CE 6701 Structural Dynamics and Earthquake Engineering Dr. P. Venkateswara Rao

CE 6701 Structural Dynamics and Earthquake Engineering Dr. P. Venkateswara Rao Associate Professor Dept. of Civil Engineering SVCE, Sriperumbudur

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 3

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 4

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 5

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 6

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 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 7

References Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 8

Vibration: Unit I Theory of Vibrations 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 9

Vibration: Unit I Theory of Vibrations 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). Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 10

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 11

Dynamic loading: Unit I Theory of Vibrations 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 12

Comparison of static loading and dynamic loading: i) In static problem: Load is constant with time. W Ex: Weight of a bridge span on bridge pilings. In dynamic problem: Loading and its response varies with time. W (t) Ex: A truck moving across the same bridge span exerts a dynamic load on the pilings. Inertia forces Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 13

Comparison of static loading and dynamic loading: ii) In static problem: Response due to static loading is displacement only. W y In dynamic problem: Response due to dynamic loading is displacement, velocity and acceleration. W (t) Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 14

Comparison of static loading and dynamic loading: iii) In static problem: Solution of static problem is only one. W y In dynamic problem: Solutions of dynamic problem are infinite and are time dependent. W (t) Dynamic analysis is more complex and time consuming than static analysis Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 15

Comparison of static loading and dynamic loading: iv) In static problem: Response calculation is done by static equilibrium. W y In dynamic problem: Response not only depends on load but also depend on inertia forces which oppose the accelerations producing them. Inertia forces are most W (t) important characteristics of a structural dynamic problem. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 16

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 17

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 18

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 19

Degrees of freedom: Unit I Theory of Vibrations 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 20

Degrees of freedom: Example for SDOF: Unit I Theory of Vibrations k 1 m x x Spring mass system Building frame Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 21

Degrees of freedom: Unit I Theory of Vibrations 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: x 1 x 2 k 1 k 2 m 1 m 2 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 22

Degrees of freedom: Unit I Theory of Vibrations 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: x 3 x 3 x 2 x 1 x 3 Cantilver beam with infinite number of degrees of freedom Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 23

F(t) Unit I Theory of Vibrations Mathematical modelling of an SDOF system: K Portal frame x To understand the dynamic behaviour of structure, it is necessary to develop their models under dynamic loads such as earthqukes, wind, blasts etc. 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 24

F(t) Unit I Theory of Vibrations Mathematical modelling of an SDOF system: K Portal frame x 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. c k m x Inertia force, F i = mx Damping force, F D = cx Spring force, F s =kx m FBD F(t) Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 25

F(t) Unit I Theory of Vibrations Mathematical modelling of an SDOF system: x Inertia force, F i = mx Damping force, F D = cx Spring force, F s =kx m FBD F(t) 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 26

F(t) Unit I Theory of Vibrations Mathematical modelling of an SDOF system: x Inertia force, F i = mx Damping force, F D = cx Spring force, F s =kx m FBD F(t) 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 27

F(t) Unit I Theory of Vibrations Mathematical modelling of an SDOF system: x Inertia force, F i = mx Damping force, F D = cx Spring force, F s =kx m FBD F(t) The elements to determine the dynamic behaviour: i) the inertia force, F i = mx Ii) the restoring force or spring force, F s =kx iii) the damping force, F D = 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) Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 28

Free vibration of undamped SDOF system: x m k Inertia force, F i = mx Spring force, F s =kx m FBD Considering the equilibrium of all forces in X- direction, the governing equation of motion for the SDOF is, F i + F s = 0 mx + kx = 0 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 29

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 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 30

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 ( d2 x dt 2) Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 31

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 ω n 2 which is an unknown parameter. Now Eq.(1) can be rewritten as, x = ω n 2 x x + ω n 2 x = 0 2 This is known as equation of motion and is second order linear differential equation. The constant ω n is yet to be determined by the analysis. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 32

