Mechanical Vibrations - Forced Vibrations

Mechanical Vibrations(ME 421)

Section – 4 Single Degree of Freedom Systems:

Harmonically Excited Vibrations

Instructor: Muhammad Haider

Book:Mechanical Vibrations, by S.S. Rao, Fifth Edition, Chapter 3

SDOF: Harmonically Excited Vibrations

Course ContentsS.No. Description

1. Basic Concepts

2. Harmonic Motion, Complex Algebra and Fourier Series

3. Single Degree of Freedom Systems: Free Vibrations

4.Single Degree of Freedom Systems: Harmonically Excited Vibrations

5.Two Degree of Freedom Systems: Natural Frequencies and Mode

Shapes

6.Two Degree of Freedom Systems: Coupling, Orthogonality and

Forced Response

7. Multi Degree of Freedom Systems

8. Lagrangian Method


Section Outline


• Introduction

• Response of an Undamped System under Harmonic Force

• Response of Damped System under Harmonic Force

• Response of a Damped System under Harmonic motion of

the Base

• Response of a Damped System under Rotating Unbalance

• Self Excitation and Stability Analysis

• A system is said to undergo forced vibration whenever external energy is supplied to the system during vibration

• External energy can be in the form of

• Applied force

• Imposed displacement excitation

• External energy may be

• harmonic

• nonharmonic but periodic

• nonperiodic or random in nature

• The response of a system to a harmonic excitation is called harmonic response

• In this section, we will limit our study to only harmonically excited external sources

Introduction


• In harmonically excited system, dynamic response of a SDOF system is analyzed for the force having form

𝐹 𝑡 = 𝐹𝑜𝑒𝑖(𝜔𝑡+𝜙)or 𝐹 𝑡 = 𝐹𝑜 cos(𝜔𝑡 + 𝜙) or 𝐹 𝑡 = 𝐹𝑜 sin(𝜔𝑡 + 𝜙)

• where 𝐹𝑜is the amplitude, 𝜔 is the frequency and 𝜙 is the phase angle of the harmonic excitation.

• The value of 𝜙 depends on the value of 𝐹(𝑡) at 𝑡 = 0 and is usually taken to be zero

• Under a harmonic excitation, the response of the system will also be harmonic

• If the frequency of excitation coincides with the natural frequency of the system, the response will be very large. This condition, is called as resonance.

Introduction


Equation of Motion

• For viscously damped spring mass system, EOM with harmonic force input becomes

𝑚 ሷ𝑥 + 𝑐 ሶ𝑥 + 𝑘𝑥 = 𝐹(𝑡)

• This equation is non-homogenous, its general solution 𝑥(𝑡) is given by the sum of the homogenous solution, 𝑥ℎ(𝑡) and the particular solution, 𝑥𝑝(𝑡)

𝑥 𝑡 = 𝑥ℎ 𝑡 + 𝑥𝑝(𝑡)

• The homogenous solution, which is the solution of the homogenous equation

𝑚 ሷ𝑥 + 𝑐 ሶ𝑥 + 𝑘𝑥 = 0

• dies out with time under each of the three possible conditions of damping

Introduction


Equation of Motion

• Eventually, general solution of the equation reduces to the particular solution 𝑥𝑝(𝑡), which

represents the steady state vibration

Introduction


Homogenous Solution

Particular Solution

Total Solution

Equation of Motion

• The part of the motion that dies out due to damping (the free-vibration part) is called transient

• The rate at which the transient motion decays depends on the values of the system parameters k, c, and m

Introduction


• For the sake of simplicity, we consider an undamped system subjected to a harmonic force, 𝐹 𝑡 = 𝐹𝑜 cos𝜔𝑡

𝑚 ሷ𝑥 + 𝑐 ሶ𝑥 + 𝑘𝑥 = 𝐹(𝑡)

becomes, 𝑚 ሷ𝑥 + 𝑘𝑥 = 𝐹𝑜 cos𝜔𝑡

• We know that the homogenous solution of this equation is given by

𝑥ℎ 𝑡 = 𝐶1 cos𝜔𝑛𝑡 + 𝐶2 sin𝜔𝑛𝑡

• where, 𝜔𝑛 = 𝑘/𝑚, is the natural frequency of the system

• Exciting force 𝐹(𝑡) is harmonic, the particular solution 𝑥𝑝 𝑡

is also harmonic and has the same frequency 𝜔

• We can assume a particular solution in the form

𝑥𝑝 𝑡 = 𝑋 cos𝜔𝑡

where 𝑋, is the maximum amplitude of 𝑥𝑝 𝑡

Response of an Undamped System Under Harmonic Force


Solution

• Putting value of 𝑥𝑝 𝑡 in the EOM, we get

−𝑚𝑋𝜔2 cos𝜔𝑡 + 𝑘𝑋 cos𝜔𝑡 = 𝐹𝑜 cos𝜔𝑡

• which becomes, −𝑚𝑋𝜔2 + 𝑘𝑋 = 𝐹𝑜• we can write

𝑋 =𝐹𝑜

𝑘 −𝑚𝜔2

• so our general solution

𝑥 𝑡 = 𝑥ℎ 𝑡 + 𝑥𝑝(𝑡)

