Vibrations

1.1 Vibrations in mechanical System Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium point. It can be desirable or undesirable. Vibration analysis in today’s world is used in a huge way to understand the health of a machine.
1.1.1 Causes of vibration in machines There are various sources of vibration in a mechanical system. some of the common ones are listed below: (a) Imbalance (b) Misalignment/ shaft run out. (c) Wear of rotating/ reciprocating parts (d) Looseness of nuts and bolts or any other fastener. (e) Natural calamities such as earthquakes.
1.1.2 The harmful effects of vibrations There are various harmful effects of vibration : (a) Noise (b) Excessive wear of rotating/ receprocating parts. (c) Cracks or failure of parts (d) Loosening of fastener
1.2 Types of Vibratory motion 1. Free or natural vibrations- When no external force acts on the body, after giving it an initial displacement, then the body is said to be under free or natural vibrations. The frequency of the free vibrations is called free or natural frequency. 2. Forced vibrations- When the body vibrates under the influence of external force, then the body is said to be under forced vibrations. The external force applied to the body is a periodic disturbing force created by unbalance. The vibrations have the same frequency as the applied force. 3. Damped vibrations- When there is a reduction in amplitude over every cycle of vibration,the motion is said to be damped vibration. This is due to the fact that a certain amount of energy possessed by the vibrating system is always dissipated in overcoming frictional resistances to the motion.
1.3 Descriptors of Vibrations Amplitude - The amplitude of vibration is the magnitude of vibration. Amplitude is measured and expressed in three ways: Displacement (in Microns) Velocity (mm/sec) and Acceleration (m/sec2 or g) Period & Frequency -The time required for the system to repeat its motion is called a “period (T)”. Frequency may be expressed in Hertz (CPS) – f in Hz = RPM / 60 Cycles per minute (CPM /RPM) – CPS * 60 Radian per second (Circular frequency) – ω = 2π f Phase & Its Importance Vibration phase can simply be defined as the moment at which event occurs. Phase is the difference in time or position between forces and/or motions. It is the instantaneous position of a vibrating part at a given instance with reference to a fixed position or another vibrating part.Phase Tells us In What Direction Is The Movement relative To Other Locations on the Machine at a Given Moment in Time.Measurement of phase and its analysis can help in the diagnosis of a machinery problem. Phase is used in balancing and is useful in diagnosing imbalance, misalignment, looseness and other causes. Resonance - When frequency of the exciting force is equal to the natural frequency of the system it is called resonance. Under such conditions the amplitude of vibration builds up dangerously. Degree of Freedom - The degree of freedom of a vibrating body or system implies the number of independent coordinates which are required to define the motion of the body or system at given instant. Simple Harmonic Motion - It is a to and fro periodic motion of a particle in which : (a) acceleration is proportional to the displacement from the meanposition. (b) Acceleration is always directed towards a fixed point which is themean equilibrium position. It can be represented by an expression having a periodic function like sine or cosine. $x= Xsin \omega t$ , where X is the amplitude.
1.4 Vibration analysis procedure Mathematical Modelling Derivation of Governing Equations Solution of the Governing Equations Interpretation of the Results Mathematical modelling 1. It represents important features of the system for the purpose of deriving the governing equations. 2. Model may be linear or non-linear depending on the behavior of the components of the system. 2. Linear models permit quick solutions.A great deal of engineering judgment is necessary to formulate the non-linear model Derivations of governing equations 1. Use principles of dynamics (i) Newton’s Second Law (ii) D’Alembert’s Principle (iii) Conservation of Energy 2. Draw free body diagrams 3. Derive equations (i) Set of ordinary differential equations -for discrete systems (ii) Set of partial differential equations –for continuous systems Modelling of vibratory system To do a mathematical analysis of a vibratory system it is important to first develop an equivalent model of the system. To make an equivalent model we use three basic elements mass, spring and dampers. These components are categorized based on whether the component forces are proportional to displacement, velocity or acceleration. Correspondingly they can be divided into components that can store and release potential energy (as a spring does), dissipate energy (as a damper does) and store or release kinetic energy ( as a mass does).
