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CE-412: Introduction to Structural Dynamics and Earthquake Engineering

MODULE 5:

RESPONSE OF S.D.O.F SYSTEMS subjected to HARMONIC EXCITATIONS

Harmonic force
A harmonic force is one whose time variation is defined by any one of the following
equations
( ) = Sin( ) or Cos( )

Where po is the amplitude or maximum value of force and ω is its frequency also
called as exciting frequency or forcing frequency; T=2π/ω, is the exciting period or
forcing period.
The equations used in this module are strictly applicable to p(t) = po sin (ωt)

Time variation of harmonic force, p(t) = po sin (ωt)

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Harmonic forces: Examples

A common source of such a sinusoidal force is unbalance in rotating machines (such

as turbines, electric motors and electric generators, fans, or rotating shafts).
When the wheels of a car are not balanced, Centrifugal harmonic forces are developed
at the mass centre of the rotating wheels. If the rotational speed of the wheels (i.e.
exciting frequency, ω) is close to the natural frequency of the car’s suspension system in
vertical direction , ωn , amplitude of vertical displacement in the car’s suspension system
increases and violent shaking occur in car.

Response of Undamped systems subjected

to Harmonic forces

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Response of undamped systems to harmonic forces

The equation of motion for harmonic vibration of undamped system is:
̈ + = Sin( )

The solution to the equation is made up of two parts.

The first part is the solution which correspond to forced vibration and is known as the
Particular Solution. The corresponding vibrations are known as Steady state vibration, for
it is present because of the applied force no matter what are the initial conditions.
The second part is the solution to the free vibration, which does not require any forcing
function, this part is known as the Complimentary solution. The corresponding vibrations
are known as Transient Vibration, which depends on the initial conditions

Particular solution: Undamped Harmonic vibrations

The particular solution of undamped vibration can be determined using following relation:

1
( ) = . Sin( ) where ≠
1−

up(t) is the displacements corresponding to the Particular solution (i.e due to forced vibration
component). is termed as frequency ratio

For the sack of simplicity, we will use rω =ω/ωn in our lectures

1
⇒ ( ) = Sin( ) where ≠
1−
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Complimentary solution: Undamped Harmonic vibrations

Complementary solution of undamped vibration is given as follows:

( ) = Cos( ) + Sin( )

uc(t) is the displacements corresponding to the Complimentary solution (i.e due to

free vibration component) and depends on initial conditions.

Complete solution: Undamped Harmonic vibrations

Complete solution is the sum of complementary solution, uc(t), and particular solution, up(t)

1
( ) = Cos( ) + Sin( )+ Sin( )
1−

The constants ‘A’ and ‘B’ are determined by imposing initial conditions i.e.,

= (0) and u̇ = u̇ (0)

̇ (0) 1
( ) = (0)Cos(ω ) + − Sin( ) + Sin ( )
1− 1−

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Complete solution of undamped harmonic vibrations

The transient vibration exist even if(0) = ̇ (0) = 0 .
In such case the complementary part of solution given on previous slide specializes to:
0
( ) = 0 ∗ Cos( ) + − Sin( )
1−

or ( ) = Sin( )
−1

The complete solution is then specialized to the following form

1
or ( ) = Sin ( )− Sin( )
1−

Amplitude of ‘Static’ deflection due to harmonic force

If the force is applied slowly then ̈ = 0 and the equation of motion under harmonic force
̈ + = ( )
po
becomes: ku  p o Sin(  t) or u st  Sin(  t)
k
The subscript “st” (standing for static) indicate the elimination of acceleration’s effect
The maximum value of static deformation, (ust )o can be interpreted as the deformation
due to force amplitude, po , if applied as a static force

For brevity we will refer to (ust)o as the static deformation

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Effect of Frequency ratio, rω, on the direction of structural displacements

1
( ) = Sin( ) can be written as:
1−

1 where =
( ) = st Sin( )
1−

It can be observed from this equation that up(t) has negative sign when frequency ratio,
rω >1 (i.e. ω>ωn), and vice versa.
1
A Graph on next slide is plotted b/w frequency ratio, rω and 1 −

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Effect of Frequency ratio, rω, on the direction of structural displacements

1
( ) = st Sin( )
1 1−
1−

=
up is positive if this term is positive
and vice versa
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Effect of Frequency ratio, rω, on the direction of structural displacements

Following observation can be made from the plot given on previous slide

When rω < 1 ( i.e ω < ωn ), the displacement is positive, indicating that up(t) and p(t)
has same directions. The displacement is said to be in phase with the applied force.

When rω > 1 ( i.e ω > ωn ), the displacement is negative , indicating that the u(t) and
p(t) has apposite directions. The displacement is said to be out of phase with the
applied force.