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 33

ii). Newton s second law of motion: 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. = Static deflection m k m W m Static equilibrium position Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 34 x x k ( + x) m W x x

ii). Newton s second law of motion: 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. = Static deflection m x k m W m Static equilibrium position x k ( + x) m x x W Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 35

ii). Newton s second law of motion: 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 = Static deflection m x k m W m Static equilibrium position x k ( + x) m x x W Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 36

ii). Newton s second law of motion: According to Newton s second law, mx = kx mx + kx = 0 x + k x = 0 3 m Compared with Eq.(2) i.e., x + ω 2 n x = 0 2 ω n 2 = k m = Static deflection m x k m W m Static equilibrium position x ω n = k m k ( + x) m x x W Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 37

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 k. E. +P. E. = 0 dt Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 38

iii). Energy method: k. E. = 1 2 mv2 = 1 2 mx 2 P. E. = 1 2 kx2 d dt 1 2 mx 2 + 1 2 kx2 = 0 1 m2x x + 1 k2xx = 0 2 2 mx + kx = 0 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 39

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 sin ω n t 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 40

iv). Rayleigh s method: x = A sin ω n t x is maximum when sin ω n t=1 x max = A x = ω n A cos ω nt Velocity x is maximum when cos ω n t = 1 x max = ω n A Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 41

iv). Rayleigh s method: So maximum K.E. at the equilibrium position= 1 2 mx max 2 = 1 2 m ω na 2 Maximum P.E. at the extreme position= 1 2 kx max 2 1 2 m ω na 2 == 1 k A 2 2 = 1 2 k A 2 ω n 2 = k m ω n = k m Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 42

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 43

iv). D Alembert s method: 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. m k x Inertia force, F i = mx Spring force, F s =kx m Spring mass system Dynamic equilibrium Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 44

iv). D Alembert s method: 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 F x = 0 mx kx = 0 mx + kx = 0 mx + kx = 0 x + k m x = 0 ω n 2 = k m ω n = k m Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 45

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 cos ω n t 2. x = B sin ω n t 3. x = A cos ω n t + B sin ω n t 4. x = A sin(ω n t + ) 5. x = A cos(ω n t + ) Where A and B are constants depending on their initial condition of the motion. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 46

Solution of the equation of motion: Solution No.1: x = A cos ω n t (1) To determine the constant A, let us use the initial condition by assuming that at time t=0, the displacement x = x 0. Substituting this in the above equation (1), we get x 0 = A cos(ω n 0) x 0 = A Hence the solution, x = x 0 cos ω n t Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 47

Solution of the equation of motion: Solution No.2: x = B sin ω n t (2) To determine the constant B, let us use the initial condition by assuming that (i) at time t=0, the displacement x = x 0. (ii) At time t=0, x = x 0 Differentiating equation (2) with respect to time, x = Bω n cos ω n t Applying initial conditions, x 0 = B ω n. B = x 0 ω n Substituting in equation (2), x = x 0 ω n sin ω n t Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 48

Solution of the equation of motion: Solution No.3: x = A cos ω n t + B sin ω n t (3) The superposition of the above two solutions is also a solution. The general solution for this second order differential equation is x = x 0 cos ω n t + x 0 ω n sin ω n t Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 49

Solution of the equation of motion: Solution No.4: x = A sin(ω n t + ) (4) By expanding sine term x = A sin ω n t cos + A cos ω n t sin (4a) But the general solution is x = x 0 cos ω n t + x 0 ω n sin ω n t By comparing Eq.(4a) with general solution, i.e. comparing coefficient of cos ω n t, x 0 = A sin (5) Comparing coefficient of sin ω n t, x 0 ω n = A cos (6) Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 50

Solution of the equation of motion: Solution No.4: x 0 = A sin (5) x 0 ω n = A cos (6) Squaring and adding Eq.(5) and Eq.(6), A 2 sin 2 + A 2 cos 2 = x 0 2 + x 0 2 A = x 0 2 + x 0 2 ω n 2 ω n 2 Dividing Eq.(5) and Eq.(6), A sin = x 0 A cos x 0 ωn Hence the phase angle, = tan 1 x 0ω n x 0 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 51

