FORCED VIBRATIONS INTRODUCTION
When a mechanical system undergoes free vibrations, an initial force (causing some displacement) is impressed upon the system, and the system is allowed to vibrate under the influence of inherent elastic properties. The system however, comes to rest, depending upon the amount of damping in the system.
In engineering situations, there are instances where in an external energy source causes vibrations continuously acting on the system. Then the system is said to undergo forced vibrations, as it vibrates due to the influence of external energy source. The external energy source may be an externally impressed force or displacement excitation impressed upon the system. The excitation may be periodic, impulsive or random in nature. Periodic excitations may be harmonic or non harmonic but periodic. The amplitude of vibrations remains almost constant. Machine tools, internal combustion engines, air compressors, etc are few examples that undergo forced vibration.
3.2 FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION
Consider a spring mass damper system as shown in Figure 3.1 excited by a sinusoidal forcing function F=Fo Sinϖt
Figure 3.1
Let the force acts vertically upwards as shown in FBD. Then the Governing Differential Equation (GDE) can be written as
m = - K - C + F x88 x8 x8
C
k
m
F = F
0Sin ωt
kx
Cx
∙
F
x
F
oF
o∫
-π m mg Kδ K x m
m + C + Kx = F --- (3.1)
is a linear non homogeneous II order differential equation whose solution is in two parts.
1. Complementary Function or Transient Response Consider the homogenous differential equation,
m + C + Kx = 0 which is incidentally the GDE of a single DOF spring mass damper-system. It has been shown in earlier discussions that for different conditions of damping, the response decays with time. Thus the response is transient in nature and therefore termed as transient response.
For an under damped system the complementary function or transient response.
xc = X1 ēζ Sin (ωdt + φ )
xc = X1 ēζ (A Sin ωdt + B Cos ωdt)--- (3.2)
2. Particular Integral or Steady State Response
This response neither builds up nor decays with time. It is steady state harmonic oscillation having frequency equal to that of excitation. It can be determined as follows.
Consider non-homogenous differential equation
m + C + K = Fo Sin ωt --- (3.3)
The particular integral or steady state response is a steady state oscillation of the same frequency ω as that of external excitation and the displacement vector lags the force vector by some angle.
Let x = X Sin (ωt - φ ) be the trial solution X: Amplitude of oscillation
φ : Phase of the displacement with respect to the exciting force (angle by which the displacement vector lags the force vector).
∴Velocity =
= ωX. Cos (ωt - φ ) = ωX Sin [90 + (ωt - φ )] Acceleration
= - ω2 X. Sin (ωt - φ ), substitute these values in GDE, (equation 3.1)
We get
-m ω2 X Sin (ωt - φ ) + Cω Sin [90 + (ωt - φ )]
+ KX Sin (ωt - φ ) = Fo Sin ωt
m ω2 X Sin (ωt - φ ) - Cω Sin [90 + (ωt - φ )]
- KX Sin (ωt - φ ) + Fo Sin ωt = 0 --- (3.4)
The four terms in the above equation represent both in magnitude and direction, the four forces namely: inertia force, damping force, spring force and impressed force, taken in
x8 x88 x8 x88 x88 x8 x8 x8 x88 ω nt ω nt x8
order, acting on the system and their sum is equal to zero. Thus they satisfy the D’Alemberts principle. Σ F = 0. Now, if vector representation as shown in Figure 3.2, is employed to denote these forces the force polygon shown in Figure 3.3 should close.
Represent the force vectors and draw the force polygon as given below.
Figure 3.2
Figure 3.3
Impressed force: Fo Sin ωt: acts at an angle ωt from the reference axis.
Displacement vector X: Lags the force vector by an angle φ and hence shown at (ωt - φ ) from the reference axis.
Reference axis
) (ωt -
φ
)
Spring force: - KX Sin (ωt - φ ): which means that the vector – KX acting at (ωt - φ ) or KX acting in opposite direction to (ωt - φ ) = at [90 + (ωt - φ )]
Damping force: - CωX [Sin (90 + (ωt - φ )]
- CωX acting at 90 + (ωt - φ ) or CωX acting in opposite direction to [90 + (ωt - φ )]
Inertia force: mω2X Sin (ωt - φ )
Vector mω2X acting at (ωt - φ )
From the force polygon, in figure 3.3
Consider the triangle OAB. OA2 = OB2 + BA2 = (CωX)2 + (KX-mω2X)2 Fo2 = X2 (Cω)2 + x2 (K-mω2)2 Fo2 = X2 [(K-mω2)2 + (Cω)2] X = --- (3.5) and tan φ = = = tan φ = ∴ φ = tan-1 --- (3.6)
If X and φ are expressed in non-dimensional form it enables a concise graphical presentation of results. Therefore, divide the numerator and the denominator by K.
∴X = tan φ =
Further, the above equations can be expressed in terms of the following quantities = Xst - Zero frequency deflection
∴Deflection of spring mass system under the steady force Fo should not be mistaken as Δst =
= F0 √(K-mω2)2 + (cω)2 OB BA Cωx KX-mω2x Cω K-mω2 Cω K-mω2 Cω K-mω2 F0 /K √ m ω2 K)2 + (1- cω Kk ( )2 Cω /K (1- mω2 K) Fo K mg K m K 1 ωn2
= Thus X =
= r = frequency ratio
--- (3.7)
where is called magnification factor, amplification factor, or amplitude ratio. M= : It is the term by which Xst is to be multiplied to get the amplitude.
ω
tanφ = = 1 -
... tanφ = --- (3.8)
Thus the steady state response xp = X Sin (ωt - φ ), in which
X and φ are as given above. Total solution x = xc + xp
For under damped conditions:
as t ∞, xc 0 i.e., the transient response dies out. Complete solution consists only
steady state response only.
∴x = X Sin (ωt - φ ) --- (3.9)
As mentioned above, the transient vibrations die out very soon and hence the system vibrates with steady response amplitudes. The behaviour of the system can be best understood by plotting frequency response curves as given below, in figure 3.4 and 3.5.
Frequency Response Curves:
Magnification Factor vs Frequency Ratio for Different amounts of Damping
C K 2ζKωn Xst √(1 - )2 + (2 ζ. ω)2 ω2 ωn2 ωn ω ωn X = Xst √ (1 - r2)2 +( 2 ζ. r)2 X Xst X Xst 2ζ ωn ω2 ωn2 2ζr 1-r2 2ζr 1-r2
(
)
Frequency Ratio r = (ω/ω
n)
ζ = 0
ζ = 0
ζ = 0.25
ζ = 0.375
ζ = 0. 5
ζ = 0.707
ζ = 1
ζ = 2
M
ag
n
if
ic
at
i
on
Fa
ct
o
r
M
=
X
/X
st
Figure 3.4
Phase lag vs frequency ratio for different amounts of damping.
Figure 3.5
The following characteristics of the magnification factor (M) can be observed.
1) For damped system (ζ =0); M as r 1.
2) Any amount of damping (ζ >0) reduces the magnification factor (M) for all values of forcing frequency.
3) For any specified value of r, a higher value of damping reduces the value of M 4) When the force is constant (r =0), M =1.
5) The amplitude of the forced vibrations becomes smaller with increasing value of forced frequency. i.e M 0,as r ∞.
6) For 0< ζ < 1/ √2 (0 < ζ <0.707), the maximum value of M occurs when r=√(1-2ζ 2)
or ω= ωn√(1-2ζ 2), which is lower than the Undamped natural frequency ωn and the
damped natural frequency ωd = ωn√(1-2 ζ 2),
7) The maximum value of X (when r=√ (1 - 2 ζ 2) is given by (X/X
st)= 1/[2 ζ√ (1-ζ2)]
and the value of X at ω= ωn is given by (X/Xst ) = 1/2 ζ
ζ = 0
0.5
1.0
1.5
2.0 2.5
3.0
ζ = 0.25
ζ = 0.5
ζ = 0.707
ζ = 1.0
ζ = 2.0
Frequency Ratio r = (ω/ω
n)
Ph
a
se
A
n
gl
e,
φ
,
8) For ζ >1/√2, the graphs of M decreases with increasing values of r.
