ce 6701 structural dynamics and earthquake engineering · pdf filece 6701 structural dynamics...

CE 6701 Structural Dynamics and Earthquake Engineering

Dr. P. Venkateswara Rao Associate Professor

Dept. of Civil Engineering

SVCE, Sriperumbudur

Unit I – Theory of Vibrations

• Difference between static loading and dynamic loading

• Degree of freedom

• Idealisation of structure as single degree of freedom system

• Formulation of Equations of motion of SDOF system

• D‟Alemberts principles

• Effect of damping

• Free and forced vibration of damped and undamped structures

• Response to harmonic and periodic forces.

Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 2

Unit II – Multiple Degree of Freedom System

• Two degree of freedom system

• Modes of vibrations

• Formulation of equations of motion of multi degree of freedom (MDOF) system

• Eigen values and Eigen vectors

• Response to free and forced vibrations

• Damped and undamped MDOF system

• Modal superposition methods.


Unit III – Elements of Seismology

• Elements of Engineering Seismology

• Causes of Earthquake

• Plate Tectonic theory

• Elastic rebound Theory

• Characteristic of earthquake

• Estimation of earthquake parameters

• Magnitude and intensity of earthquakes

• Spectral Acceleration.


Unit IV – Response of Structures to Earthquake

• Effect of earthquake on different type of structures

• Behaviour of Reinforced Cement Concrete, Steel and Prestressed Concrete Structure under earthquake loading

• Pinching effect

• Bouchinger Effects

• Evaluation of earthquake forces as per IS:1893 – 2002

• Response Spectra

• Lessons learnt from past earthquakes.


Unit V – Design Methodology

• Causes of damage

• Planning considerations / Architectural concepts as per IS:4326 – 1993

• Guidelines for Earthquake resistant design

• Earthquake resistant design for masonry and Reinforced Cement Concrete buildings

• Later load analysis

• Design and detailing as per IS:13920 – 1993.


References • 1. Chopra, A.K., “Dynamics of Structures – Theory and

Applications to Earthquake Engineering”, 4th Edition, Pearson Education, 2011.

• 2. Agarwal. P and Shrikhande. M., "Earthquake Resistant Design of Structures", Prentice Hall of India Pvt. Ltd. 2007

• 3. Paz, M. and Leigh.W. “Structural Dynamics – Theory & Computation”, 4th Edition, CBS Publishers & Distributors, Shahdara, Delhi, 2006.

• 4. Damodarasamy, S.R. and Kavitha, S. “Basics of Structural dynamics and Aseismic design”, PHI Learning Pvt. Ltd., 2012


References



• Vibration:

– 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.

– Vibration in structural systems may result from environmental sources such as wind, earthquakes and waterways.

– Earthquakes are most important due to enormous potential for damage to structures and loss of life.

– On an average every year around 10, 000 people die worldwide due to earthquakes.



• Vibration:

– Study of repetitive motion of objects relative to a stationary frame of reference or equilibrium position.

– Vibrations can occur in many directions and results in interaction of many objects.

– Motion of vibrating system is governed by the law of mechanics, and in particular by Newton’s second law of motion (F=ma).



• Basic concepts of Vibration: – Bodies having mass and elasticity are capable to vibrate.

– 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.

– At equilibrium position, whole of the elastic energy is converted into kinetic energy and the body continuous to move in in the opposite direction.

– Whole K.E. is converted into elastic or strain energy and inturn body returns to equilibrium position.

– This way, vibration motion is repeated continuously and interchange of energy takes place.

– And hence, any motion repeats itself after an interval of time is called vibration or oscillation.



• Dynamic loading:

– Dynamics: Study of forces and motions with time dependency.

– Dynamic load: Load magnitude, direction and position changes with time.

– Structural response to dynamic loading can be done by two methods:

• i) Deterministic analysis : Structural response i.e. displacement, acceleration, velocity, stress are known as a function of time.

• Ii) Non-deterministic analysis : Time variation of of vibration is not completely known.



