Mechanical Vibrations (ME 421) Section – 4 Single Degree of Freedom Systems: Harmonically Excited Vibrations Book: Mechanical Vibrations, by S.S. Rao, Fifth Edition, Chapter 3
Instructor:: Muhammad Haider Instructor
Course Contents S.No.
Description
1.
Basic Concepts
2.
Harmo armoni nicc Mo Motion tion,, Com Compl ple ex Alge Algebr bra a and and Fou Fouri rier er Ser Series ies
3.
Sing Single le Degr Degree ee of Free Freedo dom m Syst System ems: s: Free Free Vibr Vibrat atio ions ns
4. 5. 6.
Harmonically nically Excited Single Degree of Freedom Systems: Harmo Vibrations Frequencies and Mode Two Degree of Freedom Systems: Natural Frequencies Shapes Coupling, Orthogo Orthogonalit nality y and Two Degree of Freedom Systems: Coupling, Forced Response
7.
Multi Degree of Freedom Systems
8.
Lagran Lagrangia gian n Method Method
Section Outline •
Introduction
•
Response of 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
Introduction •
•
•
•
•
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 • nonperi nonperiodi odicc or random random in nature 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 (+)or cos( cos( ) or sin( sin( )
•
•
•
•
where is the amplitude, is the frequency and is the phase angle of the harmonic excitation. The value of depends on the value of () at 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 respon se 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
ሷ ሶ () •
This equation is non-homogenous, its general solution () is given by the sum of the homogenous solution, ℎ () and the particular solution, ()
ℎ () •
The homogenous solution, which is the solution of the homogenous equation
ሷ ሶ 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 () , which represents the steady state vibration Homogenous Solution
Particular Solution
Total Solution
Introduction 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
•
For the sake of simplicity, we consider an undamped system subjected to a harmonic force, cos
ሷ ሶ () ሷ cos
•
• •
•
becomes, We know that the homogenous solution of this equation is given by
ℎ cos sin where, /, is the natural frequency of the system Exciting force () is harmonic, the particular solution is also harmonic and has the same frequency We can assume a particular solution in the form
cos where , is the maximum amplitude of
Solution • Putting value of in the EOM, we get •
cos cos cos which becomes,
•
we can write
•
so our general solution
ℎ () •
becomes,
•
cos sin cos Using initial conditions, 0 and ሶ 0 ሶ
Solution •
•
we get
ሶ ;
Hence we can write
ሶ cos sin cos •
•
•
We calculated
which can be written as
/ (1 ) (1 / ) (1 )
Quantity / is the deflection of the mass under a force and is sometimes called static deflection because is a constant force
Solution •
•
•
•
Thus we get,
(−
)
That can be written as,
(− )
Quantity / 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
/ •
System response can be studied for three distinct cases
Case 1 • •
When < / < Denominator of magnification factor , is positive
•
•
(− )
Response is given by cos without change Harmonic response of the system is in phase with the external force
Case 2 • •
When > / > Denominator of magnification factor , is negative
• •
(− )
Response is given by cos The amplitude of motion is redefined to be a positive quantity as
•
• •
1
and have opposite signs and are said to be 180 o out of phase Further / → ∞ , → 0. Thus the response of the system to a harmonic force of very high frequency is close to zero
Case 3 • •
•
•
When /
(−
, )
becomes infinite
The condition is known as resonance To find response for this condition, we rewrite
ሶ cos sin cos ሶ As, cos sin − cos cos
•
Previously we have shown that
(1 ൗ ) ሶ cos cos ⇒ cos sin (1 ൗ )
Case 3 •
Since the last term takes an indefinite form when , we apply L’Hospital’s rule to evaluate the limit of this term
lim
→
cos cos cos cos lim → 1 ൗ 1 ൗ sin lim sin → 2 ൗ 2
Finally the response of the system at resonance becomes
ሶ cos sin sin 2
Case 3
ሶ cos sin sin 2
Total Response •
For(/ ) < 1,
cos( )
•
For(/ ) > 1,
cos( )
−
−+
cos
cos
Example 3.1 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, . lb. Find a) the amplitude of vibration of the plate
Problem 3.24 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:
= = 5000 N/m, a = 0.25 m, b = 0.5 m, l = 1 m, M = 5 0 k g , m = 1 0 k g , = 500 N, =1000 rpm
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
ሶ cos sin cos • For ሶ 0, above equation reduces to cos cos Τ cos cos Τ 2sin .sin 2 2
Beating Phenomenon Τ 2sin .sin 2 2
•
Let the forcing frequency be slightly less than the natural frequency
2
•
where is a small positive quantity. Then
≈
⇒ 2
•
Multiplication of above equation gives
4
•
Putting values in above equation yields following solution
•
Τ sin sin 2 Since is small, the function sin varies slowly; its period, equal to
2/ is large. •
Above solution can be seen as representing vibration with period Τ 2/ and of variable amplitude equal to sin
Beating Phenomenon Τ sin sin 2
•
• •
•
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 2/2 2/( ) Frequency of beating as, 2
•
If the forcing function is given by
cos, the EOM becomes ሷ ሶ cos
•
The particular solution is also expected to be harmonic; we assume it as
cos( ) •
where X and is amplitude and phase lag of the response i.e. displacement vector lags the force vector by ,
ሶ sin cos( ) 2 ሷ sin cos • By substituting in EOM, we get cos cos cos cos 2 •
From vector diagram of these forces, we have
⇒
;
tan−
•
Thus particular or steady state solution of the equation becomes
cos( )
•
Now consider
Which can be simplified to
1 / /
•
Putting
1 Τ
2 /
1 1 Τ
/
Where,
/
2 Τ
2
Τ
/ , We finally get 1 1
2
;
tan−
2 1
1 1
2
•
•
•
•
•
•
For ( 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, → 0 as → ∞)
1 1
2
•
•
•
For 0 < < 1/ 2, the maximum value of M occurs when 1 2 or 1 2 which can be seen to be lower than the undamped natural frequency and the damped natural frequency The maximum value of X (when 1 2 ) is given by
•
•
1 2 1
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.