The following characteristics of the phase angle φ can be observed from the graph
1) For undamped system the phase angle is 00 for 0<r<1, and 1800 for r>1. This implies
that the excitation and response are in phase for 0<r<1 and out of phase for r>1 when ζ =0.
2) For ζ >0 and 0<r<1 the phase angle is given by 00<φ <900, implying that the response
lags excitation.
3) For ζ >0 and r>1, the phase angle is given by 900<φ < 1800, implying that the
response leads excitation.
4) For ζ >0 and r=1, the phase angle is φ =900 implying that the phase difference
between the excitation and response is 900.
5) For ζ >0 and large values of r, the phase angle φ ω approaches 1800 implying that the
response and excitation are out of phase.
The damping factor ζ has a large influence on amplitude and phase angle in the region where r = 1(resonance).The phenomenon represented be frequency response curve can be further better understood by means of vector diagram as follows. Consider three different cases as (1) ω/ ωn << 1 (2) ω/ ωn = 1 (3) ω/ ωn >> 1
Case (1): ω/ ωn << 1 for which ω should be very small
At very low frequencies, when ω is very small, the inertia for m ω2x and the damping force
Cωx are very small.
Figure 3.6
This results in small values of φ as shown in fig .The impressed force F0 is almost equal and
opposite to spring force KX. Thus for very low frequencies, the phase angle tends to zero and the impressed force wholly balance the spring force
Case (2): whenω/ ωn = 1
x
φ
(ωt-
φ
)
φ
mω
2X
Kx
F
oCωX
x
Kx
F
o
Figure 3.7
With increased value of ω, the damping force Cωx and inertia force m ω2 x increase. The
phase angle also increases. If ω is increased to such an extent that phase angle φ =900, the
force polygon becomes a rectangle as shown. The spring force and inertia vectors become equal and opposite.
KX = m ω2 x
ω = √(K/m) = ωn
ω = ωn
ω / ωn =1
This is the response condition of the system during which the forcing frequency is equal to natural frequency of the system. Also the impressed force is completely balanced by the damping force. CωX= F0 X= F0 /Cω = F0/K/ Cω/K X=Xst / 2 ζ (ω/ ωn) X=Xst / 2 ζ (ω/ ωn) = 1 Xr /Xst = 1/2 ζ Xr = Amplitude at resonance
mω
2X
CωX
Case (3): when ω / ωn >>1
Figure 3.8
At very large values of ω >φ approaches 1800, the inertia force becomes very large, where
as the spring force and damping force vectors becomes negligibly small. The improved force is wholly utilized in balancing the inertia force.
φ 1800 i.e., F
o = m ω2x
X = Fo / m ω2
NUMERICAL EXAMPLES:
3.1) A machine part of mass 2.5 Kgs vibrates in a viscous medium. A harmonic exciting
force of 30 N acts on the part and causes resonant amplitude of 14mm with a period of 0.22sec. Find the damping coefficient. If the frequency of the exciting force is changed to 4Hz, determine the increase in the amplitude of forced vibration upon removal of the damper. Data: m = 2.5Kg, F0 = 30N, X = 14mm, τ = 0.225sec
Part 1: At Resonance
ωn = forcing frequency = 2π / τ = 28.56 rad/sec
At resonance: ω = ωn = 28.56 rad/sec ωn = √(K/m) = 28.56 rad/sec K = 2039 N/m Amplitude at resonance X = As ω/ ωn = 1, X = (F0/K)/2ζ = 0.014 ... ζ = 0.526 Damping coefficient = C = Cc ζ = 2m ωn ζ = 2*2.5*28.56*0.526 = 75.04 N/m/s C = 0.07504 Ns/m Fo/K √ [1 - r2] 2 + [2ζr] 2
x
mω
2X
F
o
Part (2): When f = 4 Hz
Forcing ω = 2π *fn = 25.13 rad/sec
Frequency ωn = 28.56 rad/sec, unchanged
Amplitude of vibration with damper Xa = Fo/K
√ [1 - r2] 2 + [2ζr] 2
= 0.01544m
Amplitude of vibration without damper Xb = (30/2039)/(0.2258)
= 0.0652m
Increase in Amplitude = 0.0652 – 0.0155 = 0.0497m Amplitude = 49.7mm
3.2) A body having a mass of 15 kgs, is suspended from a spring which deflects 12mm due to
the weight of the mass. Determine the frequency of free vibrations. What viscous damping force is needed to make the motion a periodic at a speed of 1mm/sec.
If when, damped to this extent, a disturbing force having a maximum value of 100N and vibrating at 6Hz is made to act on the body. Determine the amplitude of ultimate motion. Solution:
Data: m = 15Kg; F0 = 100 N; f = 6Hz; Δst = 12mm;
(a) fn = (1/2π )√(g/ Δst) = 4.55Hz
(b) The motion becomes aperiodic, when the damped frequency is zero or when it is critically damped (ζ = 1).
ω = ωn = √(g/ Δ) = 28.59 rad/sec
C = Cc = 2m ωn = 2*15*28.59 = 857 N/m/s
= 0.857 N/mm/s
Thus a force of 0.857 N is required at a rate of 1mm/s to make the motion a periodic.
(c) X = F0 √(K-mω2)2 + (cω)2 ω = 2π f = 2π *6 = 37.7 rad/sec, f0 = 100 N fn = (1/2π )(√(K/m) ... K = 12,260 N/m X = 0.00298m = 2.98mm.
Condition for peak amplitude of vibration (Expression for peak amplitude)
The frequency at which the maximum amplitude occurs can be obtained as follows.
M = ∴ X =
i.e., for a system acted upon by a known harmonic force, the amplitude depends only on (ω/ ωn). Hence for X to be maximum √ [1 – r2] 2 + [2ζr] 2 should be minimum.
∴ ∴ ([1 - r2] 2 + [2ζr] 2) = 0 2(1 - r2) 2 (-2r) + 4ζ2r = 0 2(1 - r2)+ 4ζ2r = 0 42ζ2r = 0 = 2(1 - r2) 2ζ2 = 1-r2 r2 = 1 -2ζ2 r = √1 -2 ζ2 (ω/ ωn) peak = √1 -2 ζ2 = √1 -2 ζ2 = √1 -2 ζ2 --- (3.10)
ωp = frequency at which peak amplitude occurs.
Where ωp refers to the forcing frequency corresponding to the peak amplitude. No maximum
or peak will occur when the expression within the radical sign becomes negative i.e., for ζ > or for ζ > 0.707.
= √1 -2 ζ2
and peak amplitude is given by
(X/Xst)max = 1/[2 ζ(√1- ζ)] --- (3.11)
3.3) A machine of mass 25 kgs, is placed on an elastic foundation. A sinusoidal force of
magnitude 25N is applied to the machine. A frequency sweep reveals that the maximum X Xst Xst √ [1 - r2] 2 + [2ζr] 2 dx d(ω/ ωn) dx d(r) ω ωn
( )
peak( )
ωp peak ωn 1 √2 ω p ωn( )
steady state amplitude of 1.3mm occurs when the period of response is 0.22sec. Determine the equivalent stiffness and damping ratio of the foundation.
Solution:
Data: F0 = 25N; m = 25 Kgs; Xmax = 1.3mm; τ = 0.22sec
For a linear system, the frequency of response is same as frequency of excitation. ... Excitation frequency = ω = 2π f = 2π / τ = 28.6 rad/sec
thus Xmax = occurs, when ω = 28.6 rad/s
Condition for maximum amplitude to occur: r = √1 -2 ζ2 = ω/ω n ... ω n = ω /(√1 -2 ζ2 ) = 28.6/(√1 -2 ζ2 ) ---(1) also we have, X/Xst = 1 for Xmax = r =√1 -2 ζ2 √ [1 - r2] 2 + [2ζr] 2 Xmax/Xst = 1 √ [1 – (1 -2 ζ2 )] 2 + [4ζ2(1 -2 ζ2 )] = 1 2 ζ √ (1 -ζ2 ) Xmax/(F0/K) = 1 2 ζ √ (1 -ζ2 ) Xmaxmωn2/F0 = 1 2 ζ √ (1 -ζ2 ) 25*0.013* ωn2/25 = 1 2 ζ √ (1 -ζ2 )
0.013*28.6/(√1 -2 ζ2) = 1
2 ζ √ (1 -ζ2)
1.0633/(√1 -2 ζ2) = 1
2 ζ √ (1 -ζ2)
Squaring and rearranging, ζ4 - ζ2 +0.117 = 0
Z2 – Z + 0.117 = 0 where ζ2 = Z.
