CE 6701 Structural Dynamics and Earthquake Engineering
Dr. P. Venkateswara Rao Associate Professor
Dept. of Civil Engineering
SVCE, Sriperumbudur
Unit I β Theory of Vibrations
β’ Difference between static loading and dynamic loading
β’ Degree of freedom
β’ Idealisation of structure as single degree of freedom system
β’ Formulation of Equations of motion of SDOF system
β’ DβAlemberts principles
β’ Effect of damping
β’ Free and forced vibration of damped and undamped structures
β’ Response to harmonic and periodic forces.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 2
Unit II β Multiple Degree of Freedom System
β’ Two degree of freedom system
β’ Modes of vibrations
β’ Formulation of equations of motion of multi degree of freedom (MDOF) system
β’ Eigen values and Eigen vectors
β’ Response to free and forced vibrations
β’ Damped and undamped MDOF system
β’ Modal superposition methods.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 3
Unit III β Elements of Seismology
β’ Elements of Engineering Seismology
β’ Causes of Earthquake
β’ Plate Tectonic theory
β’ Elastic rebound Theory
β’ Characteristic of earthquake
β’ Estimation of earthquake parameters
β’ Magnitude and intensity of earthquakes
β’ Spectral Acceleration.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 4
Unit IV β Response of Structures to Earthquake
β’ Effect of earthquake on different type of structures
β’ Behaviour of Reinforced Cement Concrete, Steel and Prestressed Concrete Structure under earthquake loading
β’ Pinching effect
β’ Bouchinger Effects
β’ Evaluation of earthquake forces as per IS:1893 β 2002
β’ Response Spectra
β’ Lessons learnt from past earthquakes.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 5
Unit V β Design Methodology
β’ Causes of damage
β’ Planning considerations / Architectural concepts as per IS:4326 β 1993
β’ Guidelines for Earthquake resistant design
β’ Earthquake resistant design for masonry and Reinforced Cement Concrete buildings
β’ Later load analysis
β’ Design and detailing as per IS:13920 β 1993.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 6
References β’ 1. Chopra, A.K., βDynamics of Structures β Theory and
Applications to Earthquake Engineeringβ, 4th Edition, Pearson Education, 2011.
β’ 2. Agarwal. P and Shrikhande. M., "Earthquake Resistant Design of Structures", Prentice Hall of India Pvt. Ltd. 2007
β’ 3. Paz, M. and Leigh.W. βStructural Dynamics β Theory & Computationβ, 4th Edition, CBS Publishers & Distributors, Shahdara, Delhi, 2006.
β’ 4. Damodarasamy, S.R. and Kavitha, S. βBasics of Structural dynamics and Aseismic designβ, PHI Learning Pvt. Ltd., 2012
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 7
References
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 8
Unit I β Theory of Vibrations
β’ Vibration:
β Motion of a particle or a body or a system of concentrated bodies having been displaced from a position of equilibrium, appearing as an oscillation.
β Vibration in structural systems may result from environmental sources such as wind, earthquakes and waterways.
β Earthquakes are most important due to enormous potential for damage to structures and loss of life.
β On an average every year around 10, 000 people die worldwide due to earthquakes.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 9
Unit I β Theory of Vibrations
β’ Vibration:
β Study of repetitive motion of objects relative to a stationary frame of reference or equilibrium position.
β Vibrations can occur in many directions and results in interaction of many objects.
β Motion of vibrating system is governed by the law of mechanics, and in particular by Newtonβs second law of motion (F=ma).
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 10
Unit I β Theory of Vibrations
β’ Basic concepts of Vibration: β Bodies having mass and elasticity are capable to vibrate.
β When body particles are displaced by the application of external force, the internal forces in the form of elastic energy present in the body, try to bring it to its original position.
β At equilibrium position, whole of the elastic energy is converted into kinetic energy and the body continuous to move in in the opposite direction.
β Whole K.E. is converted into elastic or strain energy and inturn body returns to equilibrium position.
β This way, vibration motion is repeated continuously and interchange of energy takes place.
β And hence, any motion repeats itself after an interval of time is called vibration or oscillation.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 11
Unit I β Theory of Vibrations
β’ Dynamic loading:
β Dynamics: Study of forces and motions with time dependency.
β Dynamic load: Load magnitude, direction and position changes with time.
β Structural response to dynamic loading can be done by two methods:
β’ i) Deterministic analysis : Structural response i.e. displacement, acceleration, velocity, stress are known as a function of time.