1.5 Springs Springs can store or release potential energy. They are generally assumed to be massless. A typical plot of the force Fs vs the elongation δ will be as below In the liner range, δ is proportional to Fs, where the constant of proportionality 'k' is equal to the slope of the curve Fs Vs δ. Hence in linear range, the relation between force and elongation δ is $F_s=k(X_1-X_2)$ Equation for potential energy stored in the spring is $V(x) =k{\delta^2\over 2}$
1.6 Dampers Dampers can dissipate energy. It is generally represented by a piston loosely fitted in a cylinder filled with oil so that the viscous fluid can flow around the piston inside the cylinder. Dampers also like springs are assumed to be massless. A dampening force $F_d$ applied at one end is balanced by a corresponding for $F_d$ at the other end. It is also assumed that the force $F_d$ causes smooth shear in the viscous fluid, so that the plot of $F_d$ Vs $\dot{\delta}$ is linear, where $\dot{\delta}=\dot{x_2}-\dot{x_1}$ . The relation between the Force $F_d$ and the velocity of separation $\dot{\delta}$ is $F_d=c\dot{\delta}=c(\dot{x_2}-\dot{x_1})$ Where the proportionality constant C is merely the slope of the curve. C is known as coefficient of viscous damping. The unit is newton.second per meter(N.s/m). The damper force is non conservative, as it depends on the velocity and not on position. Regarding damper as part of a system we can write $\dot{E} = (-c\dot{\delta})\dot{\delta}$ Where E is the total energy in the system, but the right side of the equation is negative as long as $\dot{\delta}\neq0$ and is equal to zero when $\dot{\delta}=0$ . Hence we must conclude that the system loses energy steadily, so that viscous damper dissipate energy.
1.7 Mass Since Fm = m.ax, which means that force is proportional to acceleration and hence if a graph is plotted then we will get a straight line. The energy equation is $E={mV^2\over2}$ which implies that masses store energy as velocity increases and release energy as velocity decreases.
1.8 Equivalent spring, mass and dampers There can be cases when these basic elements are used in a combination. The general procedure to solve such cases is to make a single equivalent of these components.
1.9 Spring in parallel-
1.10 Spring in series-
1.11 Equivalent Mass : In a system which has 3 mass, the equivalent mass can be calculated as below ${1\over2}m_1\dot{x_1}^2+{1\over2}m_2\dot{x_2}^2+{1\over2}m_3\dot{x_3}^2={1\over2}m_e\dot{x_e}^2$
1.12 Equivalent Dampers : Parallel dampers : $C_e=C_1+C_2$ Series dampers : ${1\over{C_e}}={1\over{C_1}}+{1\over{C_2}}$
1.13 Modelling of Mechanical systems By modelling of mechanical system we mean to represent the mechanical system in its equivalent mass, spring and damper system. There can be mutiple ways of modeliing a system, however for a good model one should always keep in mind that it should contian all the important dynamics of the acutal system.
1.14 Vibrations in single degree of freedom systems Undamped free vibration - There are various method to analyse an undamped system. Let us take the case of mass hung by a spring. it's equivalent mathematical model is shown below: based on the above mathematical model the governing equation and its solution can be found by many ways. We will solve it by (i) Newton's second law (ii) D'Alemberts principle, and (iii) Energy conservation method. Solution by Newton's 2nd law It states that the rate of change of linear momentum is proportional to the force impressed upon it and it is mathmatically written as below $\Sigma F=m{dv\over{dt}} =>m{d (\dot x)\over{dt}}$ , where x is the displacement. From the above free body diagram $mg-k(\Delta_{st}+x)=m\ddot{x}$ ,Where $\Delta{st} =$ static deflection due to the weight of the hanging mass or deflection in the absence of any vibration. Hence, $k\Delta{st} = mg$ , thus $m\ddot{x}+kx=0, \quad or \quad \ddot{x}+({k\over m})x=0$ ,- - - - -Eq 1 now if we say let $({k\over m}) = \omega_n^2,\quad where \quad \omega_n$ = natural frequency of vibration of the system. Thus comparing equation 1 with the fundamental equation of SHM we can say that $\omega_n = \sqrt {k\over m}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}\sqrt{k\over m}$ , and time period off vibration is $T={1\over f_n}=2\Pi\sqrt{m\over k}$ Solution by D. Alebert's principle Which states that => Accelerating Force + Restoring Force = 0, hence $mg-k(\Delta_{st}+x)-m\ddot{x}=0$ ,Where $\Delta{st} =$ static deflection due to the weight of the hanging mass or deflection in the absence of any vibration. Hence, $k\Delta{st} = mg$ , thus $m\ddot{x}+kx=0, \quad or \quad \ddot{x}+({k\over m})x=0$ ,- - - - -Eq 1 now if we say let $({k\over m}) = \omega_n^2,\quad where \quad \omega_n$ = natural frequency of vibration of the system. Thus comparing equation 1 with the fundamental equation of SHM we can say that $\omega_n = \sqrt {k\over m}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}\sqrt{k\over