Solving the quadratic equation ζ = 0.368, 0.93
The larger value of ζ is to be discarded because the amplitude would be maximum only for ζ < 0.707 ... take ζ = 0.368
... natural frequency ω
n = ω
√ (1 – 2(0.368)2 )
ωn = 33.5 rad/sec
stiffness of the foundation,
K = mωn2 = 25(33.5)2
= 28.05*103 N/m
3.4) A weight attached to a spring of stiffness 525 N/m has a viscous damping device. When
the weight is displaced and released, without damper the period of vibration is found to be 1.8secs, and the ratio of consecutive amplitudes is 4.2 to 1.0. Determine the amplitude and phase when the force F=2Cos3t acts on the system.
Solution:
Data: K = 525 N/m; τ = 1.8secs: x1 = 4.2; x2 = 1.0; F = F0sinωt = 2cos3t
... F 0 = 2N, ω = 3 rad/sec X = Fo/K √ [1 - r2] 2 + [2ζr] 2 ωn = 2π / τ = 3.49rad/sec δ = ln(4.2/1.0) = 1.435 ζ = δ = 0.22 √ (4π 2 + δ2) r = ω/ωn = 2/3.49 = 0.573 r2 = 0.328 X = 2/525 √ [1 – 0.328] 2 + [4*0.484*0.328] X = 5.3mm
φ = tan-1(2ζr) (1-r2) φ = tan-1(2*0.22*0.573) (1-0.328) φ = tan-1(0.375) φ = 20.560
3.5) The damped natural frequency of a system as obtained from a free vibration test is 9.8
cps. During a forced vibration test with a harmonic excitation on the same system, the frequency of vibration corresponding to peak amplitude was found to be 9.6 cps.
Determine the damping factor for the system and natural frequency. ωd = 9.8 cps, (ωp / ωn) = √1 -2ζ2 ωp = 9.6 cps. ωn = ωd/√1 -2ζ2 ∴ ωp√1 -2ζ2 /ωd = √1 -2ζ2 Solving for ζ: ζ = 0.196 ωn = ωd/ √1 -2ζ2 = 10 cps.
3.6) A reciprocating pump of mass 300 Kgs is mounted at the middle of a steel plate of
thickness 12 mm and width 500 mm and length 2.5 m damped along two edges as shown. During the operation of the pump, the plate is subjected to a harmonic excitation of F(t) = 50 cos 60 t N. Determine the amplitude of vibration of the plate.
m = 300 Kgs F0 = 50 N ω = 60 K = 192EI/l3 = 176.94*103 N/m ζ = 0 X = F0 /(K-m ω2)2 X = 6.13*10-8mm
Vibrations Due to Reciprocating and Rotating Masses
Unbalance in rotating machine is one of the common causes of vibration. The centrifugal force generated (meω 2) due to the rotation of the body is proportional to the square of the
frequency of rotation. This CF varies with speed of rotation and is different from the
2.5 m
500
12
harmonic excitation discussed in previous articles in which the maximum force is independent of frequency.
Fig. 3.9 Model of Reciprocating Machine Fig. 3.10 Model of Rotating Machine
Let: m: mass of unbalanced mass.
M: Total mass including unbalanced mass. e = eccentricity of unbalanced mass. = crank radius of reciprocating machine = stroke / 2.
The force due to the unbalanced mass is as shown in the FBD. The GDE
∴m + C + Kx = meω 2 Sin ω t
= F0 Sin ω t
where F0 = meω 2
Let the steady state response be x = X. Sin (ω t - φ )
from the previous discussion we have = where Xst = F0 / K, r = ω /ω n here F0 = meω 2. ∴X = X = x8 X Xst 1 √[1-mr2]2 + [2 ζ r]2 meω 2 /K √[1-mr2]2 + [2 ζ r]2 meω 2 M M K √[1-mr2]2 + [2 ζ r]2 MX me ω 2 M/K √[1-mr2]2 + [2 ζ r]2 ω 2 * 1 ω 2 (K/M) ω n2 x8
ω
ω
= =
--- (3.12) and
φ = tan-1 --- (3.13)
The variation of MX/me with (r = ω /ω n) for different values of ζ is shown in figure 3.11
However, the variation of φ and r remains as earlier.
Figure 3.11
The following observations can be made.
Case (i) when ω <<<ω n, say r ≈ 0
MX/me = 0 (independent of ζ, effect of damping is negligible)
Case (ii) when ω = ω n ; r = 1
∴MX/me = 1/2 ζ, dependent on ζ
if ζ = 0, MX/me = ∞, it is a case of resonance.
Case (iii) when ω >>>ω n, r>>1, r ≈ ∞.
MX/me = 1 (independent of ζ, effect of damping is negligible)
At low speed meω 2 is zero and hence the curve starts from zero. It increases with increase in
(ω /ω n) until the condition of resonance is achieved. At resonance MX/me = 1/2 ζ and thus
√[1-mr2]2 + [2 ζ r]2 MX me r2 √[1-mr2]2 + [2 ζ r]2 2 ζ r 1-r2
M
X
/
m
e
ζ = 0
ζ = 0.1
ζ = 0.15
ζ = 0.25
ζ = 0.5
ζ = 1.0
r =
ω
/
ω
n
the amplitude X is limited, by the damping present in the system. When ω /ω n is very large
MX/me approaches unity.
Numerical Examples
Unbalanced Rotating and Reciprocating Masses and Force Transmissibility.
1) A reciprocating machine of mass 75 Kgs is mounted on springs of stiffness 11.76*105 N/m
and a damper of damping factor 0.2. The slider of mass 2 Kgs within the machine has a reciprocating motion with a stroke of 0.08 m. The speed is 3000 rpm. Assuming the motion of the piston to be harmonic, determine
1. Amplitude of vibration of the machine. Solution:
M = 75 Kgs; m = 2 Kgs, K = 11.76*105 N/m.
For vibrations due to rotating unbalance Amplitude of vibration = e = stroke/2 = 0.08/2 = 0.04 m ω = 2π (3000) / 60 = 314 rad/sec. ω n = √K/m = √11.76*10.5 /75 = 125 rad/sec. ω /ω n = r = 314 /125 = 2.51 75 (X)/2(0.04) = (2.51)2 / √(1-2.52)2 + (2*0.2*2.51)2 X = 0.00125 m = 1.25 mm MX me r2 √[1- r2]2 + [2 ζ r]2
Vibration Isolation and Transmissibility
In machines vibrations are caused due to unbalanced masses. These vibrations are transmitted to the foundation upon which the machines are installed. If the transmission of vibrations to the foundations is not avoided the adjoining machines also set to vibrate.
To minimize the forces transmitted to the foundation machines are usually mounted on springs or dampers or some other vibration isolation material. Vibration isolation is measured in terms of the motion or force transmitted to the foundation. The lesser the force or motion transmitted the greater the vibration isolation.
Force Transmissibility or Transmissibility Ratio
In the case of forced vibrations, it is defined as the rating force transmitted to that impressed upon the system. it is a measure of the effectiveness of a isolating material.
For a spring mass damper system under harmonic excitation
X = --- (a) and
φ = tan-1 --- (b)
The forces are transmitted to the foundation or structure through the springs and dampers provided in the system. Thus the force transmitted to the foundation are the spring force KX and the damping force cω x. Hence the total force transmitted to the foundation is the vector sum of KX and Cω X as shown in the Figure.
Ft = Force transmitted = √(KX)2 + (Cω X)2 = √(KX)2 + (Cω X)2. (KX)2 / (KX)2 = (KX) √1 + (Cω /K)2 Xst √[1- r2]2 + [2 ζ r]2 2 ζ r 1-r2
( )
m
Impressed force
Spring force
(KX)
Damping force
(CωX)
Foundation
Substituting for X from (a)
Ft = K. √1(Cω /K)2 =
= = Transmissibility Ratio (TR)
A plot of transmissibility ratio ω /ω n is shown in figure given below.