• Comparison of static loading and dynamic loading:

– i) In static problem: Load is constant with time.

– In dynamic problem: Loading and its response varies with time.


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



– ii) In static problem: Response due to static loading is displacement only.

– In dynamic problem: Response due to dynamic loading is displacement, velocity and acceleration.


W

W (t)

y



– iii) In static problem: Solution of static problem is only one.

– In dynamic problem: Solutions of dynamic problem are infinite and are time dependent.


W

W (t)

y

Dynamic analysis is more complex and time consuming than static analysis



– iv) In static problem: Response calculation is done by static equilibrium.

– In dynamic problem: Response not only depends on load but also depend on inertia forces which oppose the accelerations producing them.


W

W (t)

y

Inertia forces are most important characteristics of a structural dynamic problem.


• Causes of dynamic effects: – Natural and manmade sources may influence the dynamic effect in

the structure.

– The most common causes are as follows: • 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.

• 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.

• iii) Support motions : Structures are often subjected to vibration due to influence of support motions.

Ex: Earthquake motion.



• Degrees of freedom: – Number of coordinates necessary to specify the position or

geometry of mass point at any instant during its vibration.

– 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.

– Each degree of freedom is having corresponding natural frequency. Therefore, a structure possesses as many natural frequencies as it has the degrees of freedom.

– For each natural frequency, the structure has its own way of vibration.

– The vibrating shape is known as characteristic shape or mode of vibration.



• Degrees of freedom: – 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.

– The block may also rotate about its own axes, these are called the three degrees of rotation.

– 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.



• Degrees of freedom:

– 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.

– 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.




– Example for SDOF:


x Spring – mass system

Building frame

x k1

m



– 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.

– Example for MDOF system:


x1 k1 k2

m1 m2

x2



– 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.

– Example for continuous system:


x3 x2 x1 x3

x3

Cantilver beam with infinite number of degrees of freedom

Unit I – Theory of Vibrations • Mathematical modelling of an SDOF system:


Portal frame

• 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)

• 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



Portal frame

• The possibility of lateral displacement is due to rigid beam/slab only.

• 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, 𝐹𝐷 = 𝑐𝑥

Inertia force, 𝐹𝑖 = 𝑚𝑥 m

K

c



• ‘m’=mass of slab and beam. Energy is stored by mass m in the form of kinetic energy.

• ‘k’ 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.

• Dashpot having damping coefficient ‘c’ represents the energy dissipation, i.e. frictional characteristics and energy losses of the frame.

• An execution force F(t) representing the external lateral force.

x F(t)

F(t)



Inertia force, 𝐹𝑖 = 𝑚𝑥

FBD

m



• Passive (inactive) elements = mass, spring, damper • Active element = excitation element, F(t) • Since the above dynamic system is divided into independent

discrete elements, this model is known as lumped parameter model.

x F(t)

F(t)




FBD

m



• The elements to determine the dynamic behaviour: • i) the inertia force, 𝐹𝑖 = 𝑚𝑥 • Ii) the restoring force or spring force, Fs =kx

• iii) the damping force, 𝐹𝐷 = 𝑐𝑥 • iv) the exciting force, F(t)

• Considering the equilibrium of all forces in X- direction, the

govrning equation of motion for the SDOF is,

𝒎𝒙 + 𝒄𝒙 + 𝐤𝐱 = 𝐅(𝐭)

x F(t)

F(t)




FBD

m

Unit I – Theory of Vibrations • Free vibration of undamped SDOF system:


x

k m



FBD

m

• Considering the equilibrium of all forces in X- direction, the governing equation of motion for the SDOF is,

• 𝐹𝑖 + 𝐹𝑠 = 0

𝒎𝒙 + 𝐤𝐱 = 𝟎

Unit I – Theory of Vibrations • Derivation of equation of motion:


Differential equation describing the motion is known as equation of motion. Methods to derive the equation of motion: i) Simple Harmonic Motion method ii) Newton’s method iii) Energy method iv) Rayleigh’s method v) D’Alembert’s principle

Unit I – Theory of Vibrations • i). Simple Harmonic motion method:


• 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.

