertia 1.8e-6m^4. Assuming the mass of the beam is small compared to the mass of the machine, what is the stiffness of the beam? 









1.1.4. Spring in combination



Most of vibrational problems include more than one spring. That means they are in some combination. In this section will talk about what combinations of springs exist and how to solve them. The solution is very simple. You need to replace this combination of springs with one which has equivalent stiffness. The equivalent stiffness is determined such that the system with a combination of springs has the same displacement x as the equivalent system when both systems are subject to the same force F. A model of one degree of freedom system consisting of a block attached to a spring of an equivalent stiffness is illustrated in next figure.
The resultant force acting on the block is : 





The springs in the system are parallel. The displacement of each spring in the system is the same, but the resultant force acting on the block is the sum of the forces developed in the parallel springs. If x is the displacement of the block, then the force developed in the i-th spring is ki*x and the resultant is




Equating the forces form leads to


 




The springs shown in next figure are in series. The force developed in each spring is the same and equal to the force acting on the block. The displacement of the block is the sum of the changes in length of the springs in the series combination. If xi is the change in length of the i-th spring then





Since the force is the same in each spring, xi=F/ki and eq becomes
 




Since the series combinations is to be replaced by a spring of an equivalent stiffness is used leading to
 



We can see similarity with electrical components because electrical circuit component can also be placed in series and parallel and effect of the combination replaced by a single component with and equivalent value. The equivalent capacitance of capacitors in parallel or series is calculated like that of springs in parallel or series. The equivalent resistance of resistors in series is the sum of the resistances, whereas the equivalent resistance of resistors in parallel is calculated by using an equation similar to Eq.(1.26).