β’ Ii) Non-deterministic analysis : Time variation of of vibration is not completely known.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 12
Unit I β Theory of Vibrations
β’ Comparison of static loading and dynamic loading:
β i) In static problem: Load is constant with time.
β In dynamic problem: Loading and its response varies with time.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 13
W
W (t)
Ex: Weight of a bridge span on bridge pilings.
Ex: A truck moving across the same bridge span exerts a dynamic load on the pilings.
Inertia forces
Unit I β Theory of Vibrations
β’ Comparison of static loading and dynamic loading:
β ii) In static problem: Response due to static loading is displacement only.
β In dynamic problem: Response due to dynamic loading is displacement, velocity and acceleration.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 14
W
W (t)
y
Unit I β Theory of Vibrations
β’ Comparison of static loading and dynamic loading:
β iii) In static problem: Solution of static problem is only one.
β In dynamic problem: Solutions of dynamic problem are infinite and are time dependent.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 15
W
W (t)
y
Dynamic analysis is more complex and time consuming than static analysis
Unit I β Theory of Vibrations
β’ Comparison of static loading and dynamic loading:
β iv) In static problem: Response calculation is done by static equilibrium.
β In dynamic problem: Response not only depends on load but also depend on inertia forces which oppose the accelerations producing them.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 16
W
W (t)
y
Inertia forces are most important characteristics of a structural dynamic problem.
Unit I β Theory of Vibrations
β’ Causes of dynamic effects: β Natural and manmade sources may influence the dynamic effect in
the structure.
β The most common causes are as follows: β’ i) Initial conditions: Initial conditions such as velocity and displacement
produce dynamic effect in the system.
Ex: Consider a lift moving up or down with an initial velocity . When the lift is suddenly stopped , the cabin begin to vibrate up and down since it posses initial velocity.
β’ ii) Applied forces: Some times vibration in the system is produced due to application of external forces.
Ex: i) A building subjected to bomb blast or wind forces
ii) Machine foundation.
β’ iii) Support motions : Structures are often subjected to vibration due to influence of support motions.
Ex: Earthquake motion.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 17
Unit I β Theory of Vibrations
β’ Degrees of freedom: β Number of coordinates necessary to specify the position or
geometry of mass point at any instant during its vibration.
β All real structures possess infinite number of dynamic degrees of freedom. Hence infinite number of coordinates are necessary to specify the position of the structure completely at any instant of time.
β Each degree of freedom is having corresponding natural frequency. Therefore, a structure possesses as many natural frequencies as it has the degrees of freedom.
β For each natural frequency, the structure has its own way of vibration.
β The vibrating shape is known as characteristic shape or mode of vibration.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 18
Unit I β Theory of Vibrations
β’ Degrees of freedom: β Consider a block as shown in figure that is free to move in 3-
dimensional space, which may move without rotation in each of the three directions X, Y, Z. These are called the three degrees of translation.
β The block may also rotate about its own axes, these are called the three degrees of rotation.
β Thus to define the position of the block in space, we need to define six coordinates, that is three for translation and three for rotation.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 19
Unit I β Theory of Vibrations
β’ Degrees of freedom:
β Depending on the independent coordinates required to describe the motion, the vibratory system is divided into the following categories.
(i) Single degree of freedom system (SDOF system)
(ii) Multiple degree of freedom system
(iii) Continuous system.
β If a single coordinate is sufficient to define
the position or geometry of the mass of the
system at any instant of time is called single or one degree of
freedom system.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 20
Unit I β Theory of Vibrations
β’ Degrees of freedom:
β Example for SDOF:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 21
x Spring β mass system
Building frame
x k1
m
Unit I β Theory of Vibrations
β’ Degrees of freedom:
β If more than one independent coordinate is required to completely specify the position or geometry of different masses of the system at any instant of time, is called multiple degrees of freedom system.
β Example for MDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 22
x1 k1 k2
m1 m2
x2
Unit I β Theory of Vibrations
β’ Degrees of freedom:
β If the mass of a system may be considered to be distributed over its entire length as shown in figure, in which the mass is considered to have infinite degrees of freedom, it is referred to as a continuous system. It is also known as distributed system.
