Basic concepts on structural dynamics

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Dr.L.V. Prasad .M Department of Civil Engineering

National Institute of Technology Silchar

E-mail: [email protected]

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What is Dynamics ?

The word dynamic simply means

“changes with time”

Dr.L.V.Prasad, Assistant Professor, Civil Engineering Dept, NITS

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Basic difference between static and dynamic loading

P P(t)

Resistance due to internal elastic forces of structure

Accelerations producing inertia forces (inertia forces form a significant portion of load equilibrated by the internal elastic forces of the structure)

Static Dynamic


In static problem: Response due to static loading is displacement only. In dynamic problem: Response due to

dynamic loading is displacement, velocity and acceleration.

05/01/2023 4Dr.L.V.Prasad, Assistant Professor, Civil Engineering Dept, NITS

Most Common Causes Dynamic Effect In The Structure

•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.

• 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.

•Support motions : Structures are often subjected to vibration due to

influence of support motions.

Ex: Earthquake motion.

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Vibration and oscillation: If motion of the structure is oscillating (pendulum) or reciprocatory along with deformation of the structure, it is termed as VIBRATION. In case there is no deformation which implies only rigid body motion, it is termed as OSCILLATION.

Free vibration: Vibration of a system which is initiated by a force which is subsequently withdrawn. Hence this vibration occurs without the external force.

Forced Vibration: If the external force is also involved during vibration, then it is forced vibration.

Basic Concepts of Structural dynamics


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Damping: All real life structures, when subjected to vibration resist it. Due to this the amplitude of the vibration gradually, reduces with respect to time. In case of free vibration, the motion is damped out eventually. Damping forces depend on a number of factors and it is very difficult to quantify them.

The commonly used representation is viscous damping

wherein damping force is expressed as Fd=C x.

where x. = velocity and C=damping constant.

Basic Concepts of Structural dynamics


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The number of independent displacement components that must be considered to represent the effects of all significant inertia forces of a structure.

Dynamic Degrees of Freedom

Depending upon the co-ordinates to describe the motion, we have

1. Single degree of freedom system (SDoF).2. Multiple degree of freedom (MDoF).3. Continuous system.


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Single Degree of Freedom: 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.Multiple degree of freedom (MDoF): 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. Continuous system: 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.

Dynamic Degrees of Freedom


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Single Degree of Freedom

Vertical translation Horizontal translation Horizontal translation Rotation


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Degrees of freedom: –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.

Multiple Degrees of Freedom


Example for MDOF system

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Continuous system: Degrees of freedom: –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:


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Mathematical model - SDOF System

Mass element ,m - representing the mass and inertial characteristic of the structure

Spring element ,k - representing the elastic restoring force and potential energy capacity of the

structure.

Dashpot, c - representing the frictional characteristics and energy losses of the structure

Excitation force, P(t) - represents the external force acting on structure.

P(t)

x

m

k

cF = m × x·· = p(t) – cx· – kx

mx·· + cx· + kx = p(t)


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Mathematical model - SDOF System

Undamped (C =0 &P(t)=0)1. Free Vibration

Damped ( C0 &P(t)=0)

Undamped (C =0 &P(t) 0)2. Forced Vibration

Damped ( C0 &P(t) 0)


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Equation of Motion - SDOF System

1.Simple Harmonic motion

2. Newtown’s Law of motion

3. Energy methods

4.Rayleights method

5.D’alembert’s method

Differential equation describing the motion is known as equation of motion.


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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.

Simple Harmonic motion method:

SHM is the simplest form of periodic motion. •In differential equation form, SHM is represented as − −−−(1) 𝑥 ∝ 𝑥


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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.


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Energy method: Conservative system: Total sum of energy is constant at all time.


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Rayleigh’s method: Maximum K.E. at the equilibrium position is equal to the maximum potential energy at the extreme position.


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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’.

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


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P(t) =0

x

m

k

mx·· + cx· + kx = p(t)

;;

;0

:

1;2

;0

;0

;0

:;;0)(

;;0

22

22

22..

..

..

iDD

D

EquationAuxiliaryTffm

km

kwherexx

xmkx

xkmx

motionofEquationionFreeVibrattp

Undampedc

tBtAtx

tBtAe

functionarycomplement

iximaginaryarerootsThe

t

sincos)(sincos

:

;:

2,1

Free Vibration of Undamped - SDOF System


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00( ) cos sinvx t x pt pt

p

Amplitude of motion

t

x

vo

22 00

vx

p

2Tp

or2

2 00( ) sin ( )vx t x pt

p

where, 0

0

tan xv p

x0

X =initial displacement

V =initial velocity0

0

t

Vo = X.

o & =

;/;

;sec/

;sincos)(

;;

;cossin

;;0@;;0

0.

0

0.

.

.