• SHM is the simplest form of periodic motion.

• In differential equation form, SHM is represented as 𝑥 ∝ −𝑥 −− −(1)

Where x is the rectilinear displacement and 𝑥 is acceleration (𝑑2𝑥

𝑑𝑡2)

Unit I – Theory of Vibrations • i). Simple Harmonic motion method:


• 𝑥 ∝ −𝑥 −− −(1) • 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.

• Let the constant proportionality be 𝜔𝑛2 which is an unknown

parameter.

• Now Eq.(1) can be rewritten as, 𝑥 = −𝜔𝑛2𝑥

𝑥 + 𝜔𝑛2𝑥 = 0 −− − 2

This is known as equation of motion and is second order linear differential equation.

• The constant 𝜔𝑛 is yet to be determined by the analysis.

Unit I – Theory of Vibrations • ii). Newton’s second law of motion:


• The equation of motion is just another form of Newton’s second law of motion.

• The rate of change of momentum is proportional to the impressed forces and takes place in the direction in which the force acts.



• Consider a spring – 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 ∆

∆= 𝑆𝑡𝑎𝑡𝑖𝑐 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛

Static equilibrium position

m

W

k (∆ + 𝑥)

𝑥 𝑥 m

x

𝑥



• A massless spring of constant stiffness k is shown in Figure.

• 𝑘 =𝑊

∆, ∴ 𝑊 = 𝑘∆

From the equilibrium position , the load W is pulled down a little by some force and then pulling force is removed.

m

𝑥

m

W

k ∆ ∆= 𝑆𝑡𝑎𝑡𝑖𝑐 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛

m

x


m

W

k (∆ + 𝑥)

𝑥 𝑥



• The load W will continue to execute vibration up and down which is called free vibration.

• Restoring force in X- direction= 𝑊 − 𝑘(∆ + 𝑥) = 𝑘∆ − 𝑘∆ − 𝑘𝑥 = −𝑘𝑥

m

𝑥

m

W


m

x


m

W

k (∆ + 𝑥)

𝑥 𝑥



• According to Newton’s second law, 𝑚𝑥 = −kx 𝑚𝑥 + kx = 0

𝑥 +𝑘

𝑚𝑥 = 0 −− − 3

Compared with Eq.(2) i.e., 𝑥 + 𝜔𝑛2𝑥 = 0 −− − 2

m

𝑥

m

W


m

x


m

W

k (∆ + 𝑥)

𝑥 𝑥

𝜔𝑛2 =

𝑘

𝑚

∴ 𝝎𝒏 =𝒌

𝒎

Unit I – Theory of Vibrations • iii). Energy method:


• Assumption: System is to be conservative one. • Conservative system: Total sum of energy is constant at all

time.

• For an undamped system: since there is no friction or damping force, the total energy of the system is partly potential and partly kinetic.

• ∴ 𝐾. 𝐸 + 𝑃. 𝐸.= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

• The time rate of change of total energy will be zero.

•𝑑

𝑑𝑡𝑘. 𝐸.+𝑃. 𝐸. = 0

Unit I – Theory of Vibrations • iii). Energy method:


• 𝑘. 𝐸.=1

2𝑚𝑣2 =

1

2𝑚𝑥 2

• 𝑃. 𝐸.=1

2𝑘𝑥2

•𝑑

𝑑𝑡

1

2𝑚𝑥 2 +

1

2𝑘𝑥2 = 0

•1

2𝑚2𝑥 𝑥 +

1

2𝑘2𝑥𝑥 = 0

• 𝒎𝒙 + 𝒌𝒙 = 𝟎

Unit I – Theory of Vibrations • iv). Rayleigh’s method:

• Assumptions:

• (1) Maximum K.E. at the equilibrium position is equal to the maximum potential energy at the extreme position.