β Example for continuous system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 23
x3 x2 x1 x3
x3
Cantilver beam with infinite number of degrees of freedom
Unit I β Theory of Vibrations β’ Mathematical modelling of an SDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 24
Portal frame
β’ To understand the dynamic behaviour of structure, it is necessary to develop their models under dynamic loads such as earthqukes, wind, blasts etc.
x F(t)
β’ Assumptions to develop mathematical model: Total mass is assumed to act at slab level, since mass of columns are
less and ignored. The beam/slab is assumed as infinitely rigid, so that the stiffness of the
structure is provided by the columns, i.e. flexibility of slab/beam is ignored.
Since beams are built monolithically within the columns, the beam column joint can be assumed as rigid as without any rotations at joint.
K
Unit I β Theory of Vibrations β’ Mathematical modelling of an SDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 25
Portal frame
β’ The possibility of lateral displacement is due to rigid beam/slab only.
β’ The model resulting from the above mentioned assumptions is called as shear building model.
x F(t)
x
k
m
FBD
F(t)
Spring force, Fs =kx
Damping force, πΉπ· = ππ₯
Inertia force, πΉπ = ππ₯ m
K
c
Unit I β Theory of Vibrations β’ Mathematical modelling of an SDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 26
β’ βmβ=mass of slab and beam. Energy is stored by mass m in the form of kinetic energy.
β’ βkβ represents combined stiffness of two columns for lateral deformation that is elastic restoring force and it stores the potential energy (internal strain energy ) due to columns.
β’ Dashpot having damping coefficient βcβ represents the energy dissipation, i.e. frictional characteristics and energy losses of the frame.
β’ An execution force F(t) representing the external lateral force.
x F(t)
F(t)
Spring force, Fs =kx
Damping force, πΉπ· = ππ₯
Inertia force, πΉπ = ππ₯
FBD
m
Unit I β Theory of Vibrations β’ Mathematical modelling of an SDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 27
β’ Passive (inactive) elements = mass, spring, damper β’ Active element = excitation element, F(t) β’ Since the above dynamic system is divided into independent
discrete elements, this model is known as lumped parameter model.
x F(t)
F(t)
Spring force, Fs =kx
Damping force, πΉπ· = ππ₯
Inertia force, πΉπ = ππ₯
FBD
m
Unit I β Theory of Vibrations β’ Mathematical modelling of an SDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 28
β’ The elements to determine the dynamic behaviour: β’ i) the inertia force, πΉπ = ππ₯ β’ Ii) the restoring force or spring force, Fs =kx
β’ iii) the damping force, πΉπ· = ππ₯ β’ iv) the exciting force, F(t)
β’ Considering the equilibrium of all forces in X- direction, the
govrning equation of motion for the SDOF is,
ππ + ππ + π€π± = π (π)
x F(t)
F(t)
Spring force, Fs =kx
Damping force, πΉπ· = ππ₯
Inertia force, πΉπ = ππ₯
FBD
m
Unit I β Theory of Vibrations β’ Free vibration of undamped SDOF system:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 29
x
k m
Spring force, Fs =kx
Inertia force, πΉπ = ππ₯
FBD
m
β’ Considering the equilibrium of all forces in X- direction, the governing equation of motion for the SDOF is,
β’ πΉπ + πΉπ = 0
ππ + π€π± = π
Unit I β Theory of Vibrations β’ Derivation of equation of motion:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 30
Differential equation describing the motion is known as equation of motion. Methods to derive the equation of motion: i) Simple Harmonic Motion method ii) Newtonβs method iii) Energy method iv) Rayleighβs method v) DβAlembertβs principle
Unit I β Theory of Vibrations β’ i). Simple Harmonic motion method:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 31
β’ If the acceleration of a particle in a rectilinear motion is always proportional to the distance of the particle from a fixed point on the path and is directed towards the fixed point, then the particle is said to be in SHM.
β’ SHM is the simplest form of periodic motion.
β’ In differential equation form, SHM is represented as π₯ β βπ₯ ββ β(1)
Where x is the rectilinear displacement and π₯ is acceleration (π2π₯
ππ‘2)
Unit I β Theory of Vibrations β’ i). Simple Harmonic motion method:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 32
β’ π₯ β βπ₯ ββ β(1) β’ The negative sign in Eq.(1) indicates the direction of motion of a
particle towards a fixed point which is opposite to the direction of displacement.
β’ Let the constant proportionality be ππ2 which is an unknown
parameter.
β’ Now Eq.(1) can be rewritten as, π₯ = βππ2π₯
π₯ + ππ2π₯ = 0 ββ β 2
This is known as equation of motion and is second order linear differential equation.
β’ The constant ππ is yet to be determined by the analysis.