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Effect of Frequency ratio, rω, on the direction of structural displacements

u
Sin( )

Structure displaces in the direction of force when ω/ωn <1

u
Sin( )

Structure displaces opposite to direction of force when ω/ωn >1

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Deformation Response factor, Rd

1 1
( ) = Sin( )= st Sin( )
1− 1−

Another mathematical form of the above mentioned equation is:

1
( ) = ( − )= ( − ) Where = =
st 1−

Rd = Deformation (or Displacement) response factor,

uo=Amplitude of dynamic displacement, and,
φ= Phase angle (or Phase Lag)
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Influence of Frequency ratio, rω, on Deformation Response factor, Rd

=1 = 2

Nearly static response . =

force may be defined as
quasi-static when rω ≤ 0.2
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Influence of Frequency ratio, rω, on Deformation Response factor, Rd

Following observation can be made from the plot

When rω is small ( i.e ω is small and thus force is ‘slowly varying’ ), Rd is only slightly
greater than 1 or in the other words amplitude of dynamic deformation, uo, is almost same as
amplitude of static deformation, (ust)o.

When rω >1 ( i.e ω is large and thus force is ‘rapidly varying’ ), Rd is less than 1 or in the
other words dynamic deformation amplitude is less than static deformation.

When → ∞ , Rd becomes zero

When rω is close to 1.0, Rd is many times larger than 1

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Phase Angle,φ
Phase angle (or Phase lag) is the amount by which Response, up(t), lags the Applied force, p(t)
For an Undamped systems subjected to Harmonic forces
0 < i. e. , r < 1
=
180 > i. e. , r > 1

ωt= φ or t= φ/ ω = φ/ (2π/T)

t= (φ/ 2π)T

Thus in an undamped system (ωn > ω or Tn<T) subjected to a harmonic force with exciting time
period , T, dynamic displacement will produce after 0 s time. i.e., at force and corresponding
displacement occur at same time
However, in an undamped system (ωn < ω or Tn>T) subjected to a harmonic force with exciting
time period , T, dynamic displacement will produce (180/2π)T =T/2 s after the corresponding force

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Problem 5.1
A video camera, of mass 2.0 kg, is mounted on the top of a bank building for surveillance. The video
camera is fixed at one end of a tubular aluminum rod whose other end is fixed to the building as
shown in Figure.
The wind-induced force acting on the video camera, is
found to be harmonic with p(t) = 25 sin 75t N. Determine
Posin 75t
the cross-sectional dimensions of the aluminum tube if the
maximum amplitude of steady state vibration of the video
camera is to be limited to 0.005 m. E Aluminum = 71 GPa

solution
m = 2 kg , (up)o= 0.005 m , E = 71* 109 N/m2
p(t) = 25 Sin (75t)
By comparing with p(t)= po Sin (ωt)
po= 25 N and ω = 75 rad /s
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= ⇒ = ……. ( )

= ⁄ = ⁄
= =

(I) ⇒ = ⇒ 0.005 = ( )

⇒ 0.005 = 25 . or = 16254 = ⁄

16254 ∗ 16254 ∗ (0.5)

=
3
=
3 ∗ 71 ∗ 10 ⇒ = 9.54 ∗ 10

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Response of Damped systems to

Harmonic forces

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Response of damped systems subjected to Harmonic forces

The equation of motion for harmonic vibration of damped system is:
̈ + ̇+ = Sin( )

This equation is to be solved subjected to initial conditions

( ) = (0) ̇ ( ) = ̇ (0)
The particular solution of this differential equation is

( ) = Sin( ) + Cos( )
Where

1− −2
= & =
1− + 2 1− + 2

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Response of damped systems subjected to Harmonic forces

The complementary solution is:

( ) = Cos( ) + Sin( )

The complete solution is: u(t) = uc(t)+up (t)

( ) = Cos( ) + Sin( ) + Sin( ) + Cos( )

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Steady state response of damped systems subjected to Harmonic forces

Transient response (difference b/w total response and steady-state response) in a damped
system diminishes after few cycles of forced vibration (see Graph) and at that stage
u =u +u =u +0=u
, ( ) = ( )+ ( )
( )

( )
st

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Deformation response factor, Rd , in damped system subjected to Harmonic forces

The Steady state deformation can be rewritten as:
( ) = Sin( − )= Sin( − )

Where = + = Tan −

Substituting the values of C and D (given on slide 22) in equations for determining Rd and φ ,
results in:
1
= =
1− + 2

2
and = Tan
1−

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Deformation response factor, Rd , in damped system subjected to Harmonic forces