• (2). The motion is assumed to be SHM, then 𝑥 = 𝐴 sin𝜔𝑛𝑡

• Where x is the displacement of the system from its mean position after time t.

• A is the maximum displacement of the system from equilibrium position to extreme position.



𝑥 = 𝐴 sin𝜔𝑛𝑡

• 𝑥 is maximum when sin𝜔𝑛𝑡=1

• 𝑥𝑚𝑎𝑥 = 𝐴

• 𝑥 = 𝜔𝑛𝐴 cos𝜔 𝑛𝑡

• Velocity 𝑥 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑤ℎ𝑒𝑛 cos𝜔𝑛𝑡 = 1

• 𝑥 𝑚𝑎𝑥 = 𝜔𝑛𝐴



So maximum K.E. at the equilibrium position=1

2𝑚𝑥 𝑚𝑎𝑥

2

=1

2𝑚 𝜔𝑛𝐴

2

Maximum P.E. at the extreme position=1

2𝑘𝑥𝑚𝑎𝑥

2

=1

2𝑘 𝐴 2

1

2𝑚 𝜔𝑛𝐴

2 ==1

2𝑘 𝐴 2

𝜔𝑛2 =

𝑘

𝑚

∴ 𝝎𝒏 =𝒌

𝒎


Unit I – Theory of Vibrations • iv). D’Alembert’s method:

To find the solution of a dynamic problem by using the methods of statics.

According to Newton’s second law, 𝐹 = 𝑚𝑎 𝐹 −𝑚𝑎 = 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 ‘ma’ 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 ‘ma’.

The imaginary force ‘ma’ is known as inertia force and the position of equilibrium is called dynamic equilibrium.



D’Alemberts principle states that ‘a system may be in dynamic equilibrium by adding to the external forces, an imaginary force, which is commonly known as the inertia force’.

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.


x

k m

Spring – mass system Dynamic equilibrium




– Using D’Alembert’s principle, to bring the body to a dynamic equilibrium position, the inertia force ‘𝑚𝑥 is to be added in the direction opposite to the direction of motion.

– Equilibrium equation is 𝐹𝑥 = 0 −𝑚𝑥 − 𝑘𝑥 = 0

− 𝑚𝑥 + 𝑘𝑥 = 0 𝑚𝑥 + 𝑘𝑥 = 0

𝑥 +𝑘

𝑚𝑥 = 0

𝜔𝑛2 =

𝑘

𝑚

𝝎𝒏 =𝒌

𝒎


Unit I – Theory of Vibrations • Solution of the equation of motion:

– The governing differential equation of motion is

𝑚𝑥 + 𝑘𝑥 = 0

It is in the form of homogeneous second order linear equation.

There are five different solutions for the above equation of motion.

1. 𝑥 = 𝐴 cos𝜔𝑛𝑡

2. 𝑥 = 𝐵 sin𝜔𝑛𝑡

3. 𝑥 = 𝐴 cos𝜔𝑛𝑡 + 𝐵 sin𝜔𝑛𝑡

4. 𝑥 = 𝐴 sin(𝜔𝑛𝑡 + ∅)

5. 𝑥 = 𝐴 cos(𝜔𝑛𝑡 + ∅)

Where A and B are constants depending on their initial condition of the motion.



– Solution No.1: 𝑥 = 𝐴 cos𝜔𝑛𝑡 −− −(1)

To determine the constant A, let us use the initial condition by assuming that at time t=0, the displacement 𝑥 = 𝑥0.

Substituting this in the above equation (1), we get 𝑥0 = 𝐴 cos(𝜔𝑛 × 0)

∴ 𝑥0 = 𝐴

Hence the solution, 𝒙 = 𝒙𝟎 𝐜𝐨𝐬𝝎𝒏𝒕



– Solution No.2: 𝑥 = B sin𝜔𝑛𝑡 −−− −(2)

To determine the constant B, let us use the initial condition by assuming that

(i) at time t=0, the displacement 𝑥 = 𝑥0.

