The document outlines the differences between static and dynamic forces, highlighting key concepts such as load variation, response types, and complexity in analysis. It details how earthquake engineering focuses on horizontal motions due to their significance and discusses measures of ground motion like peak ground acceleration and the factors that influence it. Lastly, it covers topics related to soil behavior during earthquakes, including liquefaction types and effects, as well as essential seismic design principles.

1. Static and dynamic forces

2. Write a short note on static and dynamic forces.
• Static - load is constant, Dynamic means time varying. So
time additional parameter in the dynamic loading.
• i) In static , the load is constant w.r.t. time. But incase of
dynamic loading the load is time varying in nature. Both
loading and its reactance varies w.r.t. time.
• ii) Static problem has only one response is displacement.
But dynamic problem has mainly three responses, such as
displacement, velocity and acceleration.
• iii) Static load does not produce vibration and dynamic load
produces vibration.
• iv) Static problem has only one solution whereas a dynamic
problem has infinite an infinite no. of solutions which are
time dependant in nature. Thus dynamic analysis is more
complex and time consuming than static analysis.

3. Write a short note on static and dynamic forces.
• In static, the response can be calculated by
principles of force or static equilibrium,
whereas in case of dynamic problem the
response does not depend not only upon load
but also upon inertial forces which opposes the
acceleration producing them
• Thus the total responses are calculated by
including inertia forces along with static
equilibrium. Hence inertia forces are the most
important characteristics of a structural
dynamic problem.

4. Why Horizontal motions are of most
importance for earthquake engineering
• Shaking often strongest on horizontal component:
– Earthquakes radiate larger S waves than P waves
– Decreasing seismic velocities near Earth’s surface
produce refraction of the incoming waves toward the
vertical, so that the ground motion for S waves is
primarily in the horizontal direction
• Buildings generally are weakest for horizontal
shaking

5. What are the most useful measures of
ground motion?
• PGA (peak ground acceleration)
• PGV (peak ground velocity)
• Response spectral acceleration (elastic,
inelastic) at periods of engineering interest
• Intensity (Can be related to PGA and PGV.)

6. What factors control the level of ground
motion?
• Peak ground acceleration (PGA)
• easy to measure because the response of most
instruments is proportional to ground acceleration
• liked by many engineers because it can be related to
the force on a short-period building
• convenient single number to enable rough evaluation
of importance of records
• BUT it is not a measure of the force on most buildings
• and it is controlled by the high frequency content in
the ground motion (i.e., it is not associated with a
narrow range of frequencies); records can show
isolated short-duration, high-amplitude spikes with
little engineering significance

7. What Controls the Level of Shaking?
• Magnitude
• Directivity
– Larger fault, more energy released and over a larger
area
• Distance from fault
– Shaking decays with distance
• Local site response (rock or soil)
– amplify the shaking
– Strongest shaking in rupture direction
– Pockets of higher shaking

8. 3. What is natural frequency
• Generally frequency is number of cycles per
unit time. When no external force acts on
the system after giving initial displacement
the body vibrates.
• These vibrations are called free vibrations
and their frequency is called natural
frequency .
• It is expressed as f = 2
• Where f = natural frequency
•  = angular frequency

9. 4. Give the concept of mode superposition 2010N
• For linear structures with certain types of
damping , the response in each mode of
vibration can be determined independently of
the response in the other modes. The
independent modal response can then be
combined to determine the total response.
• This is the basic concept of the mode
superposition method.

10. List any two causes of Earthquakes
• Relative movement of the plates causes stress
to build up on their boundaries. As movement
occurs, strain energy accumulates in the
vicinity of the boundaries
• The energy is eventually dissipated: either
smoothly and continuously or in a stick slip
manner that produces earthquakes.
• The sudden slip at the fault causes the
Earthquake – a violent shaking of the Earth.

11. Define Seismogram
• The instrument that measures earthquake
shaking is a seismograph.
• The variation of ground acceleration
with time recorded at a point on
ground during an earthquake is an
accelerogram (seismogram)

12. P(t)
x
m
k
c
2. Define SDOF idealization
SDOF means Single Degree of Freedom. The SDOF system
consists of a rigid block with mass m, elastic spring with
stiffness k and a viscous damper c
.
fs = kx
fd = cx
P(t)
.
(1)
(2)
f I = m x
..
2
2
;
dx d x
x x
dt dt
 
( )
mx cx kx P t
  
( )
mx cx kx P t
  
x(t) = displacement measured from position of static
equilibrium
P(t) = time varying load acts on the mass block causing
dynamic risk

13. 7. Explain site specific response spectra
• Ground shaking is amplified at “soft soil” (low
velocity) sites
• Shear-wave velocity is commonly used to
predict amplification
– VS30 ( time it takes for a shear wave to travel from a
30 m depth to the land surface, i.e., time-averaged
30-m velocity)
• Low-velocity soil site gives much higher ground
motion than rock site. Vs30 is a good predictor
of site response.

14. 2. What is meant by negative damping
• Negative damping is a special case, when the
system is such that damping adds energy to the
system instead if it being dissipated.
• In this damping the amplitude tends to
increase which lead to instability of the system.
• This type of supplying energy to the system is
known as negative damping
• It occurs in the case of wires of transmission
lines.

15. Define critical damping
• Critical damping provides the quickest
approach to zero amplitude for a damped
oscillator. With less damping (under damping)
it reaches the zero position more quickly, but
oscillates around it. With more damping (over
damping), the approach to zero is slower.
Critical damping occurs when the damping
coefficient is equal to the undammed
resonant frequency of the oscillator.

17. What is overdamping?
• Over damping of a damped oscillator will
cause it to approach zero amplitude more
slowly than for the case of critical damping.
The damping coefficient is greater than the
undamped resonant frequency .

18. What is passive damping?
• Passive damping refers to the
energy dissipation within the
structure by damping devices
such as isolator, by structural
joints, or by structural member's
internal elements.