The following observations can be made.
Case (i) when ω /ω n = 0, r = 0, TR = 1, (independent of ζ )
(ii) ω = ω n, r = 1, resonance.
TR = , dependent on ζ
If ζ = 0, TR = ∞
Case (iii) when ω /ω n = √2, Ft/F0 = 1, independent of ζ
Case (iv) when ω /ω n >>> , r ∞
Ft / F0 = TR = 0
Discussions: When ω /ω n = 0, i.e., the force is steadily applied, TR = 1, irrespective of the
amount of damping produced in the systems. When ω /ω n = 1, it is condition of resonance.
The force transmitted is infinity. If damping is used the magnitude of transmitted force is F0 K √[1- r2]2 + [2 ζ r]2 F0 √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 Ft F0 √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 √ 1+ 4 ζ 2 2 ζ
Frequency Ratio r = (ω/ω
n)
F
t/F
0ζ = 0
ζ = 0.2
ζ = 0.5
ζ = 0.6
ζ = 0.6
ζ = 0.2
ζ = 0
reduced. When ω /ω n < √2 the transmitted is always greater then the impressed force. When
ω /ω n = √2, for all the values of damping the force transmitted is equal to impressed force.
When ω /ω n > √2, the transmitted force is always less than the impressed force. it also
implies that TR decreases with decreasing values of ζ. Thus, an undamped spring is superior to a damped spring in reducing force transmissibility. But certain amount of damping is necessary for ω to pass through the resonance condition.
As seen from the above order to isolate vibrations due to external force, ω /ω n should be
very large, i.e., >√2. For a given value of ω /ω n should be very small. The static deflection
of the spring should be as high as possible. These conditions will be satisfied by materials like steel springs, rubber, cork, felt etc., which are generally used as vibration isolators.
Numerical Examples on
Unbalanced Rotating and Reciprocating Masses and Force Transmissibility.
1. A reciprocating machine of mass 75 Kgs is mounted on springs of stiffness 11.76*105 N/m
and a damper of damping factor 0.2. The slider of mass 2 Kgs within the machine has a reciprocating motion with a stroke of 0.08 m. The speed is 3000 rpm. Assuming the motion of the piston to be harmonic.
2. Amplitude of vibration of the machine. 3. Transmissibility ratio.
4. Force transmitted to the foundation. 5. Is vibration isolation achieved? If so how. Solution:
M = 75 Kgs: m = 2 Kgs, K = 11.76*105 N/m.
For vibrations due to rotating unbalance Amplitude of vibration = e = stroke/2 = 0.08/2 = 0.04 m ω = 2π (3000) / 60 = 314 rad/sec. ω n = √K/m = √11.76*10.5 /75 = 125 rad/sec. ω /ω n = r = 314 /125 = 2.51 75 (X)/2(0.04) = (2.51)2 / √(1-2.52)2 + (2*0.2*2.51)2 X = 0.00125 m = 1.25 mm Transmissibility Ratio: (TR) MX me r 2 √[1- r2]2 + [2 ζ r]2
TR = = TR = = = TR = 0.1861
Force transmitted to the foundation Also TR =
Ft = (TR) F0 = (0.1861) * meω 2
= (0.1861) * 2 * 0.04 (314)2
Ft = 1467.9 N
Vibration Isolation: Vibration isolation is achieved as only 18.6 % of the maximum shaking force (F0) is transmitted to the foundation. This is because the operating range of frequency
ratio ω /ω n = r < √2 (2.51> 1.41).
As r >>>>> √2, 0
2. A mass of 100 Kg, is mounted on a spring support having a spring stiffness of 20000 N/m and a damping coefficient of 100 NS/m. The mass is acted upon by a harmonic force of 39 N at the undamped natural frequency of the set up. Find
1. Amplitude of vibration of the mass.
2. Phase difference between the force and displacement. 3. Force transmissibility ratio.
3. A refrigerator of mass 35 Kgs operating at 480 rpm is supported on 3 springs. If only 10% of the shaking force is to be transmitted to the foundation what should be the value of K. =
assuming that no damped used ζ = 0 TR = 1/√(1- r2) = 1/ ± (1- r2) ω = 2π* 480 / 60 = 16 π rad/sec, TR = 0.1 = 0.1 = ∴± 0.1 – 0.1 (16 π/ω n)2 = 1 Ft F0 √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 √1 + (2*0.2*2.51)2 √(1-2.52)2 + (2*0.2*2.51)2 √ 1+ 1.0080 √28.09 + 2.008 √ 2.008√29.09 Ft F0 Ft F0 Ft F0 √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 Ft F0 1 ± [1- (16 π/ω n)2]
When positive sign is considered - 0.1 (16 π/ω n)2 ± 1 – 0.1
(16 π/ω n)2 = 0.9/-0.1 = - 9
∴(16 π/ω n) = √-9 which is not possible
Taking the negative sign - 0.1 + 0.1 (16 π/ω n)2 = 1
16 π/ω n = (1+0.1/0.1) = √11
∴ω n = 15.15 rad/sec.
ω n K = √K/m ∴Keq = 8.037 N/m
∴K = 8.037/3 = 2.679 N/m.
4. A machine supported symmetrically on four springs has a mass of 80 Kgs. The mass of the reciprocating mass is 2.2 Kgs which move through a vertical stroke of 100 mm with SHM. Neglecting damping, determine the combined stiffness of the springs so that the force transmitted to the foundation is 1/20th of the impressed force. the machine crank shaft rotates
at 800 rpm.
If, under actual working conditions, the damping reduces the amplitudes of successive vibrations by 30%, find (a) The force transmitted to the foundation at 800 rpm (b) The force transmitted to the foundation at resonance. (c) The amplitude of vibrations at resonance. M = 80 Kgs, m 2.2 Kg TR = 1/20 = 0.05 N = 800 rpm
e = 100/2 = 50 mm. ω = 2 πN/60 = 83.78 rad/sec. 1. In the absence of damping
TR = 1/ (r2 – 1)
0.05 = - 1 ∴ ω n = 18.28 rad/sec
ω n2 = K/m
K = Mω n2 = 26.739 N/m
2. When damping is present
δ = In (X1/x2) = In (1/1-0.3) = 2 π ζ / √1 – ζ2
1 83.78
ω n
∴ ζ = 0.0567
Ft = Force transmitted to foundation at 800 rpm.
= = 0.0563 = TR Ft = F0 * TR = TR * meω 2 = 43.47 N 3. At resonance ω /ω n = 1 TR = = 8.875 = TR ∴(Ft) Res = F0 * TR = meω n2 * TR = 2.2 * 0.05 * (18.78)2 * 8.875 (Ft)Res = 326.25 N
4. Amplitude of vibration at resonance. =
= 32.6.25/26.739 = 12.2 mm
Forced Vibrations due to Excitation of Base
Some times the base or support of a spring-mass damper system undergoes harmonic excitation as shown in figure.
Let y: denotes the displacement of the base and x: denotes the displacement of the mass from static equilibrium position at a given instant t; such that
Absolute Amplitude of the Mass (X) y = Y sin ω t Ft F0 √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 √ 1+ (2 ζ r)2 2 ζ Ft F0
Force transmitted at Resonance Stiffness
+ x
+ y
Base
y = Y Sin ω
t
M
Base excitation
Now from NSL
m = - K (x-y) – C( - ) = - Kx + Ky – C +
∴m + C + Kx = C + Ky governing differential equation. Let the steady state response
x = X Sin (ω t - φ ) = ω X cos (ω t - φ ) = ω X Sin [90 + (ω t - φ )] = - ω 2 X Sin (ω t - φ ) also y = Y Sin ω t = ω Y Cos ω t = ω Y Sin [90 + ω t]
= - ω 2 Y Sin (ω t), substituting these values in GDE
- m ω 2X Sin (ω t - φ ) + Cω X Sin [90 + (ω t - φ )] + KX Sin (ω t - φ )
= Cω Y Sin [90 ω t] + K Y Sin ω t.
m ω 2X Sin (ω t - φ ) - Cω X Sin [90 + (ω t - φ )] – KX Sin (ω t - φ )
+ Cω Y Sin [90 ω t] + K Y Sin ω t = 0
Thus Σƒ = 0. The forces can be represented as shown, and the force polygon should close. From the triangle OAB.
x88 x8 y8 x8 y8 x88 x8 y8 x8 x88 y8 y8 8
KX
CX
.