(ii) At time t=0, 𝑥 = 𝑥0

Differentiating equation (2) with respect to time, 𝑥 = 𝐵𝜔𝑛 cos𝜔𝑛𝑡

Applying initial conditions, 𝑥0 = 𝐵 𝜔𝑛.

𝐵 =𝑥 0𝜔𝑛

Substituting in equation (2), 𝒙 =𝒙 𝟎

𝝎𝒏𝐬𝐢𝐧𝝎𝒏𝒕



– Solution No.3: 𝑥 = 𝐴 cos𝜔𝑛𝑡 +B sin𝜔𝑛𝑡 −−− −(3)

The superposition of the above two solutions is also a solution.

The general solution for this second order differential equation is

𝒙 = 𝒙𝟎 𝐜𝐨𝐬𝝎𝒏𝒕 +𝒙 𝟎𝝎𝒏

𝐬𝐢𝐧𝝎𝒏𝒕



– Solution No.4: 𝑥 = 𝐴 sin(𝜔𝑛𝑡 + ∅) −−− −(4)

By expanding sine term 𝑥 = 𝐴 sin𝜔𝑛𝑡 cos ∅ + 𝐴 cos𝜔𝑛𝑡 sin ∅ −− −(4a)

But the general solution is



By comparing Eq.(4a) with general solution, i.e. comparing coefficient of cos𝜔𝑛𝑡,

𝑥0 = 𝐴 sin∅ − −(5)

Comparing coefficient of sin𝜔𝑛𝑡, 𝒙 𝟎𝝎𝒏

= A cos ∅ −− −(6)



– Solution No.4: 𝑥0 = 𝐴 sin ∅ − −(5) 𝒙 𝟎𝝎𝒏

= A cos ∅ −− −(6)

Squaring and adding Eq.(5) and Eq.(6),

𝐴2𝑠𝑖𝑛2∅ + 𝐴2𝑐𝑜𝑠2∅ = 𝑥02 +

𝑥 02

𝜔𝑛2

𝐴 = 𝑥02 +

𝑥 02

𝜔𝑛2

Dividing Eq.(5) and Eq.(6), 𝐴 sin ∅

𝐴 cos ∅=

𝑥0𝒙 𝟎𝝎𝒏

Hence the phase angle, ∅ = 𝐭𝐚𝐧−𝟏𝒙𝟎𝝎𝒏

𝒙 𝟎



– Solution No.5: 𝑥 = 𝐴 cos(𝜔𝑛𝑡 + ∅) −− −(7)

By expanding cosine term, we get 𝑥 = 𝐴 cos𝜔𝑛𝑡 cos ∅ + 𝐴 sin𝜔𝑛𝑡 sin ∅ −− −(7𝑎)

But the general solution is



By comparing Eq.(7a) with general solution, we get 𝑥0 = 𝐴 cos∅ − −(8) 𝒙 𝟎𝝎𝒏

= A sin ∅ −− −(9)

By squaring and adding Eqs. (8) and (9), we get

𝑥02 +

𝑥 02

𝜔𝑛2 = 𝐴2𝑐𝑜𝑠2∅ + 𝐴2𝑠𝑖𝑛2∅



– Solution No.5:

𝑥02 +

𝑥 02

𝜔𝑛2= 𝐴2𝑐𝑜𝑠2∅ + 𝐴2𝑠𝑖𝑛2∅

𝐴 = 𝑥02 +

𝑥 02

𝜔𝑛2

Dividing Eq.(9) and Eq.(8), 𝐴 sin ∅

𝐴 cos ∅=

𝒙 𝟎𝝎𝒏

𝑥0

Phase angle, ∅ = 𝐭𝐚𝐧−𝟏𝒙 𝟎

𝒙𝟎𝝎𝒏




Introduction: • Without damping force or frictional force the system vibrates

indefinitely with a constant amplitude at its natural frequency. • But in reality the vibration without decreasing amplitude is never

realized. • Frictional forces (or) damping forces are always present in any

physical system while undergoing motion.

• 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.