19. 1. Define viscous damping 2011N
• The concept of viscous damping is generally used to
represent the energy dissipated by the structure in
the elastic range.
• Such dissipation is due to various mechanisms such
as cracking, nonlinearity in the elastic phase of
response, interaction with non-structural elements,
soil-structure elements etc.
• The elastic viscous damping represents the
combined effect of all of the dissipation
mechanisms

20. 13b. List out some of the disastrous earthquakes 2010N
Date Location Deaths Magnitude Comments
January 23, 1556 China, Shansi 830,000 ~8
December 26,
2004
Indonesia,
Sumatra
283,106 9.1 Deaths from
earthquake & tsunami
July 27, 1976 China,
Tangshan
255,000
(official)
8.0 Estimated death toll as
high as 655,000
August 19, 1138 Syria, Aleppo 230,000
May 22, 1927 China near
Xining
200,000 8.3 Large fractures
December 22, 856 Iran,
Damghan
200,000
December 16,
1920
China, Gansu 200,000 8.6 Major fractures,
landslides
March 23, 893 Iran, Ardabil 150,000
September 1, 1923 Japan,
Kwanto
143,000 8.3 Great Tokyo fire

21. A three storied symmetrical RC school building
situated at Bhuj with the following data 2011N,2012J (16 marks)
• Plan dimension :7.2 m
• Storey height : 3.6 m
• Total Weight of beams in a storey : 135kN
• Total Weight of slab in a storey : 260kN
• Total Weight of column in a storey : 55kN
• Total Weight of walls in a storey : 540kN
• Live load : 140kN
• Weight of terrace floor : 660kN
• The structure is resting on hard rock. Determine the
total base and lateral loads at each floor levels for
5% damping using seismic coefficient method.

22. Solution
• Step 1 Determination of Natural Period
• The fundamental natural period is calculated by
assuming infill panels are provided
• T = 0.09h/√
d= 0.09x10.8/ √7.2 = 0.362 s
• Step 2 Determination of other important factors
• For T= 0.362 s damping of 5 % and for hard rock
sa/g = 2.5
• Bhuj is situated in Zone V, Zone factor z= 0.36
• Since the building is used as a school building the
Importance factor I = 1.5
• For a special moment resisting frame, the Response
Reduction Factor, R = 5

23. • Step 3 Determination of design horizontal seismic
coefficient
• The design horizontal seismic coefficient Ah=zIsa/2Rg =
0.36x1.5x2.5/2x5 = 0.135
• Step 4 Determination of seismic weight
• Weight of one storey = Total weight of
(beams+Slab+Columns+Walls+Live load)
• = (135+260+55+540+140)kN = 1130kN
• Weight of terrace floor : 660kN
• Total weight of building = (2x1130) + 660 = 2920kN
• Step 5 Determination of base shear
• Design Base shear VB= Ah W = 0.135x2920=394.2 kN

24. • Step 6 Distribution of equivalent lateral load
• The design base shear computed shall be
distributed along the height of the building as per
the following formula
• Where is calculated from base

25. Check VB = Q1+Q2+Q3
= 16.95+115.99+261.25 = 394.2 kN
Hence OK.

26. 3. Write a short note on free vibration analysis 2011N
• The vibration analysis of real system consists of the
following steps
• 1. Mathematical modeling of a real system :
Mathematical model is used to determine the nature of
the system, its features and aspects and components
involved in the real system
• 2. Formulation of governing equations: After
developing the mathematical model, the differential
equations that govern the behavior of the vibrating
system are obtained by using the law of mechanics and
principles of dynamics. Several concepts of the
principles of dynamics such as Newton’s second law of
motion, D’Alembert’s principle and the principle of
conservation of energy are most commonly used to
formulate the governing equations of motion .

27. • 3. Solutions of the governing equations of motion:
The governing equations of motion of a vibrating
system are solved to find the response of an SDOF
system lead to one ordinary differential equations
of motion in the form of second order linear
differential equation of motion.
• Vibrations of MDOF system lead to a system of
ordinary differential equations of motion.
• 4. Interpretations of results: The solution of the
equation of motion of the real physical system
gives the displacement, velocity, acceleration of
the various masses or inertia of the system

28. 8. Write short note on soil properties for dynamic
analysis 2012J
• The soil spring constant kz directly depends on
the dynamic soil shear modulus G and poisson’s
ratio .
• The unit weight s is needed to calculate the soil
density .
• If the base is resting on lose soil, then it would be
either dandified or stiffened with admixtures,
soil-cement piles or stone columns or the base
would be placed on piles.

29. Differences between Response and Design Spectra
• The response spectrum is a description, in terms
of its peak effects, of a particular ground motion.
• The design spectrum is a specification, valid for a
site or a class of sites, of design seismic forces.
• If a site falls in two different classifications, e.g.,
the site is near to a seismic fault associated with
low magnitude earthquakes and it is distant from
a fault associated with high magnitude
earthquakes, with the understanding that the
frequency contents of the two classes of events
are quite dissimilar the design spectrum should
be derived from the superposition of the two
design spectra.

31. 2. Write a short note on earthquake excitation 2011M
• The most important quantity related to
earthquake excitation is the ground acceleration.
• Ground acceleration can be recorded with an
accelerometer, basically a SDOF oscillator, with a
damping ratio  70%, whose displacements are
proportional to ground accelerations up to a
given frequency.
• Instrument records of strong ground motion first
became available in the ’30s, the first record of a
destructive ground motion being the 1940
records of El Centro earthquake.

32. 14a. Explain the concept of response spectrum. How will you
construct response spectrum for a given ground motion.
• Response spectrum introduced by M.A. Biot in 1932.
Response spectrum provides a convenient means to
summarize the peak response of all possible linear
SDOF system to a particular component of ground
motion. It also provides a practical approach to apply
the knowledge of structural dynamics to the design of
structure and development of lateral force
requirements in building codes.
• A plot of a peak value of a response quantity as a
function of a natural period ‘To’ or angular frequency
‘n’ or cyclic frequency ‘To’ is called the response
spectrum for that quantity. So such plot is for SDOF
system having fixed damping ratio ‘ξ’.