Solution of the equation of motion: Solution No.5: x = A cos(ω n t + ) (7) By expanding cosine term, we get x = A cos ω n t cos + A sin ω n t sin (7a) But the general solution is x = x 0 cos ω n t + x 0 ω n sin ω n t By comparing Eq.(7a) with general solution, we get x 0 = A cos (8) x 0 ω n = A sin (9) By squaring and adding Eqs. (8) and (9), we get x 0 2 + x 0 2 ω n 2 = A2 cos 2 + A 2 sin 2 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 52

Solution of the equation of motion: Solution No.5: x 0 2 + x 0 2 ω n 2 = A2 cos 2 + A 2 sin 2 A = x 0 2 + x 0 2 ω n 2 Dividing Eq.(9) and Eq.(8), Phase angle, = tan 1 x 0 A sin = ωn A cos x 0 x 0 ω n x 0 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 53

Damped free vibration of SDOF system: 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 54

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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 55

Damped free vibration of SDOF system: 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 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 56

Damped free vibration of SDOF system: 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. V 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 57

Damped free vibration of SDOF system: 1. Viscous Damping: Let us consider that the two plates are separated by fluid film of thickness t as shown in Figure. F t The upper plate is allowed to move parallel to the fixed plate with a velocity x. V The force F required for maintaining this velocity x of the plate is given by F = μa t x (1) = cx c=damping coefficient (N-s/m) Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 58

Damped free vibration of SDOF system: 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 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 59

Damped free vibration of SDOF system: 1. Equation of motion for viscous damping: c k m Viscous damping oscillator x Inertia force, F i = mx Damping force, F D = cx Spring force, F s =kx m 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 = e λt Where λ is a constant to be determined. This exponential function leads to algebraic equation instead of a differential equation. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 60

Damped free vibration of SDOF system: 1. Equation of motion for viscous damping: mx + cx + kx = 0 (2) c m x Inertia force, F i = mx Damping force, F D = cx Spring force, F s =kx k x = e λt x = λe λt x = λ 2 e λt Substituting the values of x, x, x in equation (2) we get, mλ 2 e λt + c λe λt + ke λt = 0 mλ 2 + cλ+k e λt = 0 The non-trivial solution is mλ 2 + cλ+k=0 λ 2 + c m λ+ k m = 0 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 61 m

Coulomb damping: 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, F s = μ s N μ s = coefficient of static friction. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 62

Coulomb damping: Kinetic friction occurs when the two bodies are undergoing relative motion and they are sliding against each other. Frictional force, F k = μ k N μ k = coefficient of dynamic friction EOM for right to left motion, mx = kx + F, for x < 0 EOM for left to right motion, mx = kx F, for x > 0 F Solution for left motion, x = A cos ω n t + B sin ω n t + k F Solution for right motion, x = C cos ω n t + D sin ω n t k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 63

Coulomb damping for first half cycle (0 t T/2): Let us consider the movement of the mass to the left in a Coulomb damping system as shown in figure. The governing differential equation is mx = kx + F mx + kx = F (1) x ω 2 + x = F n k The solution of the above equation can be written as F x = A cos ω n t + B sin ω n t + k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 64

Coulomb damping for first half cycle (0 t T/2): F x = A cos ω n t + B sin ω n t + (2) k x = x c + x p Where x c = A cos ω n t + B sin ω n t = complimentary sulution x p = F k = Partcular integral Where ω n = k m Let us assume the initial condition to determine the constants A and B (i) At t=0; x=x 0 (ii) At t=0; x = 0 x = Aω n sin ω n t + Bω n cos ω n t Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 65