• This energy transformation mechanism is called a dissipation energy. This is quite complex in nature.

Damped free vibration of SDOF system:



What is damping? • 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.

• The decreasing rate of amplitude depends upon the amount of damping.

• The damping is useful to control the amplitude of vibration.




Types (or) Nature of damping : Mainly 5types of damping • 1. Viscous damping

• 2. Coulomb damping

• 3. Structural damping

• 4. Active damping (or) Negative damping

• 5. Passive damping




1. Viscous Damping: • 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.

• This type of damping is achieved by means of device called hydraulic dashpot.

• The main components of viscous damper or dashpot are cylinder, piston and viscous fluid as shown in Figure.

• Fluid mechanics concepts are to be used here.


V



1. Viscous Damping: • Let us consider that the two plates are

separated by fluid film of thickness t as shown in Figure.

• The upper plate is allowed to move parallel to the fixed plate with a velocity 𝑥 .

• The force ‘F’ required for maintaining this velocity 𝑥 of the plate is given by

𝐹 =𝜇𝐴

𝑡𝑥 −− −(1)

= 𝑐𝑥 c=damping coefficient (N-s/m)


V

F

t



1. Viscous Damping: • 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.

• Shock absorber is the best example of the viscous damping.

• Viscous damping is largely used for system modeling since it is linear.


V

F

t



1. Equation of motion for viscous damping:


x

k

m

c




Viscous damping oscillator F. B.D.

From FBD, the governing differential equation of motion is,

𝑚𝑥 + 𝑐𝑥 + kx = 0 −−−(2) Assuming the solution may be in the form of

𝑥 = 𝑒λ𝑡 Where λ is a constant to be determined. This exponential function leads to algebraic equation instead of a differential equation.



1. Equation of motion for viscous damping:


x

k

m

c




𝑥 = 𝑒λ𝑡

𝑥 = λ𝑒λ𝑡

𝑥 = λ2 𝑒λ𝑡 Substituting the values of 𝑥, 𝑥 , 𝑥 in equation (2) we get,

𝑚λ2 𝑒λ𝑡 + c λ𝑒λ𝑡 + k𝑒λ𝑡 = 0

𝑚λ2 + cλ+k 𝑒λ𝑡 = 0

The non-trivial solution is 𝑚λ2 + cλ+k=0

λ2 +𝑐

𝑚λ+

𝑘

𝑚= 0

𝑚𝑥 + 𝑐𝑥 + kx = 0 −− −(2)



• It was named because Charles Augustin de Coulomb carried on research in mechanics.

• 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.

• Coulomb damping absorbs energy with friction, which converts that kinetic energy into thermal energy or heat.

• Static and kinetic friction occur in a vibrating system undergoing Coulomb damping.

• Static friction occurs when two bodies are stationary or undergoing no relative motion.

Frictional force, 𝐹𝑠 = 𝜇𝑠𝑁 μs = coefficient of static friction.

Coulomb damping:



• Kinetic friction occurs when the two bodies are undergoing relative motion and they are sliding against each other.

Frictional force, 𝐹𝑘 = 𝜇𝑘𝑁 μk = coefficient of dynamic friction

Coulomb damping:

EOM for left to right motion, 𝑚𝑥 = −𝑘𝑥 − 𝐹, 𝑓𝑜𝑟 𝑥 > 0

EOM for right to left motion, 𝑚𝑥 = −𝑘𝑥 + 𝐹, 𝑓𝑜𝑟 𝑥 < 0

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑙𝑒𝑓𝑡 𝑚𝑜𝑡𝑖𝑜𝑛, 𝑥 = 𝐴 cos𝜔𝑛𝑡 + B sin𝜔𝑛𝑡 + 𝐹

𝑘

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑟𝑖𝑔ℎ𝑡 𝑚𝑜𝑡𝑖𝑜𝑛, 𝑥 = 𝐶 cos𝜔𝑛𝑡 + D sin𝜔𝑛𝑡 − 𝐹

𝑘



• 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 (𝟎 ≤ 𝒕 ≤ 𝑻/𝟐):

