Basic concepts on structural dynamics

Dr.L.V. Prasad .M
Department of Civil Engineering
National Institute of Technology Silchar
E-mail: prasadsmlv@gmail.com
11/21/2016 1

What is Dynamics ?
The word dynamic simply means
“changes with time”
11/21/2016 2Dr.L.V.Prasad, Assistant Professor, Civil Engineering Dept, NITS

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.

11/21/2016 Dr.L.V.Prasad, Assistant Professor, Civil Engineering Dept, NITS 4
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.

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by a force which is subsequently withdrawn.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

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

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.

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

Single Degree of Freedom
Vertical translation Horizontal translation Horizontal translation Rotation
11/21/2016 9
Dr.L.V.Prasad, Assistant Professor, Civil Engineering Dept, NITS

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

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:

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
c
F = m × x·· = p(t) – cx· – kx
mx·· + cx· + kx = p(t)

Mathematical model - SDOF System
Undamped (C =0 &P(t)=0)
1. Free Vibration
Damped ( C0 &P(t)=0)
Undamped (C =0 &P(t)  0)
2. Forced Vibration
Damped ( C0 &P(t)  0)

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.

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 𝑥 ∝−𝑥 −−−(1)

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.

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

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

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 ‘𝑚𝑥 is to be added in the
direction opposite to the direction of motion.

P(t) =0
x
m
k
mx·· + cx· + kx = p(t)
 
 
;
;
;0
:
1;2
;0
;0
;0
:
;;0)(
;;0
22
22
22..
..
..





iD
D
D
EquationAuxiliary
T
ff
m
k
m
kwherexx
x
m
kx
xkmx
motionofEquation
ionFreeVibrattp
Undampedc









 
tBtAtx
tBtAe
functionarycomplement
ix
imaginaryarerootsThe
t




sincos)(
sincos
:
;
:
2,1



Free Vibration of Undamped - SDOF System

0
0( ) cos sin
v
x t x pt pt
p
 
Amplitude of motion
t
x
vo
2
2 0
0
v
x
p
 
  
 
2
T
p


or
2
2 0
0( ) sin ( )
v
x t x pt
p

 
   
 
where,
0
0
tan
x
v p
 
x0
X =initial displacement
V =initial velocity0
0
t
Vo = X
.
o &  = 
 
   
 
;/;
;sec/
;sincos)(
;;
;cossin
;;0@
;;0
0
.
0
0
.
.
.
0
mNkkgm
rad
m
kwhere
t
x
txtx
x
BBtx
tBtAtx
Axt















Free Vibration of Undamped - SDOF System
11/21/2016 21
=p is called circular frequency or angular
frequency of vibration (Rad/s)

Free Vibration of damped SDOF systems
km
c
mp
c
ζ
m
k
p
22


(Dimensionless parameter) - A
where,
2
0
0
2 0
mx cx kx
c k
x x x
m m
x ζpx p x
  
  
  
&& &
&& &
&& &
x
m
k
c
 is called circular frequency or angular
frequency of vibration (Rad/s)

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 2
2 0rt
e r ζpr p  
In order for this equation to be valid for all values of t,
 
2 2
2
1,2
2 0
1
r ζpr p
r p  
  
   or

Thus and are solutions and, provided r1 and r2 are different from one
another, the complete solution is
trtr 21
ee
1 2
1 2
rt r t
x 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
ζ
ζ
ζ
ζSolution depends on whether is smaller than, greater than, or equal to one.

For (Light Damping) :1 
0
0
02
1d
A x
v
B x
p



 

   
2
cos sin
1
pt
d 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 sin
1
pt o
o d o d
d
v
x t e x p t x p t
p
 


  
    
     
In other words, Eqn.B can also be written as,
where,

2
2
Damped natural period
1 Damped circular natural frequency
d
d
T
p
p p


 
  
Extremum point ( )( ) 0
cos( ) 1d
t
p t
x


 
g
Point of tangency ( )
Td = 2π / pd
xn Xn+1
t
x
2
2
Damped natural period
1 Damped circular natural frequency
d
d
d
T
p
p p


 
  

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 t
x t C e C e 
 
For (Heavy Damping)
xo
x
o t
11/21/2016 27

Such system is said to be critically damped.
1 2( ) pt pt
x 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 pt
x t x pt v t e
    
1 
2 2crC mp km 
Therefore,
cr
C
C
 
For

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
𝑘 𝑏=𝑀/Δ=3𝐸𝐼/𝐿3
Equivalent spring
stiffness,
𝑘 𝑒=𝑘 𝑏+𝑘
𝑘 𝑒 =(3𝐸𝐼/𝐿3)+k
𝑘 𝑒 =(3𝐸𝐼+𝑘𝐿3)/𝐿3
The differential equation of motion is,
𝑚𝑥 ..=−𝑘 𝑒 𝑥
The frequency of vibration,

11/21/2016
Dr.L.V.Prasad, Assistant Professor, Civil
Engineering Dept, NITS
30
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(
2
log
%52.0
22.23
12.0
22.23)58.25.4(22
/58.2
5.4
30
2
1
2














e
x
x
ratioAmplitude
decrementarthmic
c
c
xmc
srad
m
k
cr
ncr
n

Multiple degree of freedom systems
A 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 

11/21/2016
35
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/AE
AB 0 0 L 0
BC 0 0 L 0
CF - W/2 - 1/2 L WL/4
FE - W/2 - 1/2 L WL/4
DE - W/2 - 1/2 L WL/4
AD - W/2 - 1/2 L WL/4
BD + W/√2 +1/√2 √2L WL/√2
BF + W/√2 +1/√2 √2L WL/√2
BE 0 0 L 0
m
k
L
AE
K
AE
WL
AE
WL
AE
PpL
n 





414.0
1414.2
414.2
Problem 4

11/21/2016
37
THANK YOU

Basic concepts on structural dynamics

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Basic concepts on structural dynamics