33. 14a. Explain the concept of response spectrum. How will you
construct response spectrum for a given ground motion.
• A variety of response spectrum can be defined
depending on the response quantity that is plotted.
Consider the following peak response
• uo(Tn , ξ)  maxt u(t,Tn , ξ)- Deformation response spectrum
• ůo(Tn , ξ)  maxt ů(t,Tn , ξ) - Relative velocity response spectrum
• üo(Tn , ξ)  maxt ü (t,Tn , ξ) - Acceleration response spectrum
• Construction of Response Spectrum
• 1. Define ground acceleration üg(t); typically the ground
motion ordinates are defined every 0.02 sec
• 2. Select the natural vibration period and damping ratio
ξ of an SDOF system.
• 3. Compute the deformation response u(t) of SDOF due
to ground acceleration üg(t)

34. 14a. Explain the concept of response spectrum. How will you
construct response spectrum for a given ground motion.
• 3. Compute the deformation response u(t) of SDOF
due to ground acceleration üg(t)
• 4. Determine uo the peak value of u(t).
• 5. The spectral co-ordinates are
D= uo , V = (2Tn )D, and A = (2Tn )2D
• 6. Repeat steps 2 to 5 for a range of Tn and ξ values
of covering all possible systems of engineering
interests
• 7. Present the results of steps 2 to 6 graphically to
produce three separate spectra or a combined
spectrum

35. 14b. Explain the types of liquefaction and effects of
liquefaction 2012J
• Liquefaction can be divided into two main categories
flow liquefaction and cyclic mobility
• Flow liquefaction: Flow liquefaction is a phenomenon
in which the static equilibrium is destroyed by static
and dynamic loads in a soil deposit with low residual
strength.
• Residual strength is the strength of a liquefied soil.
• It occurs when the static shear stresses in the soil
exceed the shear strength of the liquefied soil.
• This will cause large deformation in the soil.
• For example, static loading can be applied by new
buildings on a slope that exerts additional forces on the
soil beneath the foundations.

36. • Earthquake blasting and pile driving are all examples of
dynamic loads that could trigger flow liquefaction.
• Once triggered, the strength of a soil susceptible to flow
liquefaction is no longer sufficient to withstand the static
stresses that were acting on the soil before the disturbance.
• Flow failures can involve the flow of considerable volume of
material, which undergo very large movements that are
actually driven by static stresses.
• Cyclic mobility: Cyclic mobility is a liquefaction phenomenon ,
triggered by cyclic loading occurring in soil deposits with static
shear stresses lower than the soil strength.
• Deformation due to cyclic mobility develop incrementally
because of static and dynamic stresses that exist during an
earthquake.
• Lateral spreading , a common result of cyclic mobility, can occur
on gentle sloping and on flat ground close to rivers and lakes.

37. Effects of liquefaction
• Typical effects of liquefaction include:
• 1. Loss of bearing strength: The ground can liquefy
and lose its ability to support structures.
• 2. Lateral spreading: The ground can slide down
very gentle slopes or stream banks riding on a
buried liquefied layer. It is mainly caused by cyclic
mobility. Lateral spreading causes damage to
foundations of buildings, pipelines, railway lines and
cause shaking of piles due to increased lateral loads.
• 3. Sand boils: Sand laden water can be ejected from
a buried liquefied layer and erupt at the surface to
form sand volcanoes; the surrounding ground often
fractures and settles.

38. • 4. Flow failures: Here earth moves down steep slopes
with large displacement and mush internal disruption
of material. This is caused by flow liquefaction.
• 5. Ground oscillation: This affects flat ground. The
liquefied sediments starts to slosh into waves as
shaking continues. Whatever is on top of the sediment
gets broken and thrown around. Cracks in the ground
open and close, and water or mud may erupt from
them.
• 6. Floatation: Light structures that are buried in the
ground (like pipelines, sewers and nearly empty fuel
tanks) can float to the surface when they are
surrounded by liquefied soil.
• 7. Settlement: Liquefied ground reconsolidates
following an earthquake, the surface may settle or
subside as shaking decreases and the underlying
liquefied soil becomes more dense.

39. 14bii) Explain any four methods to reduce liquefaction 2012J
• 1. Avoid liquefaction-susceptible soils: The first possibility is
to avoid construction on liquefaction susceptible soils.
• There are various criteria to determine the liquefaction
susceptibility of a soil. By characterizing the soil at a particular
building site according to these criteria, one can decide if the
site is susceptible to liquefaction and therefore unsuitable for
the desired structure.
• 2. Build liquefaction-resistant structures: If it is necessary to
construct on liquefaction susceptible soil due to reasons such
as space restrictions, favorable location, or other reasons, it
may be possible to make the structure liquefaction resistant
by designing foundation columns to resist the effects of
liquefaction. Structure that possesses ductility, has the ability
to accommodate large deformations, adjustable supports for
correction of differential settlements.

40. • 3. Shallow foundation aspects: It is important that all
foundation elements in a shallow foundation are tied
together to make the foundation move or settle
uniformly, thus decreasing the amount of shear forces
induced in the structural elements resting upon the
foundation.
• The well-reinforced perimeter and interior walls are
tied together to enable them to bridge over area of
local settlement and provide better resistance against
soil movements.
• A stiff foundation mat is a good type of shallow
foundation, which can transfer loads to from locally
liquefied zones to adjacent stronger ground.
• Buried utilities, such as sewage and water pipes, should
have ductile connections to the structure to
accommodate the large movements and settlements
that can occur due to liquefaction.