• becomes,

𝑥 𝑡 = 𝐶1 cos𝜔𝑛𝑡 + 𝐶2 sin𝜔𝑛𝑡 +𝐹𝑜

𝑘 −𝑚𝜔2cos𝜔𝑡

• Using initial conditions, 𝑥 𝑡 = 0 = 𝑥𝑜 and ሶ𝑥 𝑡 = 0 = ሶ𝑥𝑜



Solution

• we get

𝐶1 = 𝑥𝑜 −𝐹𝑜

𝑘 −𝑚𝜔2; 𝐶2 =

ሶ𝑥𝑜𝜔𝑛

• Hence we can write

𝑥 𝑡 = 𝑥𝑜 −𝐹𝑜

𝑘 − 𝑚𝜔2cos𝜔𝑛𝑡 +

ሶ𝑥𝑜𝜔𝑛

sin𝜔𝑛𝑡 +𝐹𝑜

𝑘 −𝑚𝜔2cos𝜔𝑡

• We calculated

𝑋 =𝐹𝑜

𝑘 −𝑚𝜔2

• which can be written as

𝑋 =𝐹𝑜

𝑘 −𝑚𝜔2=

𝐹𝑜

𝑘(1 −𝑚𝜔2

𝑘)=

𝐹𝑜/𝑘

(1 −𝜔2

𝑘/𝑚)=

𝛿𝑠𝑡

(1 −𝜔2

𝜔𝑛2)

• Quantity 𝛿𝑠𝑡 = 𝐹𝑜/𝑘 is the deflection of the mass under a force 𝐹𝑜 and is sometimes called static deflection because 𝐹𝑜 is a constant force



Solution

• Thus we get, 𝑋 =𝛿𝑠𝑡

(1−𝜔

𝜔𝑛

2)

• That can be written as, 𝑋

𝛿𝑠𝑡=

1

(1−𝜔

𝜔𝑛

2)

• Quantity 𝑋/𝛿𝑠𝑡 represents the ratio of the dynamic to the static amplitude of the motion, also called the magnification factor, amplification factor, or amplification ratio.

• The value of magnification factor is dependent on frequency ratio r =𝜔/𝜔𝑛

• System response can be studied for three distinct cases



Case 1

• When 𝟎 < 𝝎/𝜔𝑛 < 𝟏

• Denominator of magnification factor 𝑋

𝛿𝑠𝑡=

1

(1−𝜔

𝜔𝑛

2), is positive

• Response is given by 𝑥𝑝 𝑡 = 𝑋 cos𝜔𝑡

without change• Harmonic response of the system 𝑥𝑝 𝑡 is in phase with the external force



Case 2

• When 𝟎 > 𝝎/𝜔𝑛 > 𝟏

• Denominator of magnification factor 𝑋

𝛿𝑠𝑡=

1

(1−𝜔

𝜔𝑛

2), is negative

• Response is given by 𝑥𝑝 𝑡 = −𝑋 cos𝜔𝑡

• The amplitude of motion 𝑋 is redefined to be a positive quantity as

𝑋 =𝛿𝑠𝑡

𝜔𝜔𝑛

2

− 1

• 𝑥𝑝 𝑡 and 𝐹 𝑡 have opposite signs and are said to be 180o out of phase



• Further 𝝎/𝜔𝑛 → ∞ , 𝑋 → 0.• Thus the response of the system to a harmonic force of very

high frequency is close to zero

Case 3

• When 𝝎/𝜔𝑛 = 𝟏

•𝑋

𝛿𝑠𝑡=

1

(1−𝜔

𝜔𝑛

2), becomes infinite

• The condition is known as resonance • To find response for this condition, we rewrite