1 1
2
•
•
1 2 1
Conversely, if the amount of damping is known, one can make an estimate of the maximum amplitude of vibration. The value of X at by
•
For 1/
= 2,
1 2
0 when 0.
For > 1/ 2, the graph of M monotonically decreases with increasing values of r.
1 1
2
•
•
•
•
•
For an undamped system ( 0), the phase angle is 0 for 0 < < 1 and 180° for > 1, implying that the excitation and response are in phase for 0 < < 1 and out of phase for > 1 when 0 For > 0 and 0 < < 1, the phase angle is given by 0 < < 90°, implying that the response lags the excitation. For > 0 and r > 1, the phase angle is given by by 90° < < 180°, implying that the response leads the excitation. For > 0 and r 1, the phase angle is given by 90°, implying that the phase difference between the excitation and the response is 90°. For > 0 and large values of r, the phase angle approaches 180 °, implying that the response and the excitation are out of phase.
tan−
2 1
Force Vibration Vector Diagrams
≪
exciting force approximately equal to spring force
exciting force equal to damping force, and inertia force equal to spring force
cX
cX kX
mX
mX
exciting force nearly equal to inertia force
cX
kX
≫
kX
mX
Total Response Complete or general solution is given by •
ℎ
•
We know that ℎ for damped SDOF system is given by
ℎ − cos( )
•
So total response becomes
− cos( ) cos( )
•
Where,
1
•
tan−
What about and ?? To be evaluated from initial conditions for the given general • solution
0 cos cos
ሶ − cos − sin sin( )
Total Response
ሶ − cos − sin sin( ) ሶ 0 ሶ cos sin sin The solution of above equations will give
1 cos ሶ cos sin ሶ cos sin − tan ( cos )
Example 3.3 Find the total response of a single-degree-of-freedom system with m=10 kg, c = 20 N-s/m, k = 4000 N/m, =0.01m and ሶ under the following conditions: a) An external force , acts on the system with N and rad/s b) Free vibration with
Problem 3.26 Consider a spring-mass-damper system with k=4000 N/m, m = 10kg, and c = 40Ns/m. Find the steady-state and total responses of the system under the harmonic force N and the initial conditions =0.1m and ሶ
•
•
•
•
Sometimes the base or support of a spring-mass-damper system undergoes harmonic motion Let () 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 ( ) and the relative velocity between the two ends of the damper is (ሶ ሶ ) From the free-body diagram shown, we obtain the equation of motion:
ሷ (ሶ ) ሶ k ( ) 0
•
•
•
•
ሷ (ሶ ) ሶ k ( ) 0 If sin , we get ሷ ሶ k ky cሶ sin cos sin( ) − 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
sin( )
which can be written as
where
tan−
sin( )
−
•
Using trigonometric identities,
•
where
can be written as sin( )
and
•
sin( )
1 2 1 2
3 3 2 − tan− tan 1 4 1 The ratio of the amplitude of the response to that of the base motion y(t), is called the displacement transmissibility
+ − +
tan−
+ −
•
•
•
•
•
•
•
The value of is unity at r=0 and close to unity for small values of r. For an undamped system ( 0), → ∞ at resonance ( 1) The value of is less than unity ( < 1) for values of ( > 2) for any amount of damping The value of is unity for all values of at 2 For < 2, smaller damping ratios lead to larger values of On the other hand, for > 2, smaller values of damping ratio lead to smaller values of The displacement transmissibility , attains a maximum for 0 < < 1 at the frequency ratio < 1 given by
1 1 8 1 2
+ − +
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
(ሶ ) ሶ k ( ) ሷ Using sin( ), we get mω sin sin( ) where is the amplitude or maximum value of the force
•
•
transmitted to the base given by 1 2 1 2 •
The ratio
is known as the force
transmissibility.
Relative motion •
If denotes the motion of the mass relative to the base, the equation of motion becomes
ሷ ሶ k ሷ sin
•
The steady-state solution is given by
sin( )
•
where Z, the amplitude of (), can be expressed
tan−
1 2 2 − tan 1
sin( )
Example 3.4 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 . . If the vehicle speed is 20 km/hr, determine a) the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude of Y=0.05m and a wavelength of 6m
Example 3.4 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 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
•
•
•
•
•
•
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
sin
•
The equation of motion can be derived by the usual procedure
ሷ ሶ k sin
•
The solution of this equation will be identical to previously derived equation for damped forced vibration, if we replace m and by M and respectively
sin( ) •
Where,
tan−
− +
•
By defining
•
•
1
•
/ and 2
tan−
2
2 1
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 ( large), MX/me • is almost unity, and the effect of damping is negligible.
For 0
< < 1/ 2, the maximum of MX/me occurs when, 0 The solution of equation is
1 >1 12
•
Corresponding maximum value of MX/me is given by
•
•
•
1 2 1
Thus the peaks occur to the right of the resonance value of r=1 For > 1/ 2, [MX/me] does not attain a maximum. Its value grows from 0 at r=0 to 1 at → ∞ The force transmitted to the foundation due to rotating unbalanced force (F) can be found as
() ሶ •
The magnitude (or maximum value) of F can be derived as
1 4 1 4