Unit I β Theory of Vibrations β’ ii). Newtonβs second law of motion:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 33
β’ The equation of motion is just another form of Newtonβs second law of motion.
β’ The rate of change of momentum is proportional to the impressed forces and takes place in the direction in which the force acts.
Unit I β Theory of Vibrations β’ ii). Newtonβs second law of motion:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 34
β’ Consider a spring β mass system of figure which is assumed to move only along the vertical direction. It has only one degree of freedom, because its motion is described by a single coordinate x.
m m
W
k β
β= ππ‘ππ‘ππ πππππππ‘πππ
Static equilibrium position
m
W
k (β + π₯)
π₯ π₯ m
x
π₯
Unit I β Theory of Vibrations β’ ii). Newtonβs second law of motion:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 35
β’ A massless spring of constant stiffness k is shown in Figure.
β’ π =π
β, β΄ π = πβ
From the equilibrium position , the load W is pulled down a little by some force and then pulling force is removed.
m
π₯
m
W
k β β= ππ‘ππ‘ππ πππππππ‘πππ
m
x
Static equilibrium position
m
W
k (β + π₯)
π₯ π₯
Unit I β Theory of Vibrations β’ ii). Newtonβs second law of motion:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 36
β’ The load W will continue to execute vibration up and down which is called free vibration.
β’ Restoring force in X- direction= π β π(β + π₯) = πβ β πβ β ππ₯ = βππ₯
m
π₯
m
W
k β β= ππ‘ππ‘ππ πππππππ‘πππ
m
x
Static equilibrium position
m
W
k (β + π₯)
π₯ π₯
Unit I β Theory of Vibrations β’ ii). Newtonβs second law of motion:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 37
β’ According to Newtonβs second law, ππ₯ = βkx ππ₯ + kx = 0
π₯ +π
ππ₯ = 0 ββ β 3
Compared with Eq.(2) i.e., π₯ + ππ2π₯ = 0 ββ β 2
m
π₯
m
W
k β β= ππ‘ππ‘ππ πππππππ‘πππ
m
x
Static equilibrium position
m
W
k (β + π₯)
π₯ π₯
ππ2 =
π
π
β΄ ππ =π
π
Unit I β Theory of Vibrations β’ iii). Energy method:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 38
β’ Assumption: System is to be conservative one. β’ Conservative system: Total sum of energy is constant at all
time.
β’ For an undamped system: since there is no friction or damping force, the total energy of the system is partly potential and partly kinetic.
β’ β΄ πΎ. πΈ + π. πΈ.= ππππ π‘πππ‘.
β’ The time rate of change of total energy will be zero.
β’π
ππ‘π. πΈ.+π. πΈ. = 0
Unit I β Theory of Vibrations β’ iii). Energy method:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 39
β’ π. πΈ.=1
2ππ£2 =
1
2ππ₯ 2
β’ π. πΈ.=1
2ππ₯2
β’π
ππ‘
1
2ππ₯ 2 +
1
2ππ₯2 = 0
β’1
2π2π₯ π₯ +
1
2π2π₯π₯ = 0
β’ ππ + ππ = π
Unit I β Theory of Vibrations β’ iv). Rayleighβs method:
β’ Assumptions:
β’ (1) Maximum K.E. at the equilibrium position is equal to the maximum potential energy at the extreme position.
β’ (2). The motion is assumed to be SHM, then π₯ = π΄ sinπππ‘
β’ Where x is the displacement of the system from its mean position after time t.
β’ A is the maximum displacement of the system from equilibrium position to extreme position.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 40
Unit I β Theory of Vibrations β’ iv). Rayleighβs method:
π₯ = π΄ sinπππ‘
β’ π₯ is maximum when sinπππ‘=1
β’ π₯πππ₯ = π΄
β’ π₯ = πππ΄ cosπ ππ‘
β’ Velocity π₯ ππ πππ₯πππ’π π€βππ cosπππ‘ = 1
β’ π₯ πππ₯ = πππ΄
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 41
Unit I β Theory of Vibrations β’ iv). Rayleighβs method:
So maximum K.E. at the equilibrium position=1
2ππ₯ πππ₯
2
=1
2π πππ΄
2
Maximum P.E. at the extreme position=1
2ππ₯πππ₯
2
=1
2π π΄ 2
1
2π πππ΄
2 ==1
2π π΄ 2
ππ2 =
π
π
β΄ ππ =π
π
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 42
Unit I β Theory of Vibrations β’ iv). DβAlembertβs method:
To find the solution of a dynamic problem by using the methods of statics.