• The governing differential equation is 𝑚𝑥 = −𝑘𝑥 + 𝐹

𝑚𝑥 + 𝑘𝑥 = 𝐹 −−−(1) 𝑥

𝜔𝑛2+ 𝑥 =

𝐹

𝑘

The solution of the above equation can be written as

𝑥 = 𝐴 cos𝜔𝑛𝑡 + B sin𝜔𝑛𝑡 + 𝐹

𝑘




𝑥 = 𝐴 cos𝜔𝑛𝑡 + B sin𝜔𝑛𝑡 + 𝐹

𝑘−− −(2)

𝑥 = 𝑥𝑐 + 𝑥𝑝

Where 𝑥𝑐 = 𝐴 cos𝜔𝑛𝑡 + B sin𝜔𝑛𝑡 = complimentary sulution

𝑥𝑝 =𝐹

𝑘= Partcular integral

Where 𝜔𝑛 =𝑘

𝑚

Let us assume the initial condition to determine the constants A and B (i) At t=0; 𝑥=𝑥0 (ii) At t=0; 𝑥 = 0 𝑥 = −𝐴𝜔𝑛 sin𝜔𝑛𝑡 + B𝜔𝑛 cos𝜔𝑛𝑡




𝑥 = −𝐴𝜔𝑛 sin𝜔𝑛𝑡 + B𝜔𝑛 cos𝜔𝑛𝑡 −− −(3) Applying initial condition (ii) i.e., At t=0; 𝑥 = 0 in the above equation

0 = B𝜔𝑛 Since 𝜔𝑛 ≠ 0 ∴ 𝐵 = 0

Applying initial condition (i) i.e., at t=0;𝑥 = 𝑥0

𝑥0 = A +𝐹

𝑘

𝐴 = 𝑥0 −𝐹

𝑘

Hence Equation (2) can be written as

𝑥 = (𝑥0−𝐹

𝑘) cos𝜔𝑛𝑡 +

𝐹

𝑘, 0 ≤ 𝑡 ≤ 𝑇/2

This solution holds good for half cycle only.




𝑥 = (𝑥0−𝐹

𝑘) cos𝜔𝑛𝑡 +

𝐹

𝑘, 0 ≤ 𝑡 ≤ 𝑇/2 −− −(4)

We know that, 𝑇 =2𝜋

𝜔𝑛

For half cycle, 𝑇 =𝜋

𝜔𝑛

When t=𝜋

𝜔𝑛, half cycle is completed. So displacement for half the

cycle can be obtained from Eq.(4).

𝑡 =𝜋

𝜔𝑛

𝜔𝑛𝑡 = 𝜋 Substituting the value of 𝜔𝑛𝑡 in Eq.(4),

𝑥 = (𝑥0−𝐹

𝑘)cosπ +

𝐹

𝑘

𝑥 = (𝑥0−𝐹

𝑘)(−1) +

𝐹

𝑘




𝑥 = (𝑥0−𝐹

𝑘)(−1) +

𝐹

𝑘

𝒙 = − 𝒙𝟎 − 𝟐𝑭

𝒌

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.



• 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 (𝑻/𝟐 ≤ 𝒕 ≤ 𝑻):

• The governing differential equation is 𝑚𝑥 = −𝑘𝑥 − 𝐹

𝑚𝑥 + 𝑘𝑥 = −𝐹 −−−(5) 𝑥

𝜔𝑛2+ 𝑥 = −

𝐹

𝑘

The solution of the above equation can be written as

𝑥 = 𝐶 cos𝜔𝑛𝑡 + D sin𝜔𝑛𝑡 − 𝐹

𝑘




𝑥 = 𝐶 cos𝜔𝑛𝑡 + D sin𝜔𝑛𝑡 − 𝐹

𝑘−− −(6)