𝑘 −𝑚𝜔2 cos𝜔𝑛𝑡 +ሶ𝑥𝑜𝜔𝑛


𝑘 −𝑚𝜔2 cos𝜔𝑡

As, 𝑥 𝑡 = 𝑥𝑜 cos𝜔𝑛𝑡 +ሶ𝑥𝑜

𝜔𝑛sin𝜔𝑛𝑡 +

𝐹𝑜

𝑘−𝑚𝜔2 cos𝜔𝑡 − cos𝜔𝑛𝑡

• Previously we have shown that𝐹𝑜

𝑘 −𝑚𝜔2= 𝑋 =

𝛿𝑠𝑡

(1 − ൗ𝜔 𝜔𝑛

2)

⇒ 𝑥 𝑡 = 𝑥𝑜 cos𝜔𝑛𝑡 +ሶ𝑥𝑜𝜔𝑛

sin𝜔𝑛𝑡 +𝛿𝑠𝑡 cos𝜔𝑡 − cos𝜔𝑛𝑡

(1 − ൗ𝜔 𝜔𝑛

2)



Case 3

• Since the last term takes an indefinite form when 𝝎 = 𝜔𝑛, we apply L’Hospital’s rule to evaluate the limit of this term

lim𝜔→𝜔𝑛

cos𝜔𝑡 − cos𝜔𝑛𝑡

1 − ൗ𝜔 𝜔𝑛

2 = lim𝜔→𝜔𝑛

𝑑𝑑𝜔


𝑑𝑑𝜔

1 − ൗ𝜔 𝜔𝑛

2

lim𝜔→𝜔𝑛

𝑡 sin𝜔𝑡

2 ൗ𝜔𝜔𝑛2

=𝜔𝑛𝑡

2sin 𝜔𝑛𝑡

Finally the response of the system at resonance becomes

𝑥 𝑡 = 𝑥𝑜 cos𝜔𝑛𝑡 +ሶ𝑥𝑜𝜔𝑛

sin𝜔𝑛𝑡 +𝛿𝑠𝑡𝜔𝑛𝑡

2sin 𝜔𝑛𝑡



Case 3

𝑥 𝑡 = 𝑥𝑜 cos𝜔𝑛𝑡 +ሶ𝑥𝑜𝜔𝑛

sin𝜔𝑛𝑡 +𝛿𝑠𝑡𝜔𝑛𝑡

2sin 𝜔𝑛𝑡



Total Response

• For(𝜔/𝜔𝑛) < 1, 𝑥 𝑡 = 𝐴 cos(𝜔𝑛𝑡 − 𝜙) +𝛿𝑠𝑡

1−𝜔

𝜔𝑛

2 cos 𝜔𝑡

• For(𝜔/𝜔𝑛) > 1, 𝑥 𝑡 = 𝐴 cos(𝜔𝑛𝑡 − 𝜙) −𝛿𝑠𝑡

−1+𝜔

𝜔𝑛

2 cos 𝜔𝑡



A reciprocating pump, weighing 150 lb, is mounted at the middle of a steel plate of

thickness 0.5 in., width 20 in., and length 100 in., clamped along two edges as

shown in Figure. During operation of the pump, the plate is subjected to a

harmonic force, 𝑭 𝒕 = 𝟓𝟎 𝒄𝒐𝒔 𝟔𝟐. 𝟖𝟑𝟐𝒕 lb. Find

a) the amplitude of vibration of the plate

Example 3.1



Derive the equation of motion and find the steady-state response of the system

shown in Figure for rotational motion about the hinge O for the following data:

𝒌𝟏 = 𝒌𝟐 = 5000 N/m, a = 0.25 m, b = 0.5 m, l = 1 m,

M = 50 kg, m = 10 kg, 𝑭𝒐= 500 N, 𝝎=1000 rpm

Problem 3.24



Review Examples: 3.2; Practice Problems: 3.1-3.23

Section Outline


• Introduction

• Response of an Undamped System under Harmonic Force

• Beating Phenomenon

• Response of Damped System under Harmonic Force

• Response of a Damped System under Harmonic motion of

the Base

• Response of a Damped System under Rotating Unbalance

• Self Excitation and Stability Analysis

Beating Phenomenon

• If the forcing frequency is close to, but not exactly equal to, the natural frequency of the system, a phenomenon known as beating may occur

• In this kind of vibration, the amplitude builds up and then diminishes in a regular pattern