m}$ , and time period of vibration is $T={1\over f_n}=2\Pi\sqrt{m\over k}$ Solution by Energy conservation method It states that for any vibrating system the kinetic energy at equilibrium position will be maximum and potential energy 0, where as extreme positions potential energy will be maximum and kinetic energy 0. At any intermediate position the system will have both kinetic and potential energy. However at all position the sum of kinetic and potential energy will remain same or constant. Hence, K.E+P.E=Constant. Thus equating total energy at any potion we get ${1\over2}m\dot{x}^2+k[\Delta_{st}+{(0+x)\over2}]x-mgx=constant$ - - - - - -eqn2 differentiating eqn2 , we get $m\ddot{x}\dot{x}+k\Delta{st}+kx\dot{x}-mg=0$ , since $\Delta{st} =$ static deflection due to the weight of the hanging mass or deflection in the absence of any vibration and $k\Delta{st} = mg$ , we get $m\ddot{x}\dot{x}+kx\dot{x}=0$ or $m\ddot{x}+kx=0$ . This is same fundamental equation of SHM and hence $\omega_n = \sqrt {k\over m}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}\sqrt{k\over m}$ , and time period of vibration is $T={1\over f_n}=2\Pi\sqrt{m\over k}$ Thus wee see that it really does not matter which method we solve the problem wee will always reach to the same values. Developing above equation for a system is known as mathematical modeling which is required for vibration analysis or dynamic analysis of the system Governing Equation or Charecteristic Equation for spring-mass system considering the mass of the spring. Let 'L' be the length of the spring and ' $\rho$ ' be the mass of spring per unit length. Mass of spring, M = ρL, Consider a small element 'dy' of the spring at a distance 'y from a support. The mass of this element is 'dM'. $dM=\rho dy$ . Considering that the free end of the spring has moved by 'x'. The displacement at the fixed point will be zero. from similar triangles, at any point, at a distance'y', the displacement is given by 'δ' , such that: ${\delta \over y} = {x\over L}$ hence, velocity of 'dM' will be $\dot\delta$ Therefore. $\dot\delta = {\dot{x}y\over L}$ The kinetic energy of the system will be due to mass of the spring 'M' and the hanging mass 'm'. Total kinetic energy is $KE= KE_{spring}+KE_{mass}$ $KE=\int_0^L\ {1\over2}dM\dot\delta^2 +{1\over2}m\dot{x}^2$ $={1\over2}m\dot{x}^2+ \int_0^L\ {1\over2}\rho dy( {\dot{x}y\over L})^2$ , replacing the value of dM and $\dot\rho$ $={1\over2}m\dot{x}^2+ {1\over2}\rho({\dot{x}\over L})^2\int_0^L\ y^2dy$ $={1\over2}m\dot{x}^2+ {1\over2}\rho({\dot{x}\over L})^2{L^3\over 3}$ $={1\over2}m\dot{x}^2+ {1\over2}\rho({\dot{x}})^2{L\over 3}$ $={1\over2}m\dot{x}^2+ {1\over6}M({\dot{x}})^2$ , since $\rho L = M$ ,The total energy in the system is say, U and $U=K.E+P.E$ Now $P.E = {1\over 2}kx^2$ . Therefore $U={1\over 2}kx^2+{1\over 2}m\dot{x}^2+{1\over 6}M\dot{x}^2 =>U = {1\over 2}kx^2+{1\over 2}\dot{x}^2(m+{M\over 3})$ Differentiatin andd using enery method we get ${dU\over dt} = 0 =kx\dot{x}+(m +{M\over 3})\dot{x}\ddot{x}$ , or $kx+(m +{M\over 3})\ddot{x}=0 \quad => \ddot{x}+{k\over(m+{M\over 3})}x=0$ , thus we get $\omega_n = \sqrt {k\over(m+{M\over 3})}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}*\sqrt{k\over(m+{M\over 3})}$ ,
1.15 Damping Damping is the resistance offered by a body to the motion of a vibratory system. The resistance can be applied by liquid or solid internally or externally.Damping decreases amplitude of vibration and rate of decrease depends on the amount of damping.
1.16 Types of damping
1.17 Viscous Damping It is the resistance offered by viscous medium. It occurs for small velocities in lubricated surfaces,for example- a dashpot with small clearances. Damping resistance is proportional to relative velocity
1.18 Damped free vibration with viscous damping m = mass of body. kg k = stiffness of spring. N/m c= damping coefficient, N.s/m x= displacement of the body from equilibrium(mean) position, According to D'Alembert's principle $\Sigma inertia force+External forces =0$ $m\ddot{x} +c\dot{x}+kx=0$ The above equation is a linear differential equation of the second order and it's solution can be writter in the form, $x= e^{st}$ where, e = base of natural logarithms = 2.718 t = time s = a constant to be determined, Now $x=e^{st}$ , hence $\dot{x}=se^{st}$ and $\ddot{x} =s^2e^{st}$ , substituting this in the D'Alembert's equation, we get $ms^2e^{st}+cse^{st}+ke^{st}=0$ , or $ms^2+cs+k=0$ The above equation is called the characteristic equation of the system. This equation will have the solution as $S_{1,2}={-C\over 2m} \pm \sqrt (({-C\over 2m})^2-{k\over {m}})$ , this solution can alos be written as $x = C_1e^{s_1t}+C_2e^{s_2t}$ The nature of this solution depends on the term in the square roots. The three cases can be $({C\over m})^2 > 4({k \over m})$ $({C\over m})^2 = 4({k \over m})$ $({C\over m})^2 < 4({k \over m})$ Now, let the critical damping coefficient $C_c$ , also