Ky
Cy
.
x
y
x
KX
CX
.
Ky
Cy
.
x
K (x-y) C(x – y)
. .
OA2 = AB2 + BO2
F02 = (KX - mω 2X)2 + (Cω X)2
= [KX - mω 2X (KX/KX)]2 + [Cω X (KX/KX)]2
F02 = (KX)2 [(1- mω 2/K)2 + (Cω /K)2]
Also F02 = (KY)2 + (Cω Y)2 = (KY)2 [1 + Cω Y/ KY)2]
= (KY)2 [1 + Cω / K)2]
∴(KY)2 [1 + Cω / K)2] = (KX)2 [(1 - mω 2/ K)2 + (Cω / K)2]
Taking square roots and rearranging the terms KY √[1 + Cω / K)2] = KX √[1- r2]2 + [2 ζ r]2
The ratio X/Y is called displacement transmissibility. This equation is similar to that of transmissibility ratio, and all the observation and discussions are same as that discussed under transmissibility ratio and the frequency response curve given below.
Displacement transmissibility is defined are the ratio of displacement transmitted to the mass to the displacement impressed upon the base.
KX
CωX
mω
2X
KX-Cω
2X
CωY
(ωt -
φ
)
φ
KY
B
0
F
0A
X
Y
ωt
(ωt - φ
)
CωY
KX
CωX
KY
Mω
2X
X
Y
Relative Amplitude: If Z represents the relative motion of the mass with respect to the
support we have Z = x-y
∴x = (z + y)
Substituting this in the governing differential equation. m + C + Kx = C + Ky
m + C + KZ = - m
and substituting for ‘y’ from y = Y Sin ω t m + C + KZ = m ω 2Y Sin ω t
Comparing this with
m + C + Kx = meω 2 Sin ω t, of reciprocating and rotating unbalance.
The steady state relative amplitude Z and the phase angle lag φ between the excitation and relative displacement. The relative motion frequency response which is similar to reciprocating and rotating unbalance as given below, can be used in designing vibration measuring instruments. Z/ Y = r2 / √[1- r2]2 + [2 ζ r]2 φ = tan-1 [2 ζ r/ 1- r2] x88 x8 y8 y 8 8 Z 88 Z8 Z 88 Z8 x88 x8
r = (ω/ω
n)
X
/
Y
ζ = 0
ζ = 0.2
ζ = 0.5
ζ = 0.6
ζ = 0.5
ζ = 0.2
ζ = 0
Z
/Y
Force Transmitted: Force is transmitted to the base through the spring and dampers. If ‘Z’
represents the relative displacement then Force transmitted = Ft= √(KZ)2 + (Cω X)2
Ft = Z√K2 (Cω )2
The force transmitted to the base is also determined by Ft = m ω 2X
Numerical Examples on Base Excitation
1. Figure (given in problem No.3) shows a simple model of motor vehicle that can vibrate in 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 100 Km/hr. Determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude Y = 0.05 m and a wave length of 6 m.
Given m = 1200 Kg Speed 100 Km/hr
K = 400 * 103N/m Y = 0.05 m
ζ = 0.5 wave length = 6 m
= period
Model: Single degree freedom damped system base excitation.
Frequency of base excitation: ƒ = Speed/ Length of one cycle =
ƒ = 4.63 Hz.
∴Frequency of base excitation
ω = 2πƒ = 2π = 29.0087 rad/sec.
ζ = 0
ζ = 0.1
ζ = 0.15
ζ = 0.25
ζ = 0.5
ζ = 1.0
r =
ω
/
ω
n (100*1000)/3600 m/sec 6 m (100*1000) 3600 * 6 m
Natural frequency = ω n = √ K/m
= √400* 103 / 1200 = 18.2574 rad/sec.
∴Frequency ratio = r = ω /ω n = 1.5903
∴ = = 0.8493
∴Displacement amplitude of the vehicle X/Y = 0.8493 ∴ X = 0.8493 * (0.05)
X = 0.0425 m
2. A precession grinding machine is supported on an isolator that has a stiffness of 1 MN/m and a viscous damping constant of 1KN-S/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of grinding machine. Find the maximum acceptable displacement of the floor if resulting amplitude of vibration of grinding wheel is to be restricted to 10-6m. Assume that
the grinding wheel and machine are rigid bodies of total weight 5000 N.
Solution: W = 5000 N ∴m = w/g = 509.6 Kg K = 1*106 N/m W = mg;m = 5000/9.81 Kg X = 10-6m C = 103 N-S/m Y = ? y = Y Sin 10πt = ∴ω = 10 π ω = 31.4 rad/sec ω n = √K/m = √106/509.6 X Y √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 X Y √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2
y = Y sin10
π
t
.
Isolators Grinding MachineGrinding Wheel
x = X sin(ωt -
φ
)
X= 10
-6m
ω n = 44.29 rad/sec ≈ 423 rpm C = Cc∴ζ = C/Cc = 1032.mω n = 1032*509.6*44.29 Ζ = 0.0222 ω = 10 π = 31.4 rad/sec r = 31.4 / 44.29 = 0.7093 ∴ = = √1+(0.0314)2 = 1/ √(0.246 + 1.0) = 1.1166 = 1.1166 given that X = 10-6m ∴Y = X/1.1166 = 10-6/1.1166 = 8.955*10-7m Y = 8.95*10-7 mm
3. A trailer has 1000 Kg mass when fully loaded and 250 Kg when empty. The suspension has a stiffness of 350 kN/m. The damping factor is 0.5. The speed of the trailer is 100 Km/hr. The road varies sinusoidally with a wave length of 5 m. Determine the amplitude ratio of the trailer:
1.When fully loaded. 2.When empty. Data:
Mass of empty trailer = 250 Kg,
ζ = 0.50 Mass of loaded trailer = 1000 Kg, k = 350 kN/m Speed of trailer = 100 Km/hr = 100*1000/3600 = 27.77 m/sec X Y √ 1+ [2(0.0222*0.7093)] 2 √(1- 0.70932)2 + (2 ζ r)2 X Y
y = Y Sin
ω
t
Road profile
Y
m
C,K
100 km/hr
x
y
One cycle
Time period = τ
= wave length /velocity = 5/27.77 sec.
= 0.18 sec
Forcing frequency, w = 2π/ τ = 2π / 0.18 = 34.896 rad/sec.
1. Empty trailer:
Natural frequency of empty trailer
wn = √(k/m) = √(350*103/250) = 37.416 rad/sec Frequency ratio, r = w/wn = 34.896/37.416 = 0.933
The ratio of amplitude of vibration of empty trailer to that of road surface is given as = = 1.3676/0.9419 = 1.4518 X Y √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2
2. Loaded trailer:
When the trailer is fully loaded the natural frequency is given by wn = √(k/m) = √(350*103/1000) = 18.708 rad/sec Frequency ratio r = w/wn = 34.896/18.708 = 1.8653 = = 2.116 / 3.1026 = 0.6819 [X/Y]empty = 1.4518 [X/Y]loaded = 0.6819
The amplitude of vibration reduces as the mass of the system (loaded trailer) is increased. 4. An aircraft radio of mass 20 Kgs is to be isolated from engine vibrations, which is vibrating with amplitude of 0.05 mm at 500 cpm. The radio is mounted on four isolators, each having a spring scale of 31400 N/m, and damping factor of 392 NS/m.
a. What is the amplitude of vibration of the radio?
b. What is the amplitude of vibration of the radio relative to the engine. c. What is the dynamic load on each isolator due to vibration.