According to Newtonβs second law, πΉ = ππ πΉ βππ = 0
This is in the form of an equation of motion of force equilibrium in which sum of a number of force terms equals zero.
Hence, if an imaginary force which is equal to βmaβ were applied to the system in the direction opposite to the acceleration, the system could then be considered to be in equilibrium under the action of real force F and the imaginary force βmaβ.
The imaginary force βmaβ is known as inertia force and the position of equilibrium is called dynamic equilibrium.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 43
Unit I β Theory of Vibrations β’ iv). DβAlembertβs method:
DβAlemberts principle states that βa system may be in dynamic equilibrium by adding to the external forces, an imaginary force, which is commonly known as the inertia forceβ.
According to the principle, the transformation of a problem in dynamics may be reduced to one in statics.
Consider a spring-mass system in the following Figure.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 44
x
k m
Spring β mass system Dynamic equilibrium
Spring force, Fs =kx
Inertia force, πΉπ = ππ₯ m
Unit I β Theory of Vibrations β’ iv). DβAlembertβs method:
β Using DβAlembertβs principle, to bring the body to a dynamic equilibrium position, the inertia force βππ₯ is to be added in the direction opposite to the direction of motion.
β Equilibrium equation is πΉπ₯ = 0 βππ₯ β ππ₯ = 0
β ππ₯ + ππ₯ = 0 ππ₯ + ππ₯ = 0
π₯ +π
ππ₯ = 0
ππ2 =
π
π
ππ =π
π
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 45
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β The governing differential equation of motion is
ππ₯ + ππ₯ = 0
It is in the form of homogeneous second order linear equation.
There are five different solutions for the above equation of motion.
1. π₯ = π΄ cosπππ‘
2. π₯ = π΅ sinπππ‘
3. π₯ = π΄ cosπππ‘ + π΅ sinπππ‘
4. π₯ = π΄ sin(πππ‘ + β )
5. π₯ = π΄ cos(πππ‘ + β )
Where A and B are constants depending on their initial condition of the motion.
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 46
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.1: π₯ = π΄ cosπππ‘ ββ β(1)
To determine the constant A, let us use the initial condition by assuming that at time t=0, the displacement π₯ = π₯0.
Substituting this in the above equation (1), we get π₯0 = π΄ cos(ππ Γ 0)
β΄ π₯0 = π΄
Hence the solution, π = ππ ππ¨π¬πππ
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 47
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.2: π₯ = B sinπππ‘ βββ β(2)
To determine the constant B, let us use the initial condition by assuming that
(i) at time t=0, the displacement π₯ = π₯0.
(ii) At time t=0, π₯ = π₯0
Differentiating equation (2) with respect to time, π₯ = π΅ππ cosπππ‘
Applying initial conditions, π₯0 = π΅ ππ.
π΅ =π₯ 0ππ
Substituting in equation (2), π =π π
πππ¬π’π§πππ
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 48
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.3: π₯ = π΄ cosπππ‘ +B sinπππ‘ βββ β(3)
The superposition of the above two solutions is also a solution.
The general solution for this second order differential equation is
π = ππ ππ¨π¬πππ +π πππ
π¬π’π§πππ
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 49
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.4: π₯ = π΄ sin(πππ‘ + β ) βββ β(4)
By expanding sine term π₯ = π΄ sinπππ‘ cos β + π΄ cosπππ‘ sin β ββ β(4a)
But the general solution is
π = ππ ππ¨π¬πππ +π πππ
π¬π’π§πππ
By comparing Eq.(4a) with general solution, i.e. comparing coefficient of cosπππ‘,
π₯0 = π΄ sinβ β β(5)
Comparing coefficient of sinπππ‘, π πππ
= A cos β ββ β(6)
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 50
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.4: π₯0 = π΄ sin β β β(5) π πππ
= A cos β ββ β(6)
Squaring and adding Eq.(5) and Eq.(6),
π΄2π ππ2β + π΄2πππ 2β = π₯02 +
π₯ 02
ππ2
π΄ = π₯02 +
π₯ 02
ππ2
Dividing Eq.(5) and Eq.(6), π΄ sin β
π΄ cos β =
π₯0π πππ
Hence the phase angle, β = πππ§βπππππ
π π
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 51
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.5: π₯ = π΄ cos(πππ‘ + β ) ββ β(7)
By expanding cosine term, we get π₯ = π΄ cosπππ‘ cos β + π΄ sinπππ‘ sin β ββ β(7π)
But the general solution is
π = ππ ππ¨π¬πππ +π πππ
π¬π’π§πππ
By comparing Eq.(7a) with general solution, we get π₯0 = π΄ cosβ β β(8) π πππ
= A sin β ββ β(9)
By squaring and adding Eqs. (8) and (9), we get
π₯02 +
π₯ 02
ππ2 = π΄2πππ 2β + π΄2π ππ2β
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 52
Unit I β Theory of Vibrations β’ Solution of the equation of motion:
β Solution No.5:
π₯02 +
π₯ 02
ππ2= π΄2πππ 2β + π΄2π ππ2β
π΄ = π₯02 +
π₯ 02
ππ2
Dividing Eq.(9) and Eq.(8), π΄ sin β
π΄ cos β =
π πππ
π₯0
Phase angle, β = πππ§βππ π
ππππ
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 53
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 54
Introduction: β’ Without damping force or frictional force the system vibrates
indefinitely with a constant amplitude at its natural frequency. β’ But in reality the vibration without decreasing amplitude is never
realized. β’ Frictional forces (or) damping forces are always present in any
physical system while undergoing motion.