since $({C_c\over m})^2 = 4({k \over m}) =>({C\over 2m})^2 = ({k \over m})$ , Let $\zeta ={C\over C_c}$ , therefore ${c\over2m}$ can also be written as ${C\over 2m} = {C\over C_c}{C_c\over 2m}=\zeta\omega_n$ , where $\omega_n$ is the natural frequency of the systme under no damping. Thus the solution $S_{1,2}$ can alos be written as $S_{1,2} = [\zeta \pm \sqrt(\zeta^2-1)]\omega_n$ if $C>C_c \quad or \quad \zeta>1;$ System is over damped if $C=C_c \quad or \quad \zeta=1;$ System is critically damped if $C<C_c \quad or \quad \zeta<1;$ System is under damped The solution can now be again written as $x= C_1e^[\zeta + \sqrt(\zeta^2-1)]\omega_nt+C_2e^[\zeta - \sqrt(\zeta^2-1)]\omega_nt$ Let us find the constant $C_1$ and $C_2$ for overdamped, critically damped and under damped systems.
1.19 Case 1- Over damped system Displace the body through a distance $X_0$ from the equilibrium position and release without any initial velocity Thus, $x=X_0 \quad at\quad t=0, \quad therefore \quad \dot{x} = 0 \quad at \quad t=0$ , using this boundary condition in $x= C_1e^[\zeta + \sqrt(\zeta^2-1)]\omega_nt+C_2e^[\zeta - \sqrt(\zeta^2-1)]\omega_nt$ , we get $X_0 =C_1+C_2$ , Also $\dot{x} =C_1[-\zeta + \sqrt({\zeta^2-1}) ]\omega_ne^[-\zeta + \sqrt({\zeta^2-1}) ]\omega_nt+C_2[-\zeta - \sqrt({\zeta^2-1}) ]\omega_ne^[-\zeta - \sqrt({\zeta^2-1}) ]\omega_nt$ , applying boundary condition $0=C_1[-\zeta+ \sqrt(\zeta^2-1)]\omega_n+C_2[-\zeta- \sqrt(\zeta^2-1)]\omega_n$ Now, we have two equation for finding C1 & C2, which gives the values of C1 & C2 as $C_1={[\zeta + \sqrt (\zeta^2-1)]X_0\over 2 \sqrt (\zeta^2-1)}$ and $C_2={[-\zeta + \sqrt (\zeta^2-1)]X_0\over 2 \sqrt (\zeta^2-1)}$ , thus the eneral equation for motion which is overdamped is $x={X_0 \over2\sqrt (\zeta^2-1)}[[\zeta + \sqrt (\zeta^2-1)e^[-\zeta + \sqrt (\zeta^2-1)]\omega_nt+[\zeta + \sqrt (\zeta^2-1])e^[-\zeta - \sqrt (\zeta^2-1)]\omega_nt]$
1.20 Case 2 - Critically damped system $S_{1,2} =[-\zeta \pm \sqrt ( \zeta^2-1) ]\omega_n$ $S_1=S_2=\omega_n$ , therefore $x=C_1e^{\omega_nt} +C_2e^{\omega_nt}$ , now applying boundary conditions we get $C_1=X_0 \quad and \quad C_2=\omega_nX_0$ Hence , the solution for critically damped system becomes $x=X_0(1+\omega_nt)e^{ -\omega_nt}$
1.21 Case 3 - Under - damped system s1="À°+jp1À°2#!n s1="À°Àjp1À°2#!n x=C1e"À°+jp1À°2#!nt+C2e"À°Àjp1À°2#!nt x=eÀ°!nt"C1ej"p1À°2#!nt+C2eÀj"p1À°2#!nt# Applying the relationships eja=cosa+jsina eÀja=cosaÀjsina to the equation for x, we have x=e(À°!nt)"C1cosp1À°2!nt+jsinp1À°2!nt=C2cosp1À°2!ntÀjsinp1À°2!nt# x=e(À°!nt)"(C1+C2)cosp1À°2!nt+j(C1+C2)sinp1À°2!nt# The constants ( C1 + C2 ) and j(C1 - C2 ) in the above equation are real quantities which make C1 and C2 complax conjugate quantities. The equation derived above can be written in either of following three forms x=eÀ°!nt"Acosp1À°2!nt+Bsinp1À°2!nt# x=A1eÀ°!ntcos"p1À°2!nt+¶1# x=A2eÀ°!ntsin"p1À°2!nt+¶2# x_=A2p1À°2!neÀ°!ntcos"p1À°2!nt+¶2#ÀA2°!neÀ°!ntsin"p1À°2!nt+¶2# Substituting the initial conditions of equation X0=A2sin¶2 0=A2p1À°2!ncos¶2ÀA2°!nsin¶2 A2=X0p1À°2 ¶2=tanÀ1"°p1À°2# Substituting these values in the starting equation, x=X0p1À°2eÀ°!ntsin"p1À°2!nt+tanÀ1°p1À°2# This equation is of the oscillatory type and has an amplitude X0p1À°2eÀ°!nt which is seen to decay exponentially with time. Theoretically, the system will never come to rest although the amplitude of vibration may becomr infinitely small. Displacement-time plot of an under-damped system with general initial conditions Displacement-time plots of under-damped systems with zero starting velocity. tBÀtA=2Ùp1À°2!n XA=X0p1À°2eÀ°!ntA XB=X0p1À°2eÀ°!ntB Dividing one by the other XBXA=eÀ°!n(tAÀtB)=eÀ°!n(tBÀtA) ButtBÀtA=2Ùp1À°2!n ThereforeXBXA=e2Ù°=p1À°2 This is called logarithmic Decrement and is denoted dy δ. thereforeÎ=logeXBXA=2Ù°p1À°2 Let x0 represent the amplitude at partcular maxima and xn represent the amplitude after a further n cycles. This can be proved easily as below. ift=t+ntp it can be proved that lnxoxn=2nÙ°p1À° if°<0:3°then lnxox12Ù°
1.22 Forced Vibration If the system vibrates under the influence of external periodic force the vibration are known as forced vibration. Ex :Washing machine Ship propulsion system Garden water sprinkler Centrifugal pump Reciprocating engine Pendulum clock etc There are in general, three types of forcing functions commonly encountered in engineering practice 1. Periodic forcing functions. 2. Impulsive type of forcing functions. 3. Random forcing functions. The general overningg equation for a forced vibration is $M\ddot{x}+C\dot{x}+kx=F$ , where F is the external force acting on the vibrating body.
1.23 Lumped Parameters The spring constant 'k', the coefficient of viscous damping 'c' and the mass 'm' represent parameters of a system. Because they do not require spatial variables to describe their location and can be regarded as being located at discrete points, they are referred to as lumped parameters.