• The solution for beating phenomenon can be obtained by considering


𝑘 −𝑚𝜔2 cos𝜔𝑛𝑡 +ሶ𝑥𝑜𝜔𝑛


𝑘 −𝑚𝜔2 cos𝜔𝑡

• For 𝑥𝑜 = ሶ𝑥𝑜 = 0, above equation reduces to

𝑥 𝑡 = −𝐹𝑜

𝑘 − 𝑚𝜔2cos𝜔𝑛𝑡 +

𝐹𝑜𝑘 −𝑚𝜔2

cos𝜔𝑡

𝑥 𝑡 =Τ𝐹𝑜 𝑚

𝜔𝑛2 − 𝜔2



𝜔𝑛2 − 𝜔2

2 sin𝜔 + 𝜔𝑛

2𝑡. sin

𝜔𝑛 − 𝜔

2𝑡



Beating Phenomenon


𝜔𝑛2 − 𝜔2

2 sin𝜔 + 𝜔𝑛

2𝑡. sin

𝜔𝑛 − 𝜔

2𝑡

• Let the forcing frequency 𝜔 be slightly less than the natural frequency𝜔𝑛 −𝜔 = 2𝜖

• where 𝜖 is a small positive quantity. Then 𝜔𝑛 ≈ 𝜔⇒ 𝜔𝑛 + 𝜔 = 2𝜔

• Multiplication of above equation gives𝜔𝑛2 −𝜔2 = 4𝜖𝜔

• Putting values in above equation yields following solution


2𝜖𝜔sin 𝜖𝑡 sin𝜔𝑡

• Since 𝜖 is small, the function sin 𝜖𝑡 varies slowly; its period, equal to 2𝜋/𝜖 is large.

• Above solution can be seen as representing vibration with period

2𝜋/𝜔 and of variable amplitude equal to Τ𝐹𝑜 𝑚

2𝜖𝜔sin 𝜖𝑡



Beating Phenomenon


2𝜖𝜔sin 𝜖𝑡 sin𝜔𝑡

• It can also be observed that the curve will go through several cycles, while the wave goes through a single cycle

• Thus the amplitude builds up and dies down continuously. • The time between the points of zero amplitude is called the period of

beating and is given by 𝜏𝑏 = 2𝜋/2𝜖 = 2𝜋/(𝜔𝑛 − 𝜔)• Frequency of beating as, 𝜔𝑏 = 2𝜖 = 𝜔𝑛 − 𝜔



• If the forcing function is given by 𝐹 𝑡 = 𝐹𝑜 cos𝜔𝑡, the EOM becomes𝑚 ሷ𝑥 + 𝑐 ሶ𝑥 + 𝑘𝑥 = 𝐹𝑜 cos𝜔𝑡

• The particular solution is also expected to be harmonic; we assume it as𝑥𝑝 𝑡 = 𝑋 cos(𝜔𝑡 − 𝜙)

• where X and 𝜙 is amplitude and phase lag of the response i.e. displacement vector lags the force vector by 𝜙,

ሶ𝑥𝑝 𝑡 = −𝑋𝜔 sin 𝜔𝑡 − 𝜙 = 𝑋𝜔 cos(𝜔𝑡 − 𝜙 +𝜋

2)

ሷ𝑥𝑝 𝑡 = −𝑋𝜔2 sin 𝜔𝑡 − 𝜙 = 𝑋𝜔2 cos 𝜔𝑡 − 𝜙 + 𝜋

• By substituting 𝑥𝑝 in EOM, we get

𝑚𝑋𝜔2 cos 𝜔𝑡 − 𝜙 + 𝜋 + 𝑐𝑋𝜔 cos 𝜔𝑡 − 𝜙 +𝜋

2+ 𝑘𝑋 cos 𝜔𝑡 − 𝜙 = 𝐹𝑜 cos𝜔𝑡

Response of a Damped System Under Harmonic Force


• From vector diagram of these forces, we have𝐹𝑜2 = 𝑘𝑋 −𝑚𝑋𝜔2 2 + 𝑐𝑋𝜔 2

𝐹𝑜2 = 𝑋2 𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 2

⇒ 𝑋 =𝐹𝑜

𝑘 − 𝑚𝜔𝟐 2 + 𝑐𝜔 2;

𝜙 = tan−1𝑐𝜔

𝑘 − 𝑚𝜔2

• Thus particular or steady state solution of the equation becomes

𝑥𝑝 𝑡 =𝐹𝑜

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

cos(𝜔𝑡 − 𝜙)



• Now consider

𝑋 =𝐹𝑜

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

Which can be simplified to

𝑋 =𝐹𝑜

𝑘2 1 −𝑚𝜔2/𝑘 2 + 𝑐𝜔/𝑘 2

=𝐹𝑜/𝑘

1 − Τ𝜔 𝜔𝑛2 2 + 2𝜁𝑚𝜔𝑛𝜔/𝑘

2

𝑋

𝛿𝑠𝑡=

1

1 − Τ𝜔 𝜔𝑛2 2 + 2𝜁 Τ𝜔 𝜔𝑛

2

Where,

• Putting 𝑟 = 𝜔/𝜔𝑛, We finally get



𝛿𝑠𝑡 = 𝐹𝑜/𝑘 𝑐 = 𝑐𝑐𝜁 = 2𝑚𝜔𝑛𝜁 𝜔𝑛 = Τ𝑘 𝑚

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2; 𝜙 = tan−1

2𝜁𝑟

1 − 𝑟2

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2



𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2



• For (𝜁 = 0), magnification factor (M) reduces to an undamped case

• Any amount of damping reduces the M for all values of the forcing frequency.