β’ The presence of damping forces or frictional forces, form a mechanism through which the mechanical energy of the system (kinetic energy or potential energy) is transformed to other form of energy such as heat energy.
β’ This energy transformation mechanism is called a dissipation energy. This is quite complex in nature.
Damped free vibration of SDOF system:
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 55
What is damping? β’ A phenomenon in which the energy of the system is gradually
reduced or the amplitude of the vibration goes on decreasing and finally the vibration of the system is completely eliminated and the system is brought to rest is known as damping.
β’ The decreasing rate of amplitude depends upon the amount of damping.
β’ The damping is useful to control the amplitude of vibration.
Damped free vibration of SDOF system:
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 56
Types (or) Nature of damping : Mainly 5types of damping β’ 1. Viscous damping
β’ 2. Coulomb damping
β’ 3. Structural damping
β’ 4. Active damping (or) Negative damping
β’ 5. Passive damping
Damped free vibration of SDOF system:
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 57
1. Viscous Damping: β’ When a system is made to vibrate in a
surrounding viscous medium that is under the control of highly viscous fluid, the damping is called viscous damping.
β’ This type of damping is achieved by means of device called hydraulic dashpot.
β’ The main components of viscous damper or dashpot are cylinder, piston and viscous fluid as shown in Figure.
β’ Fluid mechanics concepts are to be used here.
Damped free vibration of SDOF system:
V
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 58
1. Viscous Damping: β’ Let us consider that the two plates are
separated by fluid film of thickness t as shown in Figure.
β’ The upper plate is allowed to move parallel to the fixed plate with a velocity π₯ .
β’ The force βFβ required for maintaining this velocity π₯ of the plate is given by
πΉ =ππ΄
π‘π₯ ββ β(1)
= ππ₯ c=damping coefficient (N-s/m)
Damped free vibration of SDOF system:
V
F
t
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 59
1. Viscous Damping: β’ Viscous damping is a method of converting
mechanical vibrational energy of a body into heat energy, in which a piston is attached to the body and is arranged to move through liquid in a cylinder that is attached to a support.
β’ Shock absorber is the best example of the viscous damping.
β’ Viscous damping is largely used for system modeling since it is linear.
Damped free vibration of SDOF system:
V
F
t
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 60
1. Equation of motion for viscous damping:
Damped free vibration of SDOF system:
x
k
m
c
Spring force, Fs =kx
Damping force, πΉπ· = ππ₯
Inertia force, πΉπ = ππ₯ m
Viscous damping oscillator F. B.D.
From FBD, the governing differential equation of motion is,
ππ₯ + ππ₯ + kx = 0 βββ(2) Assuming the solution may be in the form of
π₯ = πΞ»π‘ Where Ξ» is a constant to be determined. This exponential function leads to algebraic equation instead of a differential equation.