1.1 Vibrations in mechanical System

Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium point. It can be desirable or undesirable. Vibration analysis in today’s world is used in a huge way to understand the health of a machine.

1.1.1 Causes of vibration in machines

There are various sources of vibration in a mechanical system. some of the common ones are listed below:

(a) Imbalance

(b) Misalignment/ shaft run out.

(c) Wear of rotating/ reciprocating parts

(d) Looseness of nuts and bolts or any other fastener.

(e) Natural calamities such as earthquakes.

1.1.2 The harmful effects of vibrations

There are various harmful effects of vibration :

(a) Noise

(b) Excessive wear of rotating/ receprocating parts.

(c) Cracks or failure of parts

(d) Loosening of fastener

1.2 Types of Vibratory motion

1. Free or natural vibrations- When no external force acts on the body, after giving it an initial displacement, then the body is said to be under free or natural vibrations. The frequency of the free vibrations is called free or natural frequency.

2. Forced vibrations- When the body vibrates under the influence of external force, then the body is said to be under forced vibrations. The external force applied to the body is a periodic disturbing force created by unbalance. The vibrations have the same frequency as the applied force.

3. Damped vibrations- When there is a reduction in amplitude over every cycle of vibration,the motion is said to be damped vibration. This is due to the fact that a certain amount of energy possessed by the vibrating system is always dissipated in overcoming frictional resistances to the motion.

1.3 Descriptors of Vibrations

Amplitude - The amplitude of vibration is the magnitude of vibration.

Amplitude is measured and expressed in three ways:
      Displacement (in Microns)
     Velocity (mm/sec) and
     Acceleration (m/sec2 or g)

Period & Frequency -The time required for the system to repeat its motion is called a “period (T)”.

Frequency may be expressed in
Hertz (CPS) – f in Hz = RPM / 60
Cycles per minute (CPM /RPM) – CPS * 60
Radian per second (Circular frequency) – ω = 2π f

Phase & Its Importance

Vibration phase can simply be defined as the moment at which event occurs. Phase is the difference in time or position between forces and/or motions. It is the instantaneous position of a vibrating part at a given instance with reference to a fixed position or another vibrating part.Phase Tells us In What Direction Is The Movement relative To Other Locations on the Machine at a Given Moment in Time.Measurement of phase and its analysis can help in the diagnosis of a machinery problem. Phase is used in balancing and is useful in diagnosing imbalance, misalignment, looseness and other causes.

Resonance - When frequency of the exciting force is equal to the natural frequency of the system it is called resonance. Under such conditions the amplitude of vibration builds up dangerously.

Degree of Freedom - The degree of freedom of a vibrating body or system implies the number of independent coordinates which are required to define the motion of the body or system at given instant.

Simple Harmonic Motion - It is a to and fro periodic motion of a particle in which :
(a) acceleration is proportional to the displacement from the meanposition.
(b) Acceleration is always directed towards a fixed point which is themean equilibrium position.
It can be represented by an expression having a periodic function like sine or cosine. $x= Xsin \omega t$ , where X is the amplitude.

1.4 Vibration analysis procedure

Mathematical Modelling
Derivation of Governing Equations
Solution of the Governing Equations
Interpretation of the Results

Mathematical modelling
     1. It represents important features of the system for the purpose of deriving the governing equations.
     2. Model may be linear or non-linear depending on the behavior of the components of the system.
     2. Linear models permit quick solutions.A great deal of engineering judgment is necessary to formulate the non-linear model

Derivations of governing equations
     1. Use principles of dynamics
          (i) Newton’s Second Law
          (ii) D’Alembert’s Principle
          (iii) Conservation of Energy
     2. Draw free body diagrams
     3. Derive equations
          (i) Set of ordinary differential equations -for discrete systems
          (ii) Set of partial differential equations –for continuous systems

Modelling of vibratory system
To do a mathematical analysis of a vibratory system it is important to first develop an equivalent model of the system. To make an equivalent model we use three basic elements mass, spring and dampers.

These components are categorized based on whether the component forces are proportional to displacement, velocity or acceleration. Correspondingly they can be divided into components that can store and release potential energy (as a spring does), dissipate energy (as a damper does) and store or release kinetic energy ( as a mass does).

1.5 Springs

Springs can store or release potential energy. They are generally assumed to be massless. A typical plot of the force Fs vs the elongation δ will be as below

In the liner range, δ is proportional to Fs, where the constant of proportionality 'k' is equal to the slope of the curve Fs Vs δ. Hence in linear range, the relation between force and elongation δ is $F_s=k(X_1-X_2)$

Equation for potential energy stored in the spring is $V(x) =k{\delta^2\over 2}$

1.6 Dampers

Dampers can dissipate energy. It is generally represented by a piston loosely fitted in a cylinder filled with oil so that the viscous fluid can flow around the piston inside the cylinder. Dampers also like springs are assumed to be massless. A dampening force $F_d$ applied at one end is balanced by a corresponding for $F_d$ at the other end. It is also assumed that the force $F_d$ causes smooth shear in the viscous fluid, so that the plot of $F_d$ Vs $\dot{\delta}$ is linear, where $\dot{\delta}=\dot{x_2}-\dot{x_1}$ .

The relation between the Force $F_d$ and the velocity of separation $\dot{\delta}$ is $F_d=c\dot{\delta}=c(\dot{x_2}-\dot{x_1})$

Where the proportionality constant C is merely the slope of the curve. C is known as coefficient of viscous damping. The unit is newton.second per meter(N.s/m).