• For any specified value of r, a higher value of damping reduces the value of M.

• In the case of a constant force (when r=0), the value of M=1

• The reduction in M in the presence of damping is very significant at or near resonance.

• The amplitude of forced vibration becomes smaller with increasing values of the forcing frequency (that is, 𝑀 → 0 as 𝑟 → ∞)

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2



• For 0 < 𝜁 < 1/ 2, the maximum value of M occurs when

𝑟 = 1 − 2𝜁2 or 𝜔 = 𝜔𝑛 1 − 2𝜁2

• which can be seen to be lower than the undamped natural frequency and the damped natural frequency

• The maximum value of X (when

𝑟 = 1 − 2𝜁2) is given by𝑋

𝛿𝑠𝑡 𝑚𝑎𝑥

=1

2𝜁 1 − 𝜁2

• Eq can be used for the experimental determination of the measure of damping present in the system.

• In a vibration test, if the maximum amplitude of the response is measured, the damping ratio of the system can be found using Eq.

𝑋

𝛿𝑠𝑡=

1

1 − 𝑟2 2 + 2𝜁𝑟 2



𝑋

𝛿𝑠𝑡 𝑚𝑎𝑥

=1

2𝜁 1 − 𝜁2

• Conversely, if the amount of damping is known, one can make an estimate of the maximum amplitude of vibration.

• The value of X at 𝜔 = 𝜔𝑛by𝑋

𝛿𝑠𝑡 𝜔=𝜔𝑛

=1

2𝜁

• For 𝜁 = 1/ 2, 𝑑𝑀

𝑑𝑟= 0 when 𝑟 = 0.

For 𝜁 > 1/ 2, the graph of M monotonically decreases with increasing values of r.



• For an undamped system (𝜁 = 0), the phase angle is 0 for 0 < 𝑟 < 1 and 180°for 𝑟 > 1, implying that the excitation and response are in phase for 0 < 𝑟 < 1and out of phase for 𝑟 > 1 when 𝜁 = 0

• For 𝜁 > 0 and 0 < 𝑟 < 1, the phase angle is given by 0 < 𝜙 < 90°, implying that the response lags the excitation.

• For 𝜁 > 0 and r > 1, the phase angle is given by by 90° < 𝜙 < 180°, implying that the response leads the excitation.

• For 𝜁 > 0 and r = 1, the phase angle is given by 𝜙 = 90°, implying that the phase difference between the excitation and the response is 90°.

• For 𝜁 > 0 and large values of r, the phase angle approaches 180°, implying that the response and the excitation are out of phase.

𝜙 = tan−12𝜁𝑟

1 − 𝑟2



cX𝜔

mX𝜔2

𝐹𝑜

kX

𝜙

𝜔𝑡

cX𝜔

mX𝜔2

𝐹𝑜

kX

𝜙

𝜔𝑡

cX𝜔

mX𝜔2

𝐹𝑜

kX

𝜙 𝜔𝑡

Force Vibration Vector Diagrams

𝝎

𝝎𝒏≪ 𝟏

exciting force approximately equal to

spring force

𝝎

𝝎𝒏= 𝟏

exciting force equal to damping force, and inertia

force equal to springforce

𝝎

𝝎𝒏≫ 𝟏

exciting force nearly equal to inertia force



Total Response• Complete or general solution is given by

𝑥 𝑡 = 𝑥ℎ 𝑡 + 𝑥𝑝 𝑡

• We know that 𝑥ℎ 𝑡 for damped SDOF system is given by𝑥ℎ 𝑡 = 𝑋𝑜𝑒

−𝜁𝜔𝑛𝑡 cos(𝜔𝑑𝑡 − 𝜙𝑜)• So total response becomes

𝑥 𝑡 = 𝑋𝑜𝑒−𝜁𝜔𝑛𝑡 cos(𝜔𝑑𝑡 − 𝜙𝑜) + 𝑋 cos(𝜔𝑡 − 𝜙)