Part A m = 20 Kgs K = 4*31400 = 125600 N/m C = 4*392 = 1568 NS/m y = Y Sin ω t ∴ Y = 0.05 mm, ω = 2πƒ = 2π* 500/ 60 rad/sec ω = 52.5 rad/sec. ω n = √K/m = √125600/20 = 79.2 rad/sec. ω /ω n = 52.5 / 79.5 = 0.662 ζ = C/2 √Km = 0.496 = X = 0.069 mm = 6.9*10-5m Part B = X Y √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 X Y √ 1+ (2 ζ r)2 √[1- r2]2 + [2 ζ r]2 X Y
Z = 0.025 mm = 2.5*105m
Part C
Ft = Z √K2+ (Cω )2 = 2.5*10-5 √(1568*52.4)2 + 1256002
Ft = 3.8 N ∴Total force transmitted.
∴Ft on each isolator = 3.8/4 = 0.95 N
The dynamic load can also be computed using Ft = mω 2X
= (20) (52.4)2*6.9*10-5
Chapter 5
Vibration Measuring Instruments
5.1 Introduction
In practice the measurement of vibrations becomes necessary due to following reasons.
1. To determine natural frequencies, modal shapes and damping ratios. The measurement of frequencies of vibration and forces developed is necessary to design active vibration isolation systems.
2. The theoretically computed vibration characteristics of a machine or structure may be different from the actual values due to the assumptions made in the analysis. (To verify the analytical models).
3. Periodic measurement of vibration characteristics of machines and structures becomes essential to ensure adequate safety margins. (Preventive maintenance). 4. Measurement of input and resulting output vibration characteristics of a system helps
in identifying the system in terms of its mass, stiffness and damping.
5.2 Vibration Measurement Scheme
Figure 5.1 shows the basic features of a vibration measurement scheme. 1. Vibrating machine or structure.
2. Vibration transducer or pick up. 3. Signal conversion instrument. 4. Display / recording.
5. Data analysis.
Figure 5.1 Vibration Measurement Scheme
The motion of a vibrating body is converted in to an electrical signal by the vibration transducer or pick up. The transducer transforms changes in mechanical quantities such as displacement velocity, acceleration in to changes in electrical quantities such as voltage or current. (Electrodynamic pick up, electromagnetic pick up, piezo electric pick up, inductive displacement pick up, LVDT pick up, capacitive pick up). Since the output signal of a transducer is too small to be recorded directly, a signal conversion instrument is used to amplify the signal to the required value (Amplifier). The output from the signal conversion
1
2 3 4
instrument can be displayed on a display unit or stored in a computer for later use (Oscilloscope, A to D converters, milli voltmeters, computers etc., the data can then be analyzed to determine the desired vibration characteristics of the machine. Depending upon the quantity measured the vibration measuring instrument is called a vibrometer, a velocity meter, an accelerometer, a phase meter or a frequency meter. To summarise, following are the guidelines.
1. Displacement measurements may be useful for studying low frequency vibrations, where corresponding velocity and acceleration measurements are too small for practical purposes.
2. Velocity measurements may be useful at intermediate frequencies where displacement measurements are likely to be small to measure conveniently.
3. Acceleration measurements may be useful at high frequencies.
Instead of the above, vibration analyzers can also be used. Several commercial vibration analyzers are available today. They consist of a vibration pick up and an FFT (Fast Fourier Transformation) analyser, a balancing kit for phase measurement and an inbuilt computer. The pick up essentially a piezo electric type with a natural frequency of 25 kcps. (KHz). Built in double integration is also available for displacement plots. FFT converts time domain signal to a signal in frequency domain to identify the frequencies of concern.
5.3 Vibration pick ups: Seismic Instruments
The commonly used vibration pick ups are called seismic instruments. The basic element of many vibration measuring instrument is a seismic unit which is basically a spring mass-damper system mounted on a vibrating body on which measurements are to be made as shown in Figure 5.2.
Figure 5.2Seismic Unit
Depending on the frequency range utilized displacement, velocity or acceleration is indicated, by the relative motion of the suspended mass with respect to the case.
C
x
Casing
Behaviour of Seismic unit
Consider the equation of motion of spring-mass-damper system, subjected to base excitation, as shown in Figure 5.3.
Figure 5.3
mx = - C (x-y) – K (x-y)
if Z = x-y; relative displacement the equation of motion becomes mZ + CZ + KZ = mω 2Y Sin ω t
from this =
φ = tan-1 [2 ζ r/1-r2]
The parameters that influence Z/Y and φ are: (1) frequency ratio r = ω /ω n. (2) Damping
factor ζ, as shown in the Figure 5.4.
Base
M y = Y Sin ω tx = X Sin (
ω
t-
φ
)
..
. .
..
.
.
Z Y r2 √[1- r2]2 + [2 ζ r]2 Figure 5.4. Frequency response curvesZ
/
Y
ζ = 0
ζ = 0.1
ζ = 0.15
ζ = 0.25
ζ = 0.5
ζ = 1.0
r
=
ω
/
ω
nRange for Vibrometer Range for
Figure 5.4 also shows the range of frequencies corresponding to which a seismic instrument act as a vibrometer or an accelerometer. Type of instrument is determined by the useful range of frequencies with respect to the natural frequency (ω n) of the instrument. The relative
displacement Z, may represent the displacement or acceleration depending upon ω n of the
seismic unit and frequency of vibrating body, ω .
5.4 Vibrometer or Seismometer
It is an instrument with low natural frequency. Therefore,
ω >>>>> ω n
r >>>> 1, r is very large.
Z/Y ≈ 1, in particular when r > 3 Z/Y ≈ 1, (independent of ζ )
∴Z=Y
Relative displacement of the seismic mass = displacement of base.
∴Z = X-Y, X = 0, ∴Z=Y
Hence the seismic mass remains stationary. It remains undisturbed in space. The supporting casing moves the vibrating body. Thus the relative displacement between the casing and the mass is the true displacement of the casing. Like wise, the relative velocity between the casing and the mass is the true velocity of casing. Usually, the relative motion Z is converted into electric voltage. The seismic mass is a magnet moving relative to the coils fixed to the case, as shown in Figure 5.6.
Figure 5.6
The voltage generated is proportional to the rate of cutting of magnetic field. Therefore the output of the instrument is proportional to the velocity of the vibrating body. Such instruements are called velometers. A typical instrument of this kind may have a natural frequency of 1 Hz to 5 Hz and a useful range of 10 Hz to 2000 Hz. The sensitivity of such instruments may be in the range of 20 mV/cm to 350 mV/cm. Both the displacement and acceleration are available from the velocity type transducer by means of the integrator or the differeniator provided in most signal conditioner units.
0 0 0 00 0 Seismic mass
y
x
Limitation of Vibrometers
In order to have r >>>1, ω n should be very small. This means that, the mass must be very
large and the spring must have a very low stiffness. Therefore, a vibrometer is a spring-mass-damper system with a very large mass and a flexible spring. This results in bulky instrument, which is not desirable in many applications.
In practice, a vibrometer may not have a large value of r, and hence the value of Z, may not be exactly equal to Y. In such cases the true value of Y, can be computed from:
=
5.5 Accelerometer
It is an instrument with high natural frequency. When the natural frequency of the instrument is high compared to that of the vibrations to be measured, the instrument indicates acceleration.
Then
ω <<<< ω n,
r <<<<< 1,
the factor √[r – (ω /ω n)2]2 + (2 ζr)2 approaches unity.
∴ Z (ω /ω n)2.Y
(1/ω n2). ω 2Y
Hence, Z α ω 2 Y, which implies that Z is proportional to the acceleration of the vibrating
body. Thus in order to make r <<<< 1, ω n should be very large. Hence K should be very
large and m should be small. This means that, the instrument needs a small mass and spring of large stiffness. Therefore, the instrument will be very small in size and compact.
Due to their small size and high sensitivity accelerometers are preferred in vibration measurements. The acceleration measured can be integrated once or twice with the help of modern electrical circuits to obtain velocity and displacement of the system.
Thus the difference between a vibrometer and an accelerometer is in its natural frequency. In vibrometer it is very small where as in accelerometer it is very high. The principle of construction remains same.
5.6 Useful Frequency Range
The useful range of accelerometer can be seen from the following graph for different amounts of damping ζ. Useful frequency range is that range of r between which the maximum error is less than 0.01 %. The useful frequency range for un damped accelerometer is very much limited. However, with ζ = 0.7 the useful frequency range is quite large, that is, between 0 ≤ ω /ω n≤ 0.20, as shown in Figure 5.7.