The damper force is non conservative, as it depends on the velocity and not on position. Regarding damper as part of a system we can write $\dot{E} = (-c\dot{\delta})\dot{\delta}$

Where E is the total energy in the system, but the right side of the equation is negative as long as $\dot{\delta}\neq0$ and is equal to zero when $\dot{\delta}=0$ . Hence we must conclude that the system loses energy steadily, so that viscous damper dissipate energy.

1.7 Mass

Since Fm = m.ax, which means that force is proportional to acceleration and hence if a graph is plotted then we will get a straight line.

The energy equation is $E={mV^2\over2}$

which implies that masses store energy as velocity increases and release energy as velocity decreases.

1.8 Equivalent spring, mass and dampers

There can be cases when these basic elements are used in a combination. The general procedure to solve such cases is to make a single equivalent of these components.

1.9 Spring in parallel-

1.10 Spring in series-

1.11 Equivalent Mass :

In a system which has 3 mass, the equivalent mass can be calculated as below

${1\over2}m_1\dot{x_1}^2+{1\over2}m_2\dot{x_2}^2+{1\over2}m_3\dot{x_3}^2={1\over2}m_e\dot{x_e}^2$

1.12 Equivalent Dampers :

Parallel dampers : $C_e=C_1+C_2$

Series dampers : ${1\over{C_e}}={1\over{C_1}}+{1\over{C_2}}$

1.13 Modelling of Mechanical systems

By modelling of mechanical system we mean to represent the mechanical system in its equivalent mass, spring and damper system. There can be mutiple ways of modeliing a system, however for a good model one should always keep in mind that it should contian all the important dynamics of the acutal system.

1.14 Vibrations in single degree of freedom systems

Undamped free vibration - There are various method to analyse an undamped system. Let us take the case of mass hung by a spring. it's equivalent mathematical model is shown below:

based on the above mathematical model the governing equation and its solution can be found by many ways. We will solve it by
(i) Newton's second law
(ii) D'Alemberts principle, and
(iii) Energy conservation method.

Solution by Newton's 2nd law

It states that the rate of change of linear momentum is proportional to the force impressed upon it and it is mathmatically written as below

$\Sigma F=m{dv\over{dt}} =>m{d (\dot x)\over{dt}}$ , where x is the displacement.

From the above free body diagram

$mg-k(\Delta_{st}+x)=m\ddot{x}$ ,Where $\Delta{st} =$ static deflection due to the weight of the hanging mass or deflection in the absence of any vibration. Hence, $k\Delta{st} = mg$ , thus

$m\ddot{x}+kx=0, \quad or \quad \ddot{x}+({k\over m})x=0$ ,- - - - -Eq 1

now if we say let $({k\over m}) = \omega_n^2,\quad where \quad \omega_n$ = natural frequency of vibration of the system. Thus comparing equation 1 with the fundamental equation of SHM we can say that

$\omega_n = \sqrt {k\over m}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}*\sqrt{k\over m}$ , and time period off vibration is $T={1\over f_n}=2\Pi*\sqrt{m\over k}$

Solution by D. Alebert's principle

Which states that => Accelerating Force + Restoring Force = 0, hence

$mg-k(\Delta_{st}+x)-m\ddot{x}=0$ ,Where $\Delta{st} =$ static deflection due to the weight of the hanging mass or deflection in the absence of any vibration. Hence, $k\Delta{st} = mg$ , thus

$m\ddot{x}+kx=0, \quad or \quad \ddot{x}+({k\over m})x=0$ ,- - - - -Eq 1

now if we say let $({k\over m}) = \omega_n^2,\quad where \quad \omega_n$ = natural frequency of vibration of the system. Thus comparing equation 1 with the fundamental equation of SHM we can say that

$\omega_n = \sqrt {k\over m}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}*\sqrt{k\over m}$ , and time period of vibration is $T={1\over f_n}=2\Pi*\sqrt{m\over k}$

Solution by Energy conservation method

It states that for any vibrating system the kinetic energy at equilibrium position will be maximum and potential energy 0, where as extreme positions potential energy will be maximum and kinetic energy 0. At any intermediate position the system will have both kinetic and potential energy. However at all position the sum of kinetic and potential energy will remain same or constant. Hence, K.E+P.E=Constant. Thus equating total energy at any potion we get

${1\over2}m\dot{x}^2+k[\Delta_{st}+{(0+x)\over2}]x-mgx=constant$ - - - - - -eqn2

differentiating eqn2 , we get

$m\ddot{x}\dot{x}+k\Delta{st}+kx\dot{x}-mg=0$ , since $\Delta{st} =$ static deflection due to the weight of the hanging mass or deflection in the absence of any vibration and $k\Delta{st} = mg$ , we get $m\ddot{x}\dot{x}+kx\dot{x}=0$ or $m\ddot{x}+kx=0$ . This is same fundamental equation of SHM and hence

$\omega_n = \sqrt {k\over m}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}*\sqrt{k\over m}$ , and time period of vibration is $T={1\over f_n}=2\Pi*\sqrt{m\over k}$

Thus wee see that it really does not matter which method we solve the problem wee will always reach to the same values. Developing above equation for a system is known as mathematical modeling which is required for vibration analysis or dynamic analysis of the system

Governing Equation or Charecteristic Equation for spring-mass system considering the mass of the spring.

Let 'L' be the length of the spring and ' $\rho$ ' be the mass of spring per unit length.

Mass of spring, M = ρL, Consider a small element 'dy' of the spring at a distance 'y from a support. The mass of this element is 'dM'.

$dM=\rho dy$ .