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 61
1. Equation of motion for viscous damping:
Damped free vibration of SDOF system:
x
k
m
c
Spring force, Fs =kx
Damping force, πΉπ· = ππ₯
Inertia force, πΉπ = ππ₯ m
π₯ = πΞ»π‘
π₯ = Ξ»πΞ»π‘
π₯ = Ξ»2 πΞ»π‘ Substituting the values of π₯, π₯ , π₯ in equation (2) we get,
πΞ»2 πΞ»π‘ + c Ξ»πΞ»π‘ + kπΞ»π‘ = 0
πΞ»2 + cΞ»+k πΞ»π‘ = 0
The non-trivial solution is πΞ»2 + cΞ»+k=0
Ξ»2 +π
πΞ»+
π
π= 0
ππ₯ + ππ₯ + kx = 0 ββ β(2)
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 62
β’ It was named because Charles Augustin de Coulomb carried on research in mechanics.
β’ In this damping energy is absorbed constantly through sliding friction, which is developed by relative motion of the two surfaces that slide against each other.
β’ Coulomb damping absorbs energy with friction, which converts that kinetic energy into thermal energy or heat.
β’ Static and kinetic friction occur in a vibrating system undergoing Coulomb damping.
β’ Static friction occurs when two bodies are stationary or undergoing no relative motion.
Frictional force, πΉπ = ππ π ΞΌs = coefficient of static friction.
Coulomb damping:
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 63
β’ Kinetic friction occurs when the two bodies are undergoing relative motion and they are sliding against each other.
Frictional force, πΉπ = πππ ΞΌk = coefficient of dynamic friction
Coulomb damping:
EOM for left to right motion, ππ₯ = βππ₯ β πΉ, πππ π₯ > 0
EOM for right to left motion, ππ₯ = βππ₯ + πΉ, πππ π₯ < 0
ππππ’π‘πππ πππ ππππ‘ πππ‘πππ, π₯ = π΄ cosπππ‘ + B sinπππ‘ + πΉ
π
ππππ’π‘πππ πππ πππβπ‘ πππ‘πππ, π₯ = πΆ cosπππ‘ + D sinπππ‘ β πΉ
π
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 64
β’ Let us consider the movement of the mass to the left in a Coulomb damping system as shown in figure.
Coulomb damping for first half cycle (π β€ π β€ π»/π):
β’ The governing differential equation is ππ₯ = βππ₯ + πΉ
ππ₯ + ππ₯ = πΉ βββ(1) π₯
ππ2+ π₯ =
πΉ
π
The solution of the above equation can be written as
π₯ = π΄ cosπππ‘ + B sinπππ‘ + πΉ
π
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 65
Coulomb damping for first half cycle (π β€ π β€ π»/π):
π₯ = π΄ cosπππ‘ + B sinπππ‘ + πΉ
πββ β(2)
π₯ = π₯π + π₯π
Where π₯π = π΄ cosπππ‘ + B sinπππ‘ = complimentary sulution
π₯π =πΉ
π= Partcular integral
Where ππ =π
π
Let us assume the initial condition to determine the constants A and B (i) At t=0; π₯=π₯0 (ii) At t=0; π₯ = 0 π₯ = βπ΄ππ sinπππ‘ + Bππ cosπππ‘
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 66
Coulomb damping for first half cycle (π β€ π β€ π»/π):
π₯ = βπ΄ππ sinπππ‘ + Bππ cosπππ‘ ββ β(3) Applying initial condition (ii) i.e., At t=0; π₯ = 0 in the above equation
0 = Bππ Since ππ β 0 β΄ π΅ = 0
Applying initial condition (i) i.e., at t=0;π₯ = π₯0
π₯0 = A +πΉ
π
π΄ = π₯0 βπΉ
π
Hence Equation (2) can be written as
π₯ = (π₯0βπΉ
π) cosπππ‘ +
πΉ
π, 0 β€ π‘ β€ π/2
This solution holds good for half cycle only.
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 67
Coulomb damping for first half cycle (π β€ π β€ π»/π):
π₯ = (π₯0βπΉ
π) cosπππ‘ +
πΉ
π, 0 β€ π‘ β€ π/2 ββ β(4)
We know that, π =2π
ππ
For half cycle, π =π
ππ
When t=π
ππ, half cycle is completed. So displacement for half the
cycle can be obtained from Eq.(4).
π‘ =π
ππ
πππ‘ = π Substituting the value of πππ‘ in Eq.(4),
π₯ = (π₯0βπΉ
π)cosΟ +
πΉ
π
π₯ = (π₯0βπΉ
π)(β1) +
πΉ
π
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 68
Coulomb damping for first half cycle (π β€ π β€ π»/π):
π₯ = (π₯0βπΉ
π)(β1) +
πΉ
π
π = β ππ β ππ
π
This is the amplitude for the left extreme of the body. From this equation it is clear that the initial displacement is reduced by 2F/k.