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Effect of rω and ζ on the Deformation response factor, Rd

rω <<1

Following observation can be made from the plot

Damping reduce Rd for all values of frequency ratio, rω. However rate of reduction
highly depend on the magnitude of rω
If the rω is around 0.2 and below, ( i.e. force is ‘slowly varying’), Rd is only slightly
greater than 1 and thus unaffected by damping and

≅ st = /

⇒ =1
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Effect of rω and ζ on the Deformation response factor, Rd

rω >>1

If rω is high, around 2 and above,( i.e force is ‘rapidly varying’), Rd tends to zero. In other
words Rd is unaffected by damping. uo can be approximated as:

≈( ) = /

⟹ = /

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Effect of rω and ζ on the Deformation response factor, Rd

rω =1
If the rω≈ 1( i.e frequency of force is close to natural frequency), Rd is :

1 1
= = =
1− + 2 1−1 + 2 .1

1
⟹ =
2

The above equations shows that at resonance (rω≈1) , Rd is a direct function of damping

e.g., At ζ = 5% of critical damping, uo= 1/(2*0.05) (ust)o = 10 (ust)o

. i.e. Maximum dynamic displacement is 10 times the static deformation.
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Effect of rω and ζ on the Deformation response factor, Rd

rω =1
uo can be alternatively written as

1 1
= = . = .
2 2 2
2

= . =

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Deformation response factor, Rd , in damped system subjected to Harmonic forces

Rd ≈ 1 ≈ /
1 2 3
Rd =1/(2ζ)

Effect of rω and ζ on the Phase angle

φ=166.5o

ζ= 0.1

Above graph shows that Phase lag increases by increasing damping

For =1 = 0.1, = / = (166.5 ∗ 180)/(2 / )

⟹ = 0.463
Thus steady state deformation lag corresponding dynamic force by 0.925/T
seconds, where T= Forcing period

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Effect of rω and ζ on the Deformation response factor, Rd

Watch the video to study the influence of rω on variation in
Model fn = ωn/2π f= ω/2π rω w.r.t tall model=
dynamic displacements of a lightly damped elastic system
f/ fn,tall
Tall 4.00 *4.00 4/4=1.00
Medium 6.35 **6.35 6.35/4=1.59
Short 11.35 ***11.35 11.35/4=2.84
Comments:
1. Tall model resonant frequency = 4 Hz
2. When exciting frequency for tall model is 6.35 Hz,
rω = 1.59 and similarly when exciting frequency is
11.35 Hz, r ω =2.84.
3. Both the above mentioned r ω being greater than 1
(out-of-phase – de-amplification) and also greater
than 20.5 showing reduction even in static
deflection (i.e. approaching stationary position
specially at 2.84).
0.7
4. For systems with ζ ≥1/20.5 the static deflection are
0.15 continuously decays toward stationary position
1.59 2.84
, rω

Resonant Frequencies

A resonant frequency is defined as the forcing frequency at which the largest response
amplitude occurs.
Figure on next slide shows that the peaks in the frequency-response curves for
displacement, velocity, and acceleration occur at slightly different frequencies.
These resonant frequencies can be determined by setting to zero the first derivative of Rd ,
1
Rv, and Ra with respect to rω ; for < 2 they are:

For an undamped system the three resonant frequencies are identical and equal to the
natural frequency ωn of the system

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Rd

Rv

Ra

Rd ,Rv and Ra for a damped system excited by harmonic force. 35

Problem 5.2
A rotating machine (m = 100 kg) is mounted at the mid point of a 5m long, simply supported steel
beam with a square x-section. It was observed that machine, while running at a speed of 300 rpm,
exert an unbalanced harmonic force of amplitude 100 N on the beam and causes maximum steady-
state displacement of 20 mm in vertical direction.
Determine whether the beam will be safe in bending at resonance or not. Assume permissible stress
in bending is 125 Mpa. Take E= 207 Gpa.
Consider damping effect and ignore self weight of beam in calculations
100 kg
Solution
m =100 kg , po = 100 N, (up)o= 20 mm = 0.02 m

ω = 300 rev/min = 300 *2π rad/ 60 s = 10π rad/s

l= 5m, E= 207 GPa

Disp. resonant frequency, = 1−2 ……….( )

⟹ 10 = 100 1 − 2

⟹ 100 = ⁄ 1−2 ⟹ 10 = (1 − 2 ) ………….. (II)

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1
= = ⁄ = = 1−2
1− + 2
0.02 1
⇒ = , ≅( )
100 4 + [4 −8 ]

⇒ 2 ∗ 10 4 −4 =1 ⇒ 4 ∗ 10 − =1

⇒ = 2500 …..( )
ζ 1−

Eqn II ⇒ 10 = 1−2

Substituting k (from eqn III) in II 10 = 2500 ∗ 1−2

−

⟹ 4 − = 1−2 ⟹ 16 − = 1−2

Taking ζ2 =x
16 − = 1−2

After expanding and re-arranging

1565.06 − 1565.06 + 1 = 0 ⇒ = = 0.999, 0.00064

⇒ = 0.999, 0.0252 = 2.52 %

⇒ = 2500 = 2500 = 99238 /

ζ 1− 0.0252 1 − (0.0252)

= = 99238 ∗ 0.02 = 1985 N

100*9.81+ 1985 = 2966 N

2.5 m 2.5 m

Mmax M= pl/4= 2966*5/4= 3707 N.m

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( ) = /
=?