41. • 4. Deep foundation aspect: Liquefaction can cause
large lateral loads on pile foundations.
• Piles driven through a weak, potentially liquefiable, soil
layer to a stronger layer not only have to carry vertical
loads from superstructure, but must also be able to
resist horizontal loads and bending moments induced
by lateral movements if the weak layer liquefies.
• Piles of larger dimensions and/or more reinforcement
can achieve sufficient resistance.
• It is important that the piles are connected to the cap in
a ductile manner that allows some rotation to occur
without a failure of the connection.
• If the pile connection fail, the cap cannot resist
overturning moments from the superstructure by
developing vertical loads in the piles.

42. 8. What are different effects of liquefaction (any four).
2011N
• 1. Loss of bearing strength
• 2. Lateral spreading
• 3. Sand boils
• 4. Flow failures
• 5. Ground oscillation
• 6. Floatation
• 7. Settlement

43. 13b. Describe any 4 recent Indian earthquakes and explain
how properties of soil are destroyed 2011N
• 1. Koyna earthquake of 1967: An earthquake of magnitude
6.5 occurred close to the 103 m high concrete gravity dam at
Koyna.
• Prior to this earthquake, this area has been considered
aseismic.
• But after construction of dam, the seismic activity increased
significantly because of filling and emptying of reservoir in
1962.
• This earthquake caused widespread damage such as killing
about 200 persons and injuring more than1500 people.
• This earthquake is considered as an example for the reservoir-
induced seismicity.
• This earthquake led to the revision of Indian seismic zone
map wherein the area around Koyna was changed from Zone I
to Zone IV.

44. 2. Uttarkashi earthquake of 1991
• An earthquake of magnitude 6.6 struck the districts
of uttarkashi, Tehri and Chamoli in the state of Uttar
Pradesh on October 20, 1991.
• About 768 persons lost their lives, with about 5,066
injured.
• Maximum peak ground acceleration of about 0.31
was recorded at Uttarkashi .
• Many four storey building in Uttarkashi with RC
frame and infill walls sustained the earthquake.
However some of the RC building collapsed.

45. 3. Killari (Latur) Earthquake of 1993
• A magnitude of 6.4 earthquake was felt on
September 30, 1993 at Killari in Latur district killing
about 81,000 persons .
• Prior to this event this area was considered as a
seismic and placed in lowest seismic zone (zone I)
by the Indian code (IS1893-1894).
• This event led to several studies, giving a new
perspective to seismic hazard assessment in the
peninsular India.
• It also led to strengthening of the seismic network,
upgrading several existing facilities.

46. 4. Jabalpur earthquake of 1997
• An earthquake of magnitude 6 struck the Jabalpur
city on May 22, 1997.
• Several RC frame buildings with open first storey
were damaged due to failure of ground storey
columns.
• However, numerous RC frame buildings of three to
four story's with brick infill walls performed well,
because this brick walls acted as shear walls.
• Thus this earthquake helped to understand the
response of various types of structures offering
valuable guidelines in the design and construction
of earthquake resistant structures.

47. 5. Bhuj earthquake of 2001
• A powerful earthquake of magnitude 7.7 struck the kutch region
of Gujarat on the 26th of January 2001.
• The earthquake was felt over a large part of the country and as far
as Nepal, Delhi, Calcutta, Bombay and Chennai.
• The greatest damage occurred in the region of kutch.
• At the end of March, the official estimate of causalities was
20,000.
• The number injured reported to be 1,66,000 of which 20,700
suffered serious injury.
• It is estimated that about 3,70,000 houses and huts were
completely destroyed while another 9,31,000 partially destroyed.
• The total financial loss is reported as Rs. 21,300 crores.
• A number of hospital buildings, telephone exchange buildings, civil
administration buildings, and water service buildings were
damaged seriously hampering the rescue and relief operations.
• Many building collapsed because of improper detailing.
• The earthquake caused extensive ground movement, cracking,
liquefaction and lateral spreading in the region of kutch.

48. 13b. Explain the lessons learnt from the past earthquakes 2011M
• For RC framed structures:
• Avoid soft story ground floors-often the columns are
damaged by the cyclic displacements between the moving
soil and upper part of the building.
• Avoid soft storey upper floors
• Avoid short columns. These attract large shear and failure.
• Avoid open ground floor and discontinuous columns.
• Provide suitable foundation based on the soil condition.
• Adopt seismic resistant design for the buildings which are
located in the zone of seismicity.
• For Masonry building:
• Poor wall connections at corners often lead to failure of wall.
• Infill walls with large openings leading to extensive damage.
• Rigid roof slab, excessive load and improper distribution can
lead to roof failure.

49. What is the importance of ductility 2010N
• Concept of earthquake design philosophy is achieved
by introducing ductility at the predetermined
positions in the structure.
• This will enable the structure to absorb energy during
earthquakes to avoid sudden collapse of the structure.
• The term ductility is defined as the ability of a
structure to undergo inelastic deformations beyond
the initial yield deformations without increase in load
resistance.
• The magnitude of earthquake forces induced in a
structure will mainly depend on ductility of the
structure.

50. Importance of ductility
• By introducing the ductility in earthquake resistant
buildings, they have the ability to reverse large
lateral deformation before failure during an
earthquake and withstand earthquake effects with
some damage but without collapse.
• This is beneficial to the users of the structure as in
case of earthquake if the structure is to collapse, it
will undergo large deformation before failure.
• This gives warning to the occupants and provides
sufficient time for taking preventive measures.
• Hence it is important for the structure to behave as
a ductile structure.