• Where,

• What about 𝑋𝑜 and 𝜙𝑜??• To be evaluated from initial conditions for the given general

solution𝑥 0 = 𝑥0 = 𝑋0 cos 𝜙0 + 𝑋 cos𝜙

ሶ𝑥 𝑡

= 𝑋0 −𝜁𝜔𝑛𝑒−𝜁𝜔𝑛𝑡 cos 𝜔𝑑𝑡 − 𝜙𝑜 − 𝑒−𝜁𝜔𝑛𝑡𝜔𝑑 sin 𝜔𝑑𝑡 − 𝜙𝑜

− 𝑋𝜔 sin(𝜔𝑡 − 𝜙)

𝜔𝑑 = 1 − 𝜁2𝜔𝑛 𝑋 =𝐹𝑜

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2


𝑘 − 𝑚𝜔2



Total Responseሶ𝑥 𝑡

= 𝑋0 −𝜁𝜔𝑛𝑒−𝜁𝜔𝑛𝑡 cos 𝜔𝑑𝑡 − 𝜙𝑜 − 𝑒−𝜁𝜔𝑛𝑡𝜔𝑑 sin 𝜔𝑑𝑡 − 𝜙𝑜

− 𝑋𝜔 sin(𝜔𝑡 − 𝜙)ሶ𝑥 0 = ሶ𝑥0 = −𝜁𝜔𝑛𝑋0 cos𝜙0 + 𝜔𝑑𝑋0 sin𝜙0 + 𝜔𝑋 sin𝜙

The solution of above equations will give

𝑋0 = 𝑥0 − 𝑋 cos𝜙 2 +1

𝜔𝑑2 𝜁𝜔𝑛𝑥0 + ሶ𝑥 − 𝜁𝜔𝑛𝑋 cos𝜙𝑜 − 𝜔𝑋 sin𝜙 2

12

𝜙0 = tan−1𝜁𝜔𝑛𝑥0 + ሶ𝑥 − 𝜁𝜔𝑛𝑋 cos𝜙0 − 𝜔𝑋 sin𝜙

𝜔𝑑(𝑥0 − 𝑋 cos𝜙)

Find the total response of a single-degree-of-freedom system with m=10 kg, c = 20

N-s/m, k = 4000 N/m, 𝒙𝟎=0.01m and ሶ𝒙𝟎 = 𝟎 under the following conditions:

Example 3.3


a) An external force 𝑭 𝒕 = 𝑭𝟎 𝒄𝒐𝒔 𝝎𝒕, acts on the system with 𝑭𝟎 = 𝟏𝟎𝟎N and 𝝎 = 𝟏𝟎rad/s

b) Free vibration with 𝑭 𝒕 = 𝟎


Consider a spring-mass-damper system with k=4000 N/m, m = 10kg, and c = 40N-

s/m. Find the steady-state and total responses of the system under the harmonic

force 𝑭 𝒕 = 𝟐𝟎𝟎 𝒄𝒐𝒔𝟏𝟎𝒕 N and the initial conditions 𝒙𝟎=0.1m and ሶ𝒙𝟎 = 𝟎

Problem 3.26



Response of a Damped System Under the Harmonic Motion of Base


• Sometimes the base or support of a spring-mass-damper system undergoes harmonic motion

• Let 𝑦(𝑡) denote the displacement of the base and x(t) the displacement of the mass from its static equilibrium position at time t.

• Then the net elongation of the spring is (𝑥 − 𝑦) and the relative velocity between the two ends of the damper is ( ሶ𝑥 − ሶ𝑦)

• From the free-body diagram shown, we obtain the equation of motion:𝑚 ሷ𝑥 + 𝑐( ሶ𝑥 − ሶ𝑦)+ k (𝑥 − 𝑦)= 0



𝑚 ሷ𝑥 + 𝑐( ሶ𝑥 − ሶ𝑦)+ k (𝑥 − 𝑦)= 0• If 𝑦 𝑡 = 𝑌 sin𝜔𝑡, we get

𝑚 ሷ𝑥 + 𝑐 ሶ𝑥+ k 𝑥 = ky + c ሶ𝑦 = 𝑘𝑌 sin𝜔𝑡 + 𝑐𝜔𝑌 cos𝜔𝑡= 𝐴 sin(𝜔𝑡 − 𝛼)

• where

• This shows that giving excitation to the base is equivalent to applying a harmonic force of magnitude A to the mass.