= 48 ⟹ = = 99238 ∗ ∗ ∗
= 1.248 ∗ 10

= = = 1.248 ∗ 10
ℎ = 0.0622

.
( ) = / = 3707 ∗ ( )/(1.248 ∗ 10 )

( ) = 92.38 < = 125 MPa

⇒ Beam will be safe in bending at resonance

Solution (Problem 5.2)- Short cut method

Since for structural systems is very low, therefore Disp. resonant frequency, ≈ =1

1
= =
1− + 2

0.02 1 1
⇒ = =
100 [1 − 1] + [4 ] 2

1 2500
⇒ 2 ∗ 10 = ⇒ =
2

= = / ⟹ 10 = 2500
100 = ( ∗ 100)

⟹ 100 = ⟹ = 25/100 = 0.0253

2500 2500
⇒ = = = 98776 /
0.0253

Rest of procedure is same

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Number of cycles required by a harmonic force to develop full resonance

Following relation between peak displacement at full resonance, uo, and peak displacement
at resonance after j vibration cycles, uj , exist for underdamped systems

= 1−

The relation is plotted on next slide for ζ = 0.02, 0.05 and 0.10.
It can be observed that lower the damping ratio, larger are the number of cycles required
to reach a certain percentage of uo .
 For example, the number of cycles required to reach full resonance, uo , is 50, 20 and 9
for for ζ = 0.02, 0.05 and 0.10, respectively.
 While number of cycles for 95% of uo is 24, 10 and 5 for ζ = 0.02, 0.05 and 0.10,
respectively. 41

Number of cycles required by a harmonic force to develop full resonance

0.95

ζ Cycles for full Cycles for 95% disp

resonance at resonance
0.10 9 5
0.05 20 10
0.02 50 24

j= No. of cycles

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Problem 5.3
The speed of rotating machine in problem 5.2 is so adjusted that it exceeds resonant frequency after
completing 5 revolutions at ω = ωn .
Determine the amplitude of dynamic displacement and static force for design.

= 1− ⟹
20
∗ . ∗
=1−

⟹ = 10.97 100*9.81+ 1092 = 2073 N

= = 99238 ∗ 0.011 = 1092 N

2.5 m 2.5 m

Mmax

Exercise 5.1
1. When the person stands in the centre of the floor system shown, he
causes a Deflection of 0.2 in. of floor under his feet. He walks (or runs
quickly) in the same area , how many steps per second would cause the
floor to vibrate with the greatest vertical amplitude
Ans: 7 steps per second

2. A fixed-fixed steel beam, of length 5 m, width 0.5 m, and thickness 0.1 m, carries an electric
motor of mass 75 kg and speed 1200 rpm at its mid-span, as shown in Figure. A rotating force of
magnitude Po = 5000 N is developed due to the unbalance in the motor.
Find the amplitude of steady-state vibrations by disregarding the mass of the beam. What
will be the amplitude if the mass of the beam is considered by replacing with a mass lumped at the
center equal to 25% of actual distributed mass. Take E = 207 Gpa and γ = 76.5 kN/m3 for steel

Ans: 4.15*10-4 m and 11.5*10-4 m

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3. If the electric motor of Problem 2 is to be mounted at the free end of a steel cantilever beam of
length 5 m , and the amplitude of vibration is to be limited to 0.5 cm, find the necessary thickness
of the beam. Include the mass of the beam in the computations
Ans: 0.62 m

4. A machine mounted on an isolated foundation is subjected to a harmonic force. The amplitude

of displacement is found to be 20 mm at resonance and 10 mm at a frequency 0.75 times the
resonant frequency. Find the damping ratio of the isolation system.
Ans: 12.3 %
5. The steel frame shown in figure supports a rotating machine which exerts
a horizontal force at the girder level, p(t)=2000 Sin 5.3t lb. Assuming 5% of
critical damping, determine amplitude of the dynamic displacement and
corresponding equivalent static force.
Take E= 29,000 ksi and I= 69.2 in4 . Assume that flexural stiffness of girder
is too high as compared to connected column
Ans: 2.05 in, 4223 lb

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