51. What are different types of damping?
• 1. Viscous damping
• 2. Coulomb damping
• 3. Structural damping
• 4. Active damping or Negative damping
• 5. Passive damping

52. What is viscous damping
• When the 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.
• The damping force in this case is proportional
to the velocity.
• Where C is damping constant
x
C
Fv



53. What is Coulomb damping
• Dry friction occurs when the motion of a body
is on a dry surface. The damping force is given
by
• Fc = N
• Where  - coefficient of kinematic friction
• N – Normal reaction

54. What are different types of vibrations
• 1) Free and Forced vibration.
• 2) Linear and Non-linear vibration
• 3) Damped and Undammed vibrations
• 4) Deterministic and Random vibrations
• 5) Longitudinal, Transverse and Torsional vibrations

55. What is response of the system
• Response is defined as the magnitude and distribution
of the resulting forces and displacements in a system
due to vibration.
• When the dynamic load is applied to the structure it
produces displacement, velocity, acceleration and also
develops stresses, strains and reaction etc.,
• Hence , the response of a system indicates any of the
effects produced by the dynamic load.
• The motion is due to initial condition is known as Free
response. When the motion is due to applied forces is
known as Forced response.

56. Equation of motion for viscous damping
• Let us consider a mass m attached to the one end of
• a linear spring having stiffness of k and also
connected by means of a dashpot or damper that
provides a viscous damping with a damping
coefficient ‘c’ as shown in figure. The force provided
by a dashpot is equal to .
• From the FBD, the governing differential equation
of motion can be written as
• (1)
m
k
c
fs = kx
fd = cx
.
f I = m x
..
x
c
0


 kx
x
c
x
m 



57. • Assuming solution ,may be in the form of
• Where  is a constant to be determined. This exponential
function leads to algebraic equation
• Substituting the values of in Eq. (1) we get
• (2)
• The above equation is known as characteristics equation
with two roots
t
e
x 

t
e
x 
2



t
e
x 



x
and
x
x 


,
0
0
0
)
(
0
2
2
2
2












m
k
m
c
k
c
m
e
k
c
m
ke
e
c
e
m
t
t
t
t













58. (3)
• The radical term is
• The roots of the characteristic equation falls
into one of the following cases
m
k
m
c
m
c
m
km
c
m
c
m
km
c
m
c
km
c
m
m
c
m
km
c
m
c
m
k
m
c
m
c
4
2
1
2
4
2
1
2
4
2
1
2
2
4
1
2
4
2
4
2
2
2
2
2
,
1
2
2
2
,
1
2
2
2
2
,
1
2
2
,
1








































m
k
m
c 4
2
2


59. • 1. If the discriminant is positive (greater than zero), the
system is termed as over damping system. The roots of the
characteristic equation are purely real and distinct
corresponding to simple exponentially decaying motion.
• 2. If the discriminant is negative, the system is termed as
under damped system. The roots of the characteristic
equation are complex conjucates, corresponding to
oscillatory motion with an exponential decay in amplitude.
• 3. If the discriminant is equal to zero, the system is termed
as critically damped system. The roots of the characteristic
equation are equal corresponding to simple decaying
motion.

60. Equation of motion for critically damped system
• When the discriminant value become zero, the special value of
damping present in the system is called as critical damping
coefficient.
• (4)
• Thus equation (4) gives the relation between critical damping
coefficient and natural frequency.
• The ratio of actual damping to the critical damping coefficient is
called as damping ratio. It is denoted by a symbol . It can be written
as
• (5)
n
c
c
m
km
c
km
m
m
k
c
m
k
m
c

2
2
4
4
0
4
2
2
2
2






c
c
c



61. • Damping ratio is one of the important parameter in damping.
• For a critical damping, the discriminate must be equal to zero that
means c tends to cc . Thus
•
• That is =1 for a critically damped system. In which roots of the
characteristic equation (3). Let  and  are the roots of the equation
• Whenever the roots are equal, the general solution is
• (6)
• In which c/2m can be rearranged as
1

c
c
c
m
c
2
, 



 
 
Bt
A
e
t
x
Bt
A
e
t
x
t
m
c
t





2
)
(
)
( 

62. • Substituting in equation (6)
• For critical damping system  =1 . Thus
• This is the general solution for a critically damped system.
• Critical damping may be defined as the minimum amount
of damping for which the system will not vibrate initially,
but it will return to equilibrium position.
n
c
c
c
c
m
km
c
c
c
km
m
c
c
c
m
c
m
c








2
2
2
2
2
2
n
m
c


2
 
Bt
A
e
t
x t
n

 
)
(
 
Bt
A
e
t
x t
n

 
)
(

63. • This will result in non periodic motion (i.e., simple
decay). The displacement decays to a negligible
level after one natural period T.

64. Equation of motion for over damped system (>1)
• In over damped system, the damping coefficient is higher than the amount
of critical damping that is c>cc . Hence,  and  are real and distinct, they
are
• Multiplying and dividing by n we get
m
k
m
c
m
c



 2
2
4
2
, 

 
 
 
1
1
2
2
2
4
2
4
2
,
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2






































































n
c
c
c
n
c
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
c
c
c
c
c
c
c
m
k
m
m
k
m
c
m
c
m
k
m
k
m
c
m
c
m
k
m
c
m
c




65. • Thus the general solution for an overdamped
system is
• (7)
• The motion of an overdamped system is non-
periodic, regardless of the initial conditions. If the
damping is higher, it will take long time to decay
from an initial disturbance.
 

