Z Y
r2
Figure 5.7. Useful frequency range
Thus an instrument with a natural frequency of 100 Hz has a useful frequency range of 0 to 20 Hz with negligible error. (Up to 20 Hz the error is less than 0.01%). Figure 5.8 shows accelerometers.
Figure 5.8. Accelerometers
Numerical Examples on Vibration Measuring Instruments
1. A vibrometer having a natural frequency of 4 rad/sec and ζ = 0.2 is attached to a structure that executes harmonic motion. If the difference between the maximum and minimum recorded value is 8 mm, find the amplitude of vibration of structure when its frequency is 40 rad/sec. ω n = 4 rad/sec, ζ = 0.2 Z = Relative amplitude = 8/2 = 4 mm ω = 40 rad/sec r = ω /ω n = 40/4 = 10 = = 1.0093 Z/Y = 1.0093 ∴ Y = 3.9631 mm
2. A vibrometer has a natural frequency of 10 cps and has a damping ratio of 0.7. It is used, by mistake, to measure vibrations of a fan base at an exciting frequency of 180 rpm. The measured vibration velocity of the fan base is 3 mm/s. What is the actual velocity of the fan base?
For a vibrometer, =
In the present case, ω n = 10 cps = 62.8 rad/sec.
Exciting frequency ω = 180 rpm = 18.84 rad/sec. Hence
r = ω /ω n = 0.3
(Z/Y) = 0.09/(0.8281 + 0.1764) = 0.089 Hence Y = Z/0.089 = 33.6 mm/s.
It may be noted that the actual velocity is beyond permissible limits, whereas what is read is well below the permissible limit. Hence one should be very careful in selecting the proper instrument.
3. A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm. If the natural frequency of the instrument is 5 Hz and if it shows 0.004 cm determine the displacement, velocity and acceleration assuming no damping.
ƒn = 5 Hz ; ω n = 2πƒ n = 10π rad/sec = 31.4 rad/sec N = 120 rpm ω = 2π N/60 = 12.56 rad/sec r = ω /ω n = 12.56/31.4 = 0.4 Z Y r2 √[1- r2]2 + [2 ζ r]2 Z 8 mm 4 mm Mean Z Y r2 √[1- r2]2 + [2 ζ r]2
Z = 0.004 cm = 0.0004 mm
For seismic instruments
= , ζ = 0
= =
∴Displacement Y = Z(1-r2) / r2 = 0.021 cm
Velocity V = ω Y = 2π N/60* 0.021 = 0.26 cm/sec Acceleration a = ω 2Y = ω (ω Y) = 3.265 cm/sec2
4. A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz. If the lowest frequency that can be measured is 40 Hz, find the value of damping factor.
Solution:
Data: ω n = 5Hz, ω = 40 Hz, error = 2%
r = ω /ω n = 40/5 = 8
Z/Y = 1.02 (since the error is 2%) = (1.02)2 = 82/(1-64)2 + (16 ζ)2 ζ = 0.35 Z Y r2 √[1- r2]2 + [2 ζ r]2 Z Y r 2 √[1- r2]2 r2 [1- r2] Z Y r2 √[1- r2]2 + [2 ζ r]2
Session-VIII (6.5.05) BKS
5.6 Useful Frequency Range of Vibration Measuring Instruments:
Useful frequency range is that range of r between which the maximum error is less than 0.01 %.
Vibrometer: As evident from Figure 5.4 for value of ζ far greater than 1, Z/Y ≈1 for any value of ζ. This means that the mass remain un disturbed in space. Hence the relative displacement between the casing and the mass is the true displacement of casing. Similarly the relative velocity between the casing and the mass is the true velocity of the casing. The instruments which have natural frequency ω n such that r >>>> 1, can read displacement or
velocity directly. They are vibrometers and velometers, respectively.
The accuracy of these instruments depend upon the amount of damping and frequency ratio at which they are used. Figure 5.6 shows the values of ζ close to 1, plotted against Z/Y. It is seen that when ζ varies from 0.6 to 0.7 the percentage error in Z as compared to Y is less than 4. For ζ = 0.707, and r >>>1. Z ≈ Y and thus the error is less than 0.01%. Thus there is a lower cut-off frequency for a vibrometer, beyond which it gives readings with error less than 0.01% (concept demonstrated in numerical examples)
Figure 5.7. Useful frequency range for vibrometer
Accelerometer:
= = r2λ
where λ = = amplitude distortion factor.
Therefore Z/Y = (ω 2/ω n2) λ Z/Y Z Y r 2 √[1- r2]2 + [2 ζ r]2 1 √[1- r2]2 + [2 ζ r]2
Z = Y (ω 2/ω
n2) λ
For r <<<< 1, λ approaches unity.
Therefore, Z α ω 2 Y, (i.e., relative displacement is proportional to the acceleration). The
amplitude distortion factor λ should remain constant over the desired range of frequency of the accelerometer.
Thus for r <<< 1, ζ = 1/√2 = 0.7, λ would remain constant at ≈ 1, in the range
λ = 1/√[1+r4]≈ 1, i.e., 0 < r < 0.25.
The useful range of accelerometer can be seen from the following graph in Fig 5.8 for different amounts of damping ζ. The useful frequency range for undamped accelerometer is very much limited. However, with ζ = 0.7 the useful frequency range is quite large, that is, between 0 ≤ ω /ω n≤ 0.25, as shown in Figure 5.8.
Figure 5.8. Useful frequency range for accelerometer
Thus an instrument with a natural frequency of 100 Hz has a useful frequency range of 0 to 20 Hz with negligible error. (Up to 20 Hz the error is less than 0.01%). Figure 5.9 shows accelerometers.
Figure 5.9. Accelerometers
Numerical Examples on Vibration Measuring Instruments
1. A vibrometer having a natural frequency of 4 rad/sec and ζ = 0.2 is attached to a structure that executes harmonic motion. If the difference between the maximum and minimum recorded value is 8 mm, find the amplitude of vibration of structure when its frequency is 40 rad/sec. ω n = 4 rad/sec, ζ = 0.2 Z = Relative amplitude = 8/2 = 4 mm ω = 40 rad/sec r = ω /ω n = 40/4 = 10 = = 1.0093 Z/Y = 1.0093 ∴ Y = 3.9631 mm
2. A vibrometer has a natural frequency of 10 cps and has a damping ratio of 0.7. It is used, by mistake, to measure vibrations of a fan base at an exciting frequency of 180 rpm. The measured vibration velocity of the fan base is 3 mm/s. What is the actual velocity of the fan base? For a vibrometer, = Z Y r2 √[1- r2]2 + [2 ζ r]2 Z 8 mm 4 mm Mean Z Y r2 √[1- r2]2 + [2 ζ r]2
In the present case, ω n = 10 cps = 62.8 rad/sec.
Exciting frequency ω = 180 rpm = 18.84 rad/sec. Hence
r = ω /ω n = 0.3, Z = 3 mm/sec
(Z/Y) = 0.09/(0.8281 + 0.1764) = 0.089 Hence Y = Z/0.089 = 3/0.089 = 33.6 mm/s.
It may be noted that the actual velocity is beyond permissible limits, whereas what is read is well below the permissible limit. Hence one should be very careful in selecting the proper instrument.
3. A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm. If the natural frequency of the instrument is 5 Hz and if it shows 0.004 cm determine the displacement, velocity and acceleration assuming no damping.
ƒn = 5 Hz ; ω n = 2πƒ n = 10π rad/sec = 31.4 rad/sec
N = 120 rpm ω = 2π N/60 =
r = ω /ω n = 2π N/60*10π = 0.4 r = 0.4
Z = 0.004 cm = 0.0004 mm
For seismic instruments
= , ζ = 0
= =
∴Displacement Y = Z(1-r2) / r2 = 0.021 cm
Velocity V = ω Y = 2π N/60* 0.021 = 0.26 cm/sec Acceleration a = ω 2Y = ω (ω Y) = 3.265 cm/sec2
4. A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz. If the lowest frequency that can be measured is 40 Hz, find the value of damping factor.