• Steady state/particular solution is given by

𝑥𝑝 𝑡 =𝐹𝑜

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

sin(𝜔𝑡 − 𝜙1)

which can be written as

𝑥𝑝 𝑡 =𝑌 𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

sin(𝜔𝑡 − 𝜙1 − 𝛼)

where 𝜙1 = tan−1𝑐𝜔

𝑘−𝑚𝜔2

𝐴 = 𝑌 𝑘2 + 𝑐𝜔 2 𝛼 = 𝑡𝑎𝑛−1 −𝑐𝜔

𝑘



𝑥𝑝 𝑡 =𝑌 𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2

sin(𝜔𝑡 − 𝜙1 − 𝛼)

• Using trigonometric identities, 𝑥𝑝 𝑡 can be written as

𝑥𝑝 𝑡 = 𝑋 sin(𝜔𝑡 − 𝜙)

• where

𝑋

𝑌=

𝑘2 + 𝑐𝜔 2

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 2=

1 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

and

𝜙 = tan−1𝑚𝑐𝜔3

𝑘 𝑘 −𝑚𝜔2 + 𝜔𝑐 2 = tan−12𝜁𝑟3

1 + 4𝜁2 − 1 𝑟2

• The ratio of the amplitude of the response 𝑥𝑝 𝑡 to that of the base

motion y(t), 𝑋

𝑌= 𝑇𝑑 is called the displacement transmissibility



𝑋

𝑌= 𝑇𝑑 =

1+ 2𝜁𝑟 2

1−𝑟2 2+ 2𝜁𝑟 2 𝜙 = tan−12𝜁𝑟3

1+ 4𝜁2−1 𝑟2



𝑋

𝑌= 𝑇𝑑 =

1+ 2𝜁𝑟 2

1−𝑟2 2+ 2𝜁𝑟 2

• The value of 𝑇𝑑 is unity at r=0 and close to unity for small values of r.

• For an undamped system (𝜁 = 0), 𝑇𝑑 →∞ at resonance (𝑟 = 1)

• The value of 𝑇𝑑 is less than unity (𝑇𝑑 <

1) for values of (𝑟 > 2) for any amount of damping 𝜁

• The value of 𝑇𝑑is unity for all values of

𝜁 at 𝑟 = 2

• For 𝑟 < 2, smaller damping ratios lead to larger values of 𝑇𝑑

• On the other hand, for 𝑟 > 2, smaller values of damping ratio lead to smaller values of 𝑇𝑑

• The displacement transmissibility 𝑇𝑑 , attains a maximum for 0 < 𝜁 < 1 at the frequency ratio 𝑟 = 𝑟𝑚 < 1 given by

𝑟𝑚 =1

2𝜁1 + 8𝜁2 − 1



Force Transmitted• A force F, is transmitted to the base or support due to the

reactions from the spring and the dashpot, which can be written as

𝐹 = 𝑐( ሶ𝑥 − ሶ𝑦)+ k (𝑥 − 𝑦)= −𝑚 ሷ𝑥• Using 𝑥𝑝 𝑡 = 𝑋 sin(𝜔𝑡 − 𝜙), we get

𝐹 = mω2𝑋 sin 𝜔𝑡 − 𝜙 =𝐹𝑇sin(𝜔𝑡 − 𝜙)• where 𝐹𝑇 is the amplitude or maximum value of the force

transmitted to the base given by

𝐹𝑇𝑘𝑌

= 𝑟21 + 2𝜁𝑟 2

1 − 𝑟2 2 + 2𝜁𝑟 2

12

• The ratio 𝐹𝑇

𝑘𝑌is known as the force

transmissibility.



Relative motion• If 𝑧 = 𝑥 − 𝑦 denotes the motion of the mass relative to the base, the equation of

motion becomes𝑚 ሷ𝑧 + 𝑐 ሶ𝑧+ k 𝑧= −𝑚 ሷ𝑦 = 𝑚𝜔2𝑌 sin𝜔𝑡

• The steady-state solution is given by

𝑧 𝑡 =𝑚𝜔2𝑌 sin(𝜔𝑡 − 𝜙1)

𝑘 − 𝑚𝜔2 2 + 𝑐𝜔 212

= 𝑍 sin(𝜔𝑡 − 𝜙1)

• where Z, the amplitude of 𝑧(𝑡), can be expressed

𝑍 =𝑚𝜔2𝑌

𝑘 −𝑚𝜔2 2 + 𝑐𝜔 212

𝑍

𝑌=

𝑟2

1 − 𝑟2 2 + 2𝜁𝑟 2

𝜙1 = tan−1𝑐𝜔

𝑘 −𝑚𝜔2= tan−1

2𝜁𝑟

1 − 𝑟2

Figure shows a simple model of a motor vehicle that can vibrate in the vertical

direction while traveling over a rough road. The vehicle has a mass of 1200 kg. The

suspension system has a spring constant of 400 kN/m and a damping ratio of 𝜻 =

𝟎. 𝟓. If the vehicle speed is 20 km/hr, determine

Example 3.4


a) the displacement amplitude of the vehicle. The road surface varies sinusoidallywith an amplitude of Y=0.05m and a wavelength of 6m