 






 






 







 


t
t
t
t
t
t
t
n
n
n
n
n
n
n
Be
Ae
e
x
Be
Ae
Be
Ae
x
1
1
1
1
2
2
2
2














67. Equation of motion for an underdamped system (<1)
• For an underdamped system, the discriminant is negative.
The discriminant is negative only when the damping
coefficient is less than critical damping coefficient. Thus the
roots of the characteristic equations are complex (or
imaginary) they are
• Therefore the solution of an underdamped system is
• It can also be written as (8)
 
1
1
,
2
2


















n
n
n













 






 

t
i
t
t
i
t n
n
n
Be
Ae
e
x
2
2
1
1 




 
t
i
t
i
t d
d
n
Be
Ae
e
x 

 




68. • Where ; d is called damped natural frequency.
• Equation (8) can be written as
• In which the term causes vibration.
• Damping period
• For this case, the system will vibrate at the damped natural frequency
which is a function of natural undamped frequency (n) and damping
ratio. The natural damped frequency is less than the undamped
frequency but for many practical cases the damping ratio is relatively
small and hence the difference is negligible.
2
1 

 
 n
d
 
 
 








motion
Periodic
decay
l
exponentia
sin
sin
cos












t
Xe
x
t
B
t
A
e
x
d
t
d
d
t
n
n
 

 
t
d
sin
2
1
2
2








n
d
d
T
t
n
Xe
Amplitude 



70. A harmonic motion has maximum velocity of 6 m/s and it
has a frequency of 12 cps. Determine its amplitude, its
period and maximum acceleration.
• Given v = 6 m/s
• f = 12 cps
• =2πf=2  12=75.398 rad/s
• Period T= 1/f= 1/12 = 0.08 sec
• A = 79.6 mm

71. 11. b(ii) A 200 kg machine is placed at the end of 1.8 m long steel
(F=210x109 N/m2) cantilever beam. The machine is observed to
vibrate with a natural frequency of 21 Hz. What is the moment of
inertia of the beam cross section about its neutral axis?
• 1.8 m 200 kg
• f = 21 Hz
•  = 2f = 6.28x 21= 131.9 rad/sec
4
5
9
3
6
3
3
6
2
10
22
.
3
10
210
3
8
.
1
10
48
.
3
3
3
/
10
48
.
3
200
9
.
131
9
.
131
m
E
kL
I
L
EI
k
m
N
k
m
k
m
k
n


















72. A mass of 1 kg is suspended by a spring having stiffness of 600 N/m. The
mass is displaced downward from its equilibrium position by a distance
of 0.01m. Find a) Equation of motion of the system b) Natural
frequency of the system c) The response of the system as a function of
time d) Total energy of the system.
• Given details m = 1 kg
K= 600 N/m
st = 0.01m
• The equation of motion is given by
• The natural frequency is given by
• Response of the system is given by
Hz
f
s
rad
m
k
n
n
898
.
3
2
/
49
.
24
1
600








0
600
1
0




x
x
kx
x
m




)
sin( 
 
 t
A
x n

73. • Amplitude
• Phase angle
• Response
• The total energy is equal to the maximum kinetic energy or
maximum potential energy.
• We know that
m
x
x
A
n
01
.
0
)
01
.
0
( 2
2
0
2
0 













 
2
tan
0
49
.
24
01
.
0
tan
tan 1
1
0
0
1 

 







 







 


x
x n

 
 
2
49
.
24
sin
01
.
0 

 t
x
 
   
m
N
KE
PE
m
N
A
k
mv
KE
m
N
kx
PE
n
03
.
0
03
.
0
01
.
0
49
.
24
1
2
1
2
1
2
1
03
.
0
01
.
0
600
2
1
2
1
max
max
2
2
2
max
2
2
max
















74. A damped free vibration test is conducted to determine the dynamic
properties of a one storey building. The mass of the building is 10,000
kg. Initial displacement of the building is 0.702 cm. maximum
displacement on the first cycle is 0.53 and period of the displacement
cycle is 1.7 s. Determine the effective weight, undamped frequency,
logarithmic decrement, damping ratio, damping coefficient, damped
frequency, and the amplitude after the 6 cycles. Nov.2011
• Given details
• m = 10000 kg T= 1.7 s n=6 x0=0.702
• Effective weight of the building:
• Undamped frequency
• We know that
s
T
n 695
.
3
7
.
1
28
.
6
2





m
N
k
k
m
k
n
/
10
365
.
1
10000
695
.
3
5







2
2
5
2
2
4
7
.
1
10
3
.
1
81
.
9
4
2










gkT
W
gk
W
T

75. Effective weight W = 93.36 kN
• We know
m
s
N
c
c
km
c
c
c
e
e
x
x
c
/
59
.
3251
10
365
.
1
10000
2
044
.
0
2
044
.
0
28
.
6
28
.
0
2
2
28
.
0
324
.
1
53
.
0
702
.
0
5
1
0






























76.  
 
cm
x
e
e
x
x
e
x
x
s
rad
n
n
d
n
d
1308
.
0
702
.
0
/
69
.
3
044
.
0
1
695
.
3
1
6
28
.
0
6
6
0
6
0
2
2



















77. A vibrating system consists of a mass 5 kg spring of stiffness
120N/m and a damper with a damping coefficient of 5 N-s/m.
Determine a) the damping factor, b) natural frequency of damped
vibration, c) logarithmic decrement, d) the ratio of two successive
amplitude and e) the number of cycles after which the initial
amplitude is reduced to 25%
• Given details
• M = 5 kg k = 120 N/m c= 5 N-s/m
• Critical damping coefficient
• Natural frequency
• a) Damping factor
b) Natural damped frequency
m
s
N
km
cc /
49
5
120
2
2 




s
rad
m
k
n /
9
.
4
5
120




102
.
0
49
5



c
c
c

  s
rad
n
d /
87
.
4
102
.
0
1
9
.
4
1
2
2




 



78. • c) Logarithmic decrement
d)The ratio between two consecutive amplitude say
• e) Number of cycles after the reduction of 25%
•
 
64
.
0
102
.
0
1
102
.
0
2
1
2
2
2










2
1
2
1
64
.
0
2
1
2
1
896
.
1
ln
x
x
x
x
e
x
x
e
x
x






2
1
x
x
cycles
cycles
x
x
x
x
x
x
n
x
x
n
n
n
n
2
166
.
2
4
ln
64
.
0
1
4
ln
64
.
0
1
4
ln
1
ln
1
1
1
1
1
1