𝑥 = 𝑥𝑐 + 𝑥𝑝

Where 𝑥𝑐 = 𝐶 cos𝜔𝑛𝑡 + D sin𝜔𝑛𝑡 = complimentary sulution

𝑥𝑝 = −𝐹

𝑘= Partcular integral

Where 𝜔𝑛 =𝑘

𝑚

Let us assume the initial condition to determine the constants C and D (i) At t=0; 𝑥=𝑥0 (ii) At t=0; 𝑥 = 0 𝑥 = −𝐶𝜔𝑛 sin𝜔𝑛𝑡 + D𝜔𝑛 cos𝜔𝑛𝑡




𝑥 = −𝐶𝜔𝑛 sin𝜔𝑛𝑡 + D𝜔𝑛 cos𝜔𝑛𝑡 −− −(7) Applying initial condition (ii) i.e., At t=0; 𝑥 = 0 in the above equation

0 = D𝜔𝑛 Since 𝜔𝑛 ≠ 0 ∴ 𝐷 = 0

Applying initial condition (i) i.e., at t=𝜋

𝜔𝑛;𝑥 = −𝑥0 + 2

𝐹

𝑘

𝐶 = 𝑥0 − 3𝐹

𝑘

Hence Equation (2) can be written as

𝑥 = (𝑥0−3𝐹

𝑘) cos𝜔𝑛𝑡 −

𝐹

𝑘, 𝑇/2 ≤ 𝑡 ≤ 𝑇

This solution holds good for second half cycle only.




𝑥 = (𝑥0−3𝐹

𝑘) cos𝜔𝑛𝑡 −

𝐹

𝑘, 𝑇/2 ≤ 𝑡 ≤ 𝑇 −− −(8)

We know that, 𝑇 =2𝜋

𝜔𝑛

When t=2𝜋

𝜔𝑛, second half cycle is completed. So displacement for the

second half the cycle can be obtained from Eq.(8).

𝑡 =2𝜋

𝜔𝑛

𝜔𝑛𝑡 varies from 𝜋 𝑡𝑜 2𝜋 Substituting the value of 𝜔𝑛𝑡 in Eq.(4),

𝑥 = (𝑥0−3𝐹

𝑘)cos2π −

𝐹

𝑘

𝒙 = (𝒙𝟎−𝟒𝑭

𝒌)




𝒙 = − 𝒙𝟎 − 𝟐𝑭

𝒌

𝒙 = 𝒙𝟎 − 𝟒𝑭

𝒌

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 – Theory of Vibrations • Example 1:


• 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 – Theory of Vibrations • Solution:

– This stiffness is parallel to 𝑘.

– Equivalent spring stiffness, 𝑘𝑒 = 𝑘𝑏 + 𝑘

=3𝐸𝐼

𝐿3+ k

=3𝐸𝐼 + 𝑘𝐿3

𝐿3

The differential equation of motion is, 𝑚𝑥 = −𝑘𝑒𝑥


• Stiffness due to applied mass M is

𝑘𝑏 =𝑀

∆=

3𝐸𝐼

𝐿3

m

k

L

Unit I – Theory of Vibrations • Solution:

𝑚𝑥 = −𝑘𝑒𝑥 𝑚𝑥 + 𝑘𝑒𝑥 = 0

𝑚𝑥 +3𝐸𝐼 + 𝑘𝐿3

𝐿3𝑥 = 0

𝒙 +𝟑𝑬𝑰 + 𝒌𝑳𝟑

𝒎𝑳𝟑𝒙 = 𝟎

The frequency of vibration, 𝑓 =1

2𝜋

𝑘𝑒

𝑚

∴ 𝒇 =𝟏

𝟐𝝅

𝒌𝑳𝟑 + 𝟑𝑬𝑰

𝒎𝑳𝟑


m

k

L

Unit I – Theory of Vibrations • Example 2:


• Find the natural frequency of the system as shown in Figure. Take

𝑘1 = 𝑘2 = 2000 𝑁/𝑚, 𝑘3 = 3000𝑁

𝑚 and m= 10 kg.

𝑘3

m=10 kg

𝑘1 𝑘2

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