A heavy machine, weighing 3000 N, is supported on a resilient foundation. The

static deflection of the foundation due to the weight of the machine is found to be

7.5 cm. It is observed that the machine vibrates with an amplitude of 1 cm when the

base of the foundation is subjected to harmonic oscillation at the undamped

natural frequency of the system with an amplitude of 0.25 cm. Find

Example 3.4


a) the damping constant of the foundation,

b) the dynamic force amplitude on the base, and

c) the amplitude of the displacement of the machine relative to the base


Response of a Damped System Under Rotating Unbalance


• Unbalance in rotating machinery is one of the main causes of vibration

• A simplified model of such a machine is shown in Fig

• The total mass of the machine is M, and there are two eccentric masses m/2 rotating in opposite directions with a constant angular velocity

• The centrifugal force due to each mass will cause excitation of the mass M.

• We consider two equal masses m/2 rotating in opposite directions in order to have the horizontal components of excitation of the two masses cancel each other.

• However, the vertical components of excitation add together and act along the axis of symmetry AA



• If the angular position of the masses is measured from a horizontal position, the total vertical component of the excitation is always given by

𝐹 𝑡 = 𝑚𝑒𝜔2 sin𝜔𝑡• The equation of motion can be derived

by the usual procedure𝑀 ሷ𝑥 + 𝑐 ሶ𝑥+ k 𝑥=𝑚𝑒𝜔2 sin𝜔𝑡

• The solution of this equation will be identical to previously derived equation for damped forced vibration, if we replace m and 𝐹0 by M and 𝑚𝑒𝜔2

respectively𝑥𝑝 𝑡 = 𝑋 sin(𝜔𝑡 − 𝜙)

• Where, 𝑋 =𝑚𝑒𝜔2

𝑘−𝑚𝜔2 2+ 𝑐𝜔 2


𝑘 −𝑀𝜔2



• By defining 𝜁 = 𝑐/𝑐𝑐 and 𝑐𝑐 = 2𝑀𝜔𝑛

𝑀𝑋

𝑚𝑒=

𝑟2

1 − 𝑟2 2 + 2𝜁𝑟 2

𝜙 = tan−12𝜁𝑟

1 − 𝑟2

• All the curves begin at zero amplitude. The amplitude near resonance is markedly affected by damping.

• Thus if the machine is to be run near resonance, damping should be introduced purposefully to avoid dangerous amplitudes.

• At very high speeds (𝜔 large), MX/me is almost unity, and the effect of damping is negligible.

• For 0 < 𝜁 < 1/ 2, the maximum of

MX/me occurs when, 𝑑

𝑑𝑟

𝑀𝑋

𝑚𝑒= 0

• The solution of equation is

𝑟 =1

1 − 2𝜁2> 1



• Corresponding maximum value of MX/me is given by𝑀𝑋

𝑚𝑒 𝑚𝑎𝑥=

1

2𝜁 1 − 𝜁2

• Thus the peaks occur to the right of the resonance value of r=1

• For 𝜁 > 1/ 2, [MX/me] does not attain a maximum. Its value grows from 0 at r=0 to 1 at 𝑟 → ∞

• The force transmitted to the foundation due to rotating unbalanced force (F) can be found as 𝐹 𝑡 = 𝑘𝑥 𝑡 + 𝑐 ሶ𝑥(𝑡)

• The magnitude (or maximum value) of F can be derived as

𝐹 = 𝑚𝑒𝜔21 + 4𝜁2𝑟2

1 − 𝑟2 2 + 4𝜁2𝑟2

An electric motor of mass M, mounted on an elastic foundation, is found to vibrate

with a deflection of 0.15 m at resonance. It is known that the unbalanced mass of

the motor is 8% of the mass of the rotor due to manufacturing tolerances used, and

the damping ratio of the foundation is 𝜻 = 𝟎. 𝟎𝟐𝟓. Determine the following

Example 3.6


a) the eccentricity or radial location of the unbalanced mass (e),

b) the peak deflection of the motor when the frequency ratio varies from resonance, and

c) the additional mass to be added uniformly to the motor if the deflection of the motor at resonance is to be reduced to 0.1 m.


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Mechanical Vibrations - Forced Vibrations

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