79. A damper offers resistance 0.08 N at a constant velocity 0.06m/s. The
damper is used with a spring of stiffness equal to 12N/m. Determine
the damping ratio and frequency of the system when the mass of the
system is 0.3 kg
• Given details
• Damping force
• k= 12N/m m= 0.3 kg
• Damping ratio
• Hence the system is underdamped.
N
x
c
F 08
.
0

  s
m
x /
06
.
0


m
s
N
km
c
m
s
N
c
N
x
c
c /
79
.
3
3
.
0
12
2
2
/
33
.
1
06
.
0
08
.
0
08
.
0










 
1
35
.
0
79
.
3
33
.
1




c
c
c


80. The damped frequency
A single degree of freedom system having a mass of 2.5 kg is
set into motion with viscous damping and allowed to
oscillate freely. The frequency of oscillation is found to be
20 Hz and measurement of the amplitude of vibration
shows two successive amplitudes to be 6mm and 5.5mm.
Determine the viscous damping coefficient.
Given details
m = 2.5 kg fd= 20 Hz x1 =6mm x2 =5.5mm
s
rad
m
k
n /
32
.
6
40
3
.
0
12





  s
rad
n
d /
92
.
5
35
.
0
1
32
.
6
1
2
2




 



81. Logarithmic decrement
• For small value of ,
• Damped natural frequency
• For small value of ,
•
087
.
0
5
.
5
6
ln
ln
2
1



x
x

014
.
0
28
.
6
087
.
0
2
2









s
rad
fd
d /
6
.
125
20
28
.
6
2 


 

6
.
125
014
.
0
1
6
.
125
1
1
2
2
2













d
n
n
d
n
d 
 
m
N
k
m
k n
m
k
n
m
k
n
/
10
394
.
0
6
.
125
5
.
2 5
2
2
2












82. • A spring mass system with Coulomb damping has a mass of 10 kg
attached to a spring of stiffness 1200 N/m. If the coefficient of
friction is 0.03, calculate a) The frequency of free vibration b)
Number of cycles corresponding to 50% reduction in amplitude if the
initial amplitude is 7 cm; and c) The time taken to achieve 50%
reduction.
• Given details
• m = 10 kg k= 1200 N/m =0.03
• Frictional force F= mg = 0.03x10x9.81 = 2.943 N
m
s
N
km
c
km
c
c
c
c
/
41
.
6
5
.
2
10
394
.
0
014
.
0
2
2
2
5












Hz
f
s
rad
n
m
k
n
74
.
1
28
.
6
95
.
10
2
/
95
.
10
10
1200










83. Amplitude after 50% reduction is half of the initial
amplitude which is equal to 3.5 cm
• Time taken to achieve 50% reduction= 4T=4x(2/10.95)
• =2.316 s
•
mm
k
F
81
.
9
10
81
.
9
•
1200
943
.
2
4
4
cycle
per
amplitude
in
Reduction
3







cycles
4
56
.
3
10
81
.
9
035
.
0
reduction
50%
in
completed
be
to
Cycles
• 3



 

84. An SDO F system is subjected to free vibration with an initial velocity
V0 without any initial displacement. Determine subsequent motion
of the system for the three damping ratios = 2.5, = 1.0 and = 0.1
• Case 1 = 2.5>1. Hence overdamping
• The given initial conditions are (i) At t=0; x =0
(ii) At t=0; = Vo
Applying initial conditions (i) i.e., At t=0; x =0
0 = A+B  A = -B
t
t
t
t
t
t
n
n
n
n
n
n
Be
Ae
x
Be
Ae
Be
Ae
x










79
.
4
2087
.
0
1
5
.
2
5
.
2
1
5
.
2
5
.
2
1
1
2
2
2
2







 







 







 







 




















x


85. Applying initial conditions (ii) i.e., At t=0; = Vo
0 = A+B  A = -B
• The subsequent motion of the system is
   
n
t
n
t n
n
Be
Ae
x 
 

79
.
4
2087
.
0 79
.
4
2087
.
0



 


x

 
n
o
n
o
n
n
o
n
n
o
V
B
V
B
B
B
V
B
A
B
A
V






218
.
0
5813
.
4
79
.
4
2087
.
0
79
.
4
2087
.
0












 
 
t
t
n
n
n
e
e
V
x 


79
.
4
2087
.
0
0
218
.
0 




86. Case 2 =1;
Since =1, the system is critically damped
• Applying initial conditions
• At t= 0; x=0
• 0=A
• (ii) At t=0; = Vo
• Since A=0; Vo =B
• The subsequent motion of the system is
 
 
Bt
A
e
x
Bt
A
e
x
t
t
n
n









 1

x

   
 
  B
A
V
e
te
B
Ae
x
Bte
Ae
x
n
t
n
t
t
n
t
t
n
n
n
n
n

















 




0

t
n
te
V
x 

 0

87. Case 3: =0.1;
Since  <0.1, the system is underdamped
• The equation of motion is
• where
•
• Applying initial condition
• At t= 0; x=0
• 0 = A
• At t=0; = Vo =0.995nB
• The subsequent equation of motion is
n
V
B

0
005
.
1


 
t
B
t
A
e
x d
d
t
n



sin
cos 
 
 
 
 
t
B
t
A
e
x
t
B
t
A
e
x
n
n
n
n
t
n
n
t
n
n
n
d
n
n













995
.
0
cos
995
.
0
995
.
0
sin
995
.
0
995
.
0
sin
995
.
0
cos
995
.
0
1
.
0
1
1
2
2













x

t
V
e
x n
n
t
n



sin
005
.
1 0
1
.
0



88. A one kg mass is suspended by a spring having a
stiffness of 1N/mm. Determine the natural frequency
and static deflection of the spring.
• Given details: k = 1N/mm = 1000N/m
• m= 1 kg
• Natural frequency
• Static deflection st
• We know that
Hz
f
s
rad
m
k
n
n
03
.
5
2
/
62
.
31
1
1000








mm
g
s
rad
g
n
st
st
n
81
.
9
62
.
31
81
.
9
/
62
.
31
2
2










89. A system vibrating with a natural frequency of 6 Hz starts with an initial
amplitude (x0) of 2 cm and an initial velocity ( ) of 25cm/s. Determine the
natural period, amplitude, maximum velocity, maximum acceleration and
phase angle. Also write the equation of motion of a vibrating system.
• Given details: f = 6 Hz x0 = 2 cm = 25 cm/s
• The natural period is given by
• The amplitude of motion
• where n = 2f = 2x6 = 37.7 rad/s
• The maximum velocity of a system is
0
x