Considering that the free end of the spring has moved by 'x'. The displacement at the fixed point will be zero. from similar triangles, at any point, at a distance'y', the displacement is given by 'δ' , such that:

${\delta \over y} = {x\over L}$

hence, velocity of 'dM' will be $\dot\delta$

Therefore. $\dot\delta = {\dot{x}y\over L}$

The kinetic energy of the system will be due to mass of the spring 'M' and the hanging mass 'm'.

Total kinetic energy is $KE= KE_{spring}+KE_{mass}$

$KE=\int_0^L\ {1\over2}dM\dot\delta^2 +{1\over2}m\dot{x}^2$

$={1\over2}m\dot{x}^2+ \int_0^L\ {1\over2}\rho dy( {\dot{x}y\over L})^2$ , replacing the value of dM and $\dot\rho$

$={1\over2}m\dot{x}^2+ {1\over2}\rho({\dot{x}\over L})^2\int_0^L\ y^2dy$

$={1\over2}m\dot{x}^2+ {1\over2}\rho({\dot{x}\over L})^2{L^3\over 3}$

$={1\over2}m\dot{x}^2+ {1\over2}\rho({\dot{x}})^2{L\over 3}$

$={1\over2}m\dot{x}^2+ {1\over6}M({\dot{x}})^2$ , since $\rho L = M$ ,The total energy in the system is say, U and $U=K.E+P.E$

Now $P.E = {1\over 2}kx^2$ . Therefore

$U={1\over 2}kx^2+{1\over 2}m\dot{x}^2+{1\over 6}M\dot{x}^2 =>U = {1\over 2}kx^2+{1\over 2}\dot{x}^2(m+{M\over 3})$

Differentiatin andd using enery method we get

${dU\over dt} = 0 =kx\dot{x}+(m +{M\over 3})\dot{x}\ddot{x}$ , or

$kx+(m +{M\over 3})\ddot{x}=0 \quad => \ddot{x}+{k\over(m+{M\over 3})}x=0$ , thus we get

$\omega_n = \sqrt {k\over(m+{M\over 3})}$ , frequency of vibration $f_n={\omega \over {2\Pi}}={1\over{2\Pi}}*\sqrt{k\over(m+{M\over 3})}$ ,

1.15 Damping

Damping is the resistance offered by a body to the motion of a vibratory system. The resistance can be applied by liquid or solid internally or externally.Damping decreases amplitude of vibration and rate of decrease depends on the amount of damping.

1.16 Types of damping

1.17 Viscous Damping

It is the resistance offered by viscous medium. It occurs for small velocities in lubricated surfaces,for example- a dashpot with small clearances. Damping resistance is proportional to relative velocity

1.18 Damped free vibration with viscous damping

m = mass of body. kg
k = stiffness of spring. N/m
c= damping coefficient, N.s/m
x= displacement of the body from equilibrium(mean) position,

According to D'Alembert's principle

$\Sigma inertia force+External forces =0$

$m\ddot{x} +c\dot{x}+kx=0$

The above equation is a linear differential equation of the second order and it's solution can be writter in the form, $x= e^{st}$

where, e = base of natural logarithms = 2.718
t = time
s = a constant to be determined,

Now $x=e^{st}$ , hence $\dot{x}=se^{st}$ and $\ddot{x} =s^2e^{st}$ , substituting this in the D'Alembert's equation, we get

$ms^2e^{st}+cse^{st}+ke^{st}=0$ , or $ms^2+cs+k=0$

The above equation is called the characteristic equation of the system. This equation will have the solution as

$S_{1,2}={-C\over 2m} \pm \sqrt (({-C\over 2m})^2-{k\over {m}})$ , this solution can alos be written as $x = C_1e^{s_1t}+C_2e^{s_2t}$

The nature of this solution depends on the term in the square roots. The three cases can be

$({C\over m})^2 > 4({k \over m})$
$({C\over m})^2 = 4({k \over m})$
$({C\over m})^2 < 4({k \over m})$

Now, let the critical damping coefficient $C_c$ , also since $({C_c\over m})^2 = 4({k \over m}) =>({C\over 2m})^2 = ({k \over m})$ ,

Let $\zeta ={C\over C_c}$ , therefore ${c\over2m}$ can also be written as ${C\over 2m} = {C\over C_c}{C_c\over 2m}=\zeta\omega_n$ , where $\omega_n$ is the natural frequency of the systme under no damping. Thus the solution $S_{1,2}$ can alos be written as

$S_{1,2} = [\zeta \pm \sqrt(\zeta^2-1)]\omega_n$

if $C>C_c \quad or \quad \zeta>1;$ System is over damped
if $C=C_c \quad or \quad \zeta=1;$ System is critically damped
if $C<C_c \quad or \quad \zeta<1;$ System is under damped

The solution can now be again written as

$x= C_1e^[\zeta + \sqrt(\zeta^2-1)]\omega_nt+C_2e^[\zeta - \sqrt(\zeta^2-1)]\omega_nt$

Let us find the constant $C_1$ and $C_2$ for overdamped, critically damped and under damped systems.