Coulomb damping for first half cycle (0 t T/2): x = Aω n sin ω n t + Bω n cos ω n t (3) Applying initial condition (ii) i.e., At t=0; x = 0 in the above equation 0 = Bω n Since ω n 0 B = 0 Applying initial condition (i) i.e., at t=0;x = x 0 x 0 = A + F k A = x 0 F k Hence Equation (2) can be written as x = (x 0 F k ) cos ω F nt +, 0 t T/2 k This solution holds good for half cycle only. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 66

Coulomb damping for first half cycle (0 t T/2): x = (x 0 F k ) cos ω nt + F k, 0 t T/2 (4) We know that, T = 2π ω n For half cycle, T = π ω n When t= π ω n, half cycle is completed. So displacement for half the cycle can be obtained from Eq.(4). t = π ω n ω n t = π Substituting the value of ω n t in Eq.(4), x = (x 0 F k )cosπ + x = (x 0 F k )( 1) + F k F k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 67

Coulomb damping for first half cycle (0 t T/2): x = (x 0 F k )( 1) + F k x = x 0 2 F 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 68

Coulomb damping for second half cycle (T/2 t T): Let us consider the movement of the mass to the right in a Coulomb damping system as shown in figure. The governing differential equation is mx = kx F mx + kx = F (5) x ω 2 + x = F n k The solution of the above equation can be written as F x = C cos ω n t + D sin ω n t k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 69

Coulomb damping for second half cycle (T/2 t T): F x = C cos ω n t + D sin ω n t (6) k x = x c + x p Where x c = C cos ω n t + D sin ω n t = complimentary sulution x p = F k = Partcular integral Where ω n = k m Let us assume the initial condition to determine the constants C and D (i) At t=0; x=x 0 (ii) At t=0; x = 0 x = Cω n sin ω n t + Dω n cos ω n t Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 70

Coulomb damping for second half cycle (T/2 t T): x = Cω n sin ω n t + Dω n cos ω n t (7) Applying initial condition (ii) i.e., At t=0; x = 0 in the above equation 0 = Dω n Since ω n 0 D = 0 Applying initial condition (i) i.e., at t= π ω n ;x = x 0 + 2 F k C = x 0 3 F k Hence Equation (2) can be written as x = (x 0 3 F k ) cos ω F nt, T/2 t T k This solution holds good for second half cycle only. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 71

Coulomb damping for second half cycle (T/2 t T): x = (x 0 3 F k ) cos ω nt We know that, T = 2π ω n F k, T/2 t T (8) When t= 2π, second half cycle is completed. So displacement for the ω n second half the cycle can be obtained from Eq.(8). t = 2π ω n ω n t varies from π to 2π Substituting the value of ω n t in Eq.(4), x = (x 0 3 F k )cos2π x = (x 0 4 F k ) F k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 72

Coulomb damping for second half cycle (T/2 t T): x = x 0 2 F k x = x 0 4 F 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. Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 73

Coulomb damping for second half cycle (T/2 t T): Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 74

Example 1: Unit I Theory of Vibrations 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 L k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 75

Solution: Unit I Theory of Vibrations Stiffness due to applied mass M is k b = M = 3EI L 3 L m k This stiffness is parallel to k. Equivalent spring stiffness, k e = k b + k = 3EI L 3 3EI + kl3 = L 3 The differential equation of motion is, mx = k e x + k Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 76

Solution: Unit I Theory of Vibrations mx = k e x mx + k e x = 0 3EI + kl3 mx + L 3 x = 0 x + 3EI + kl3 ml 3 x = 0 L m k The frequency of vibration, f = 1 2π k e m f = 1 2π kl 3 + 3EI ml 3 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 77

Example 2: Unit I Theory of Vibrations Find the natural frequency of the system as shown in Figure. Take k 1 = k 2 = 2000 N/m, k 3 = 3000 N and m= 10 kg. m k 1 k 2 m=10 kg k 3 Dr. P.Venkateswara Rao, Associate Professor, SVCE,Sriperumbudur 78