0
x

s
f
T 167
.
0
6
1
1



2
0
2
0 









n
x
x
A


s
cm
A
x n /
44
.
79
7
.
37
11
.
2
max 


 

cm
A 11
.
2
7
.
37
25
2
2
2










90. • The maximum acceleration of a system is
• Phase angle
• Equation of motion is
rad
x
x n
25
.
1
"
23
'
39
71
25
7
.
37
2
tan
tan 1
0
0
1







 







 
 



)
25
.
1
7
.
37
sin(
11
.
2
)
sin( 


 t
t
A
x n 

2
2
2
max /
76
.
2994
7
.
37
11
.
2 s
cm
A
x n 


 



91. A vertical cable 3 m long has a cross sectional area of 50 kN.
What will be the natural period and natural frequency of the
system?
• Given details: A= 4 cm2 w = 50 kN
• Stiffness
• Natural frequency
• Frequency
kg
g
w
m 8
.
5096
81
.
9
10
50 3




2
6
/
10
1
.
2 cm
kg
E 

2
6
/
10
1
.
2 cm
kg
E 

m
N
cm
kg
L
AE
k
/
10
746
.
2
981
28000
/
28000
300
10
1
.
2
4
7
6









s
T
s
rad
m
k
n
n
085
.
0
41
.
73
28
.
6
2
/
41
.
73
8
.
5096
10
746
.
2 7










Hz
T
f 76
.
11
085
.
0
1
1




92. A cantilever beam 3m long supports a mass of 500 kg at
its upper end. Find the natural period and natural
frequency. E = &I = 1300 cm4 .
Stifness for a cantilever beam
Natural frequency
Natural period
2
6
/
10
1
.
2 cm
kg
E 

 
m
N
cm
kg
L
EI
k
/
10
97
.
2
981
303
/
303
300
1300
10
1
.
2
3
3
5
3
6
3










Hz
f
s
rad
m
k
n
n
88
.
3
28
.
6
37
.
24
2
/
37
.
24
500
10
97
.
2 5










s
f
T
n
26
.
0
88
.
3
1
2
1







93. A cantilever beam AB of length L is attached to a spring k and
mass m as shown in figure. a) Form the equation of motion
and b) Find the expression for the frequency of motion
• A)Equation of motion
• Stiffness due to the applied mass M is k1
• The stiffness k1 is acting parallel to k
• Equivalent spring stiffness ke = k1 +k
3
1
3
L
EI
k 
3
3
3
3
3
L
kL
EI
k
L
EI
ke





94. • The differential equation of motion is
• b) The frequency of vibration is
0
3
0
3
0
3
3
3
3






 








 







x
m
L
kL
EI
x
x
L
kL
EI
x
m
x
k
x
m
x
k
x
m
e
e








3
3
3
2
1
2
1
mL
kL
EI
m
k
f e 





95. Find the natural frequency of the system as shown in figure.
Take k1 = k2 = 2000 N/m k3 =3000 N/m and m = 10 kg
• Two springs k1 and k2 are in parallel
• Equivalent stiffness ke1= k1 +k2= 2000+2000
• = 4000 N/m
• Again this equivalent spring is parallel to k3
• Equivalent stiffness ke = ke1 + k3 =4000+3000
• =7000 N/m
s
f
s
rad
m
k
n
e
n
21
.
4
28
.
6
46
.
26
2
/
46
.
26
10
7000










96. Consider the system shown in figure If k1 = 1500 N/m,
k2 = 2000 N/m, k3 =3000 N/m and k4 = k5 = 500 N/m, find the
mass if the system has a natural frequency of 10 Hz.
• Given details
• k1 = 1500 N/m, k2 = 2000 N/m,
• k3 =3000 N/m k4 = k5 = 500 N/m
• f = 10 Hz
• The springs k1, k2, and k3 are in series.
• Their equivalent stiffness
m
N
k
k
k
k
k
e
e
/
67
.
666
6000
2
4
3
3000
1
1500
1
2000
1
1
1
1
1
1
3
2
1
1











97. • The two lower springs k4 and k5 are connected in parallel, so
their equivalent stiffness
• ke2= k4 +k5 = 500+500 1000 N/m
• Again these two equivalent springs are in parallel
• ke = ke1 + ke2 =666.67+1000= 1666.67 N/m
• But
s
rad
f
f
n
n
/
83
.
62
10
2
2
2











 
kg
k
m
m
k
m
k
n
e
n
n
52
.
26
83
.
62
67
.
1666
2
2
2









99. Determine the natural
frequencies and mode shapes
for the shear building shown in
figure.

100. Let m3 =2
m2 =1.5
m1 =1
• Free body diagram for the equivalent system are
• (1)
• (2)
• (3)
• Writing the Eqns. (1), (2) and (3) into a matrix form,
  0
0
2
2
3
3
2
3
3
3
2
2
2
3
3
3
3








x
k
x
k
k
x
m
x
k
x
k
x
k
x
m




  0
0
3
2
3
2
1
1
1
2
2
2
1
1
1
3
2
2
2
2
2











x
k
x
k
k
x
k
x
m
x
k
x
k
x
k
x
k
x
m




0
2
1
1
1
1
1 

 x
k
x
k
x
m 


101. 0
0
0
0
0
0
0
0
0
3
2
1
2
1
2
2
2
1
1
1
1
3
2
1
3
2
1
















































x
x
x
k
k
k
k
k
k
k
k
k
x
x
x
m
m
m