1.19 Case 1- Over damped system

Displace the body through a distance $X_0$ from the equilibrium position and release without any initial velocity

Thus,

$x=X_0 \quad at\quad t=0, \quad therefore \quad \dot{x} = 0 \quad at \quad t=0$ , using this boundary condition in

$x= C_1e^[\zeta + \sqrt(\zeta^2-1)]\omega_nt+C_2e^[\zeta - \sqrt(\zeta^2-1)]\omega_nt$ , we get

$X_0 =C_1+C_2$ , Also

$\dot{x} =C_1[-\zeta + \sqrt({\zeta^2-1}) ]\omega_ne^[-\zeta + \sqrt({\zeta^2-1}) ]\omega_nt+C_2[-\zeta - \sqrt({\zeta^2-1}) ]\omega_ne^[-\zeta - \sqrt({\zeta^2-1}) ]\omega_nt$ , applying boundary condition

$0=C_1[-\zeta+ \sqrt(\zeta^2-1)]\omega_n+C_2[-\zeta- \sqrt(\zeta^2-1)]\omega_n$

Now, we have two equation for finding C1 & C2, which gives the values of C1 & C2 as

$C_1={[\zeta + \sqrt (\zeta^2-1)]X_0\over 2 \sqrt (\zeta^2-1)}$ and $C_2={[-\zeta + \sqrt (\zeta^2-1)]X_0\over 2 \sqrt (\zeta^2-1)}$ , thus the eneral equation for motion which is overdamped is

$x={X_0 \over2\sqrt (\zeta^2-1)}[[\zeta + \sqrt (\zeta^2-1)e^[-\zeta + \sqrt (\zeta^2-1)]\omega_nt+[\zeta + \sqrt (\zeta^2-1])e^[-\zeta - \sqrt (\zeta^2-1)]\omega_nt]$

1.20 Case 2 - Critically damped system

$S_{1,2} =[-\zeta \pm \sqrt ( \zeta^2-1) ]\omega_n$

$S_1=S_2=\omega_n$ , therefore $x=C_1e^{\omega_nt} +C_2e^{\omega_nt}$ , now applying boundary conditions we get

$C_1=X_0 \quad and \quad C_2=\omega_nX_0$

Hence , the solution for critically damped system becomes

$x=X_0(1+\omega_nt)e^{ -\omega_nt}$

1.21 Case 3 - Under - damped system

s1="À°+jp1À°2#!n

s1="À°Àjp1À°2#!n

x=C1e"À°+jp1À°2#!nt+C2e"À°Àjp1À°2#!nt

x=eÀ°!nt"C1ej"p1À°2#!nt+C2eÀj"p1À°2#!nt#

Applying the relationships

eja=cosa+jsina

eÀja=cosaÀjsina

to the equation for x, we have

x=e(À°!nt)"C1cosp1À°2!nt+jsinp1À°2!nt=C2cosp1À°2!ntÀjsinp1À°2!nt#

x=e(À°!nt)"(C1+C2)cosp1À°2!nt+j(C1+C2)sinp1À°2!nt#

The constants ( C1 + C2 ) and j(C1 - C2 ) in the above equation are real quantities which make C1 and C2 complax conjugate quantities.

The equation derived above can be written in either of following three forms

x=eÀ°!nt"Acosp1À°2!nt+Bsinp1À°2!nt#

x=A1eÀ°!ntcos"p1À°2!nt+¶1#

x=A2eÀ°!ntsin"p1À°2!nt+¶2#

x_=A2p1À°2!neÀ°!ntcos"p1À°2!nt+¶2#ÀA2°!neÀ°!ntsin"p1À°2!nt+¶2#

Substituting the initial conditions of equation

X0=A2sin¶2

0=A2p1À°2!ncos¶2ÀA2°!nsin¶2

A2=X0p1À°2

¶2=tanÀ1"°p1À°2#

Substituting these values in the starting equation,

x=X0p1À°2eÀ°!ntsin"p1À°2!nt+tanÀ1°p1À°2#

This equation is of the oscillatory type and has an amplitude

X0p1À°2eÀ°!nt

which is seen to decay exponentially with time. Theoretically, the system will never come to rest although the amplitude of vibration may becomr infinitely small.

Displacement-time plot of an under-damped system with general initial conditions

Displacement-time plots of under-damped systems with zero starting velocity.

tBÀtA=2Ùp1À°2!n

XA=X0p1À°2eÀ°!ntA

XB=X0p1À°2eÀ°!ntB

Dividing one by the other

XBXA=eÀ°!n(tAÀtB)=eÀ°!n(tBÀtA)

ButtBÀtA=2Ùp1À°2!n

ThereforeXBXA=e2Ù°=p1À°2

This is called logarithmic Decrement and is denoted dy δ.

thereforeÎ=logeXBXA=2Ù°p1À°2

Let x0 represent the amplitude at partcular maxima and xn represent the amplitude after a further n cycles. This can be proved easily as below.

ift=t+ntp

it can be proved that

lnxoxn=2nÙ°p1À°

if°<0:3°then

lnxox12Ù°

1.22 Forced Vibration

If the system vibrates under the influence of external periodic force the vibration are known as forced vibration.
Ex :Washing machine Ship propulsion system Garden water sprinkler Centrifugal pump Reciprocating engine Pendulum clock etc

There are in general, three types of forcing functions commonly encountered in engineering practice

1. Periodic forcing functions.
2. Impulsive type of forcing functions.
3. Random forcing functions.

The general overningg equation for a forced vibration is

$M\ddot{x}+C\dot{x}+kx=F$ , where F is the external force acting on the vibrating body.

1.23 Lumped Parameters

The spring constant 'k', the coefficient of viscous damping 'c' and the mass 'm' represent parameters of a system. Because they do not require spatial variables to describe their location and can be regarded as being located at discrete points, they are referred to as lumped parameters.

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