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 69
β’ Let us consider the movement of the mass to the right in a Coulomb damping system as shown in figure.
Coulomb damping for second half cycle (π»/π β€ π β€ π»):
β’ The governing differential equation is ππ₯ = βππ₯ β πΉ
ππ₯ + ππ₯ = βπΉ βββ(5) π₯
ππ2+ π₯ = β
πΉ
π
The solution of the above equation can be written as
π₯ = πΆ cosπππ‘ + D sinπππ‘ β πΉ
π
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 70
Coulomb damping for second half cycle (π»/π β€ π β€ π»):
π₯ = πΆ cosπππ‘ + D sinπππ‘ β πΉ
πββ β(6)
π₯ = π₯π + π₯π
Where π₯π = πΆ cosπππ‘ + D sinπππ‘ = complimentary sulution
π₯π = βπΉ
π= Partcular integral
Where ππ =π
π
Let us assume the initial condition to determine the constants C and D (i) At t=0; π₯=π₯0 (ii) At t=0; π₯ = 0 π₯ = βπΆππ sinπππ‘ + Dππ cosπππ‘
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 71
Coulomb damping for second half cycle (π»/π β€ π β€ π»):
π₯ = βπΆππ sinπππ‘ + Dππ cosπππ‘ ββ β(7) Applying initial condition (ii) i.e., At t=0; π₯ = 0 in the above equation
0 = Dππ Since ππ β 0 β΄ π· = 0
Applying initial condition (i) i.e., at t=π
ππ;π₯ = βπ₯0 + 2
πΉ
π
πΆ = π₯0 β 3πΉ
π
Hence Equation (2) can be written as
π₯ = (π₯0β3πΉ
π) cosπππ‘ β
πΉ
π, π/2 β€ π‘ β€ π
This solution holds good for second half cycle only.
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 72
Coulomb damping for second half cycle (π»/π β€ π β€ π»):
π₯ = (π₯0β3πΉ
π) cosπππ‘ β
πΉ
π, π/2 β€ π‘ β€ π ββ β(8)
We know that, π =2π
ππ
When t=2π
ππ, second half cycle is completed. So displacement for the
second half the cycle can be obtained from Eq.(8).
π‘ =2π
ππ
πππ‘ varies from π π‘π 2π Substituting the value of πππ‘ in Eq.(4),
π₯ = (π₯0β3πΉ
π)cos2Ο β
πΉ
π
π = (ππβππ
π)
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 73
Coulomb damping for second half cycle (π»/π β€ π β€ π»):
π = β ππ β ππ
π
π = ππ β ππ
π
In the first half cycle the initial displacement is reduced by 2F/k. In the second half cycle when the body moves to the right, the initial displacement will be reduced by 2F/k. So in one complete cycle, the amplitude reduces by 4F/k. But the natural frequency of the system remains unchanged in coulomb damping.
Unit I β Theory of Vibrations
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 74
Coulomb damping for second half cycle (π»/π β€ π β€ π»):
Unit I β Theory of Vibrations β’ Example 1:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 75
β’ A cantilever beam AB of length L is attached to a spring k and mass M as shown in Figure. (i) form the equation of motion and (ii) Find an expression for the frequency of motion.
m
k
L
Unit I β Theory of Vibrations β’ Solution:
β This stiffness is parallel to π.
β Equivalent spring stiffness, ππ = ππ + π
=3πΈπΌ
πΏ3+ k
=3πΈπΌ + ππΏ3
πΏ3
The differential equation of motion is, ππ₯ = βπππ₯
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 76
β’ Stiffness due to applied mass M is
ππ =π
β=
3πΈπΌ
πΏ3
m
k
L
Unit I β Theory of Vibrations β’ Solution:
ππ₯ = βπππ₯ ππ₯ + πππ₯ = 0
ππ₯ +3πΈπΌ + ππΏ3
πΏ3π₯ = 0
π +ππ¬π° + ππ³π
ππ³ππ = π
The frequency of vibration, π =1
2π
ππ
π
β΄ π =π
ππ
ππ³π + ππ¬π°
ππ³π
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 77
m
k
L
Unit I β Theory of Vibrations β’ Example 2:
Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 78
β’ Find the natural frequency of the system as shown in Figure. Take
π1 = π2 = 2000 π/π, π3 = 3000π
π and m= 10 kg.
π3
m=10 kg
π1 π2