Solution:
Data: ω n = 5Hz, ω = 40 Hz, error = 2%
r = ω /ω n = 40/5 = 8
Z/Y = 1.02 (since the error is 2%) Z Y r2 √[1- r2]2 + [2 ζ r]2 Z Y r 2 √[1- r2]2 r2 [1- r2]
=
(1.02)2 = 82/(1-64)2 + (16 ζ)2 ζ = 0.35
5. A commercial vibration pick-up has a natural frequency of 5.75 Hz and a damping factor of 0.65. What is the lowest frequency beyond which the amplitude can be measured with in (a) 1% error (b) 2% error.
Part (a).
= Error = [Z-Y]/Y*100
1 = (Z/Y-1)100
Given that error is to be 1% ∴Z/Y = 1+0.01 = 1.01
∴Z = 1.01 times Y
∴ = 1.01,
Simplification leads to 0.02 r4 – 0.31r2 + 1 = 0
giving r = 3.30 and 2.02
These are the two values at which Z/Y = 1.01 in between these two values Z/Y will be greater than 1.01, as shown.
∴ The lowest value of r beyond which the amplitude can be measured with in 1% error is r = 3.30.
∴ω /ω n = ƒ/ƒn = r = 3.30
∴ƒ = 3.30 * 5.75 = 10 Hz
ƒ = 10 Hz. Part (b).
When the error is 2% Z/Y = 1.02
When solved for r, for the given value of damping, we get imaginary value of r2. This means
that, for ζ = 0.65, the curve Z/Y v/s r, does not go as high as Z/Y = 1.02. Thus to get the frequency for 2% error we have to consider Z/Y = 0.98 Solving for r
r = 1.55
ƒ = 8.9 Hz.
6. Specify the lowest frequency of a vibrometer that can be measured with 1% error, if its natural frequency is 4 Hz and damping ratio is 0.2
Z Y r2 √[1- r2]2 + [2 ζ r]2 Z Y r2 √[1- r2]2 + [2 ζ r]2 r2 √[1- r2]2 + [2 ζ r]2 r2 √[1- r2]2 + [2 ζ r]2 Z/Y 1.00 1.01 ζ = 0.65
*
*
1.00 2.02 3.00 rζ = 0.2 ƒn = 4 Hz = 1.1 = on simplification r4 – 2r2 + 1 + 4 ζ2r2 = r4 / (1.01)2 error = 1% ∴Z/Y = 1.01 We get r4-93.38r2 + 50.75 = 0 Solving, we get r = 9.635 and 0.739
These are the two values at which Z/Y = 1.01. In between these two values Z/Y will be greater than 1.01 at r1 = 0.739 ƒ/ƒn = r1, ƒ = 2.9 Hz. at r2 = 9.635 ƒ/ƒn = r2, ƒ = 38.54 Hz.
The lowest frequency beyond which the amplitude can be measured with 1% error is 38.54 Hz.
7. An accelerometer is made with a crystal of natural frequency 20 kHz. The damping ratio of accelerometer is found to be 0.71. Determine the upper cut off frequency of the accelerometer for 1% accuracy.
= ƒn = 20 kHz ζ = 0.71 error = 1 % 1.01 = , imaginary value 0.99 = solving r = 0.3667
∴upper cut-off frequency f = r* fn = 0.3667*20k Hz = 7.335 k Hz. Z Y r2 √[1- r2]2 + [2 ζ r]2 r2 √[1- r2]2 + [2 ζ r]2 Z Y r2 √[1- r2]2 + [2 ζ r]2 r2 √[1- r2]2 + [2 ζ r]2 r2 √[1- r2]2 + [2 ζ r]2 Z/Y 1.01 1.00 0.739 1.00 r 9.63
*
*
8. A device used to measure torsional acceleration consists of a ring having a moment of inertia of 0.049 Kg-m2 connected to a shaft by a spiral spring having a scale of 0.98 Nm-/
rad, and a viscous damper having a constant of 0.11 Nm-sec/rad. When the shaft vibrates with a frequency of 15 cpm, the relative amplitude between the ring and the shaft is found to be 20. What is the maximum acceleration of the shaft?
Solution: Data: J = 0.049 Kg-m2 C t = 0.11 Nm-sec/rad Kt = 0.98 Nm/rad ω = 2π*15/60 = π/2 rad/sec ω n = √Kt/J = 4.47 rad/sec ζ = Ct/2√KtJ = 0.11 / 2√0.98*0.049 = 0.25 θz = 20 , = 2/57.3 = 0.0349 rad ω /ω n = 0.352 ∴ = θy = 0.253 rad
∴Maximum acceleration of the shaft = ω 2 θ y
= (π/2)2 * 0.253 = 0.62 rad/sec.
9. A vibrometer having a mass of 10 Kgs is used to measure the vibration amplitude of a machine which is vibrating with a frequency of 1000 cpm. If the error in the reading of the dial indicator is not to be more than 3% of the actual amplitude of the vibrating machine determine the stiffness of the vibrometer spring?
10. Show that an undamped seismic instrument will show the true response at a frequency ratio r = 1/√2.
Solution: Z/Y = r2/(1-r2), ζ = 0,
For true response Z=Y ∴ r2 /(1-r2) = 1
2r2 = 1 or r = 1/√2
(ω /ω n)2
√[1- r2]2 + [2 ζ r]2
θz
Session-IX (10.5.05) BKS
5.7 Whirling of Shafts
In many practical applications such as turbines, compressors, electric motors and pumps, a heavy rotor is mounted on a light weight flexible shaft that is supported between bearings. The mass centre of rotor do not coincide with the centre line of the shaft. Thus there will be unbalance in the rotor due to manufacturing errors. When the shaft rotates centrifugal force is induced on the shaft, which makes it to bend in the direction of eccentricity of rotor. In addition to this other effects such as stiffness and damping of the shaft, hyrtersis damping, gyroscopic effects, and fluid friction in bearings also cause the shaft to bend. This bending further increases eccentricity and hence the centrifugal force. This effect is cumulative and ultimately the shaft may even fail. The extent to which the shaft bends depends upon the eccentricity of the rotor mass and speed of the shaft.
At certain rotational speeds the shaft tends to vibrate violently in transverse direction. At these speeds the shaft has a tendency to bow-out and whirl in a complicated manner as shown in Figure 5.10 and 5.11.
Figure 5.10 Whirling of Shaft
Figure 5.11 Whirling of Shaft
O C G O C G X
e
Bearing centre
line
Undeflected Position Deflected Positionω
ω
Bent up shaft axis
Bearing centre line
Rotor or Disc
Bearing Bearing
This phenomenon is called whirling or whipping of shafts and the corresponding speeds are referred as whirling or whipping or critical speeds of shafts. These critical speeds are found to coincide with the natural frequencies of lateral (transverse) vibrations of the shaft.
The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage. Eg: The rotor blades of a turbine may come in contact with stator blades. Larger shaft deflections produce larger bearing reactions, which may lead to bearing failure. The amplitude build up is a time dependent phenomenon and therefore, it is very dangerous to continue to run the shaft at it critical speed.
The whirling motion of a shaft consists of two components of motion as shown in Figure 5.12.
a. Spinning of the shaft along with rotor about the bent up shaft axis.
b. Rotation of plane A made by the centre line of the bearings and bent up-shaft, about the centre line of the bearings.
Figure 5.12 Whirling of Shaft
The rotation of plane A, which is generally referred as whirling, may take place in the same sense as that of spinning of the shaft or in the opposite sense. Further the speed of whirling may or may not be equal to the speed of spinning of the shaft. When the whirling speed is equal to the speed of rotation of shaft it is called “synchronous whirl”.
5.7.1 Critical speed of a shaft with a single rotor (with out damping):
Consider a shaft on which a rotor in symmetrically located between two bearings. The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions.
1. Shaft is light and flexible. 2. Gravity effects are negligible. 3. Friction at shaft centre is small. 4. Damping due to air is neglected. Let m: mass of the disc in Figure 5.10.
Rotation of plane A
ω
plane Aω
Bent up shaft axis
Bearing centre line
Rotor or Disc
Bearing Bearing
Plane A Plane APlane A