0

mNkkgm

radmkwhere

txtxtx

xBBtx

tBtAtx

Axt

Free Vibration of Undamped - SDOF System


=p is called circular frequency or angular frequency of vibration (Rad/s)

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Free Vibration of damped SDOF systems

kmc

mpcζ

mkp

22

(Dimensionless parameter) - A

where,

2

0

0

2 0

mx cx kx c kx x xm m

x ζpx p x

x

m

k

c


is called circular frequency or angular frequency of vibration (Rad/s)

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Solution of Eq.(A) may be obtained by a function in the form x = ert where r is a constant to be determined. Substituting this into (A) we obtain,

2 22 0rte r ζpr p

In order for this equation to be valid for all values of t,

2 2

21,2

2 0

1

r ζpr p

r p

or



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Thus and are solutions and, provided r1 and r2 are different from one another, the complete solution is

trtr 21 ee

1 21 2

r t r tx c e c e

The constants of integration c1 and c2 must be evaluated from the initial conditions of the motion.

Note that for >1, r1 and r2 are real and negative

for <1, r1 and r2 are imaginary and

for =1, r1= r2= -p

ζζζ

ζSolution depends on whether is smaller than, greater than, or equal to one.



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For (Light Damping) :1

0

0021d

A xvB xp

2

cos sin

1

ptd d

d

x t e A p t B p t

p p

‘A’ and ‘B’ are related to the initial conditions as follows

(B)

2

cos sin1

pt oo d o d

d

vx t e x p t x p tp

In other words, Eqn.B can also be written as,

where,



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2

2 Damped natural period

1 Damped circular natural frequency

d

d

Tp

p p

Extremum point ( )( ) 0cos( ) 1d

tp t

x

Point of tangency ( ) Td = 2π / pd

xn Xn+1

t

x

2

2 Damped natural period

1 Damped circular natural frequency

dd

d

Tp

p p


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Such system is said to be over damped or super critically damped.

1

i.e., the response equation will be sum of two exponentially decaying curve

In this case r1 and r2 are real negative roots.

( ) ( )1 2( ) t tx t C e C e

For (Heavy Damping)

xo

x

o t


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Such system is said to be critically damped.

1 2( ) pt ptx t C e C te

1

The value of ‘c’ for which Is known as the critical coefficient of damping

With initial conditions,

0 0( ) 1 ptx t x pt v t e

1

2 2crC mp km

Therefore,cr

CC

For


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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.

Stiffness due to applied mass M is 𝑘𝑏= /Δ=3 /𝑀 𝐸𝐼 𝐿3

Equivalent spring stiffness, 𝑘𝑒=𝑘𝑏+ 𝑘 𝑘 𝑒 =(3 /𝐸𝐼 𝐿3)+k

𝑘 𝑒 =(3 +𝐸𝐼 𝑘𝐿3)/𝐿3

The differential equation of motion is,

𝑚𝑥 ..=−𝑘 𝑒 𝑥The frequency of vibration,


05/01/2023 Dr.L.V.Prasad, Assistant Professor, Civil Engineering Dept, NITS

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Problem 2: Calculate the natural angular frequency of the frame shown in figure. Compute also natural period of vibration. If the initial displacement is 25 mm and initial velocity is 25 mm/s what is the amplitude and displacement @t =1s.

In this case, the restoring force in the form of spring force is provided by AB and CD which are columns. The equivalent stiffness is computed on the basis that the spring actions of the two columns are in parallel.


Problem 2


Problem 3: Following data are given for a vibrating system with viscous damping mass m=4.5 kg, stiffness k= 30 N/m and damping C=0.12 Ns/m. Determine the logarithmic decrement, ratio of any 2 successful amplitudes.

033.1

033.0)1(

2log

%52.022.23

12.022.23)58.25.4(22

/58.25.4

30

2

1

2

exx

ratioAmplitude

decrementarthmic

cc

xmc

sradmk

cr

ncr

n


Multiple degree of freedom systemsA multi degrees of freedom (dof) system is one, which requires two or more coordinates to describe its motion.

These coordinates are called generalized coordinates when they are independent of each other and equal in number to the degrees of freedom of the system


Two degree of freedom systems

)( 212111..

1 xxkxkxm

231222..

2 )( xkxxkxm


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Problem 4: A pedestal bridge platform is truss supported as shown in Fig. by neglecting the self weight of the truss , estimate the frequency of vibration of the truss by idealizing a simple spring-mass system. Assume that are of cross section and young's modulus are same for all members.


Member Force (P) Unit force (p) Length (l) Ppl/AEAB 0 0 L 0BC 0 0 L 0CF - W/2 - 1/2 L WL/4FE - W/2 - 1/2 L WL/4DE - W/2 - 1/2 L WL/4AD - W/2 - 1/2 L WL/4BD + W/√2 +1/√2 √2L WL/√2BF + W/√2 +1/√2 √2L WL/√2BE 0 0 L 0

mk

LAEK

AEWL

AEWL

AEPpL

n

414.0

1414.2

414.2

Problem 4


37

THANK YOU

Basic concepts on structural dynamics

Engineering

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