CHAPTER 3
HARMONICALLY EXCITED VIBRATIONS
Introduction • Forced vibration – external energy is supplied during vibration by applied force or imposed displacement excitation. Harmonic excitation –
harmonic response
Suddenly applied non-periodic excitation-
transient response.
F(t) = F0 (+) = F0 cos ( ) = F0 sin ( )
HARMONICALL HARMONICALLY Y EXCITED EXCITED VIBRATIONS VIBRATIONS Equation of Motion mሷ + c ሶ + kx = F(t)
(1)
Non homogeneous differential equation, with solution x(t) = xh (t) + xp (t) Homogeneous solution
Particular solution
• Homogeneous solution represents free damped vibrations and will die out for all cases of damping and initial conditions t o particular solution • Hence the general solution of (1) reduces to representing steady state vibration and will be present as long as forcing function is present
HARMONICALLY EXCITED VIBRATIONS Homogeneous, particular and general solution
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
• Consider an undamped system subjected to harmonic force mሷ + kx = F0 cos (2) • Homogeneous solution is assumed as (3) x(t) = C1 cos ωn t + C2sin ωn t • As F(t) is harmonic , particular solution , xp (t) is also assumed as harmonic xp (t) = X cos (4) X – maximum amplitude • Substituting (4) in (2) X=
= − −
(5)
• Total solution is
x(t) = C1 cos
ωn t + C2sin ωn t +
cos (6)
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
• Initial conditions: x(t) = x 0 (displacement at t = 0)
ሶ (t) = ሶ 0
(velocity at t = 0) • Substitute initial conditions in (6) ሶ , C = 2 − ωn
C1 = x 0 -
(7)
• Total solution is
•
ሶ ) cos t + ( ) sin t + cos ω ω x (t) = (x - 0 n n − − ωn Let , r = Magnification factor = = = − −
(8)
(9)
HARMONICALLY EXCITED VIBRATIONS Magnification Factor Case 1: When 0 < r <1, denominator of (9) is positive and response is given by (4)
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
M=
−
Case 2: When r > 1, denominator of (9) is negative and steady state response can be expressed as xp (t) = - X cos
(10)
Amplitude X is redefined as X=
−
1
(11)
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
Case 3: When r =
= 1, amplitude X given by (9) or (11) becomes infinite
leading to resonance . Rewriting (8) as x(t) = x 0 cos
ሶ
ωn t + ( ω ) sin ωn t + n
cos −cos −
(12)
Last term in (12) is infinite when r = 1.
Applying L’ Hospital’s rule
lim →
cos −cos −
=
lim
→
(
ൗ cos −cos ) ൗ [− ]
=
sin
ωn t
(13)
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
The response of the system at resonance becomes x(t) = x 0 cos
ሶ
ωn t + ( ω ) sin ωn t + n
sin
ωn t
(14)
HARMONICALLY EXCITED VIBRATIONS Total Response: Another representation x(t) = A cos (ωn t –φ) +
−
cos
for r < 1
(15)
cos 1
for r >1
(16)
x(t) = A cos (ωn t –φ) +
−
A and φ are determined by initial conditions
HARMONICALLY EXCITED VIBRATIONS Response of Damped System Under Harmonic Force
EoM of a system is mሷ + c ሶ + kx = F0 cos
(17)
Particular solution of (17) is also assumed as harmonic xp (t) = X cos
( – φ)
(18)
Substituting (18) in (17), X [ ( k- m 2 ) cos ( – φ) - c sin ( – φ) ] = F0 cos Using trigonometric entities and equating coefficient of cos obtain X=
[(k− m2 )2 + c 2) ]1/2
, φ = tan−
c k− m 2
(19)
and sin , we
HARMONICALLY EXCITED VIBRATIONS Response of Damped System Under Harmonic Force
HARMONICALLY EXCITED VIBRATIONS Response of Damped System Under Harmonic Force
X=
[(k− m2 )2 + c 2) ]1/2
, φ = tan−
c k− m 2
(19)
Divide numerator and denominator by k and using expressions for ωn , ζ, and r, we get Magnification factor / Amplification factor / Quality Factor M=
=
− + ζ
,
φ = tan−
ζ −
(20)
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
M=
− + ζ
φ = tan−
ζ −
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
• When ζ =0, equation (20) reduces to (9). • For r = 0, M=1 • Reduction in M in the presence of damping is very significant at or near resonance.
• For 0 < ζ < √ , maximum value of M occurs when r = 1 2ζ2 • The maximum value of X when r = 1− 2ζ2 X ( )max = (21) δst 2 2 ζ 1− ζ
M=
− + ζ
HARMONICALLY EXCITED VIBRATIONS M=
Magnification Factor
• Experimental determination of the measure of damping is obtained from value of X when ω=ωn (
• For ζ = • For ζ >
√2
,
X )= δst 2 ζ dM
= 0 when r = 0
, graph of M monotonically
√2
decreases with increasing values of r
(22)
− + ζ
HARMONICALLY EXCITED VIBRATIONS Magnification Factor Phase Angle: 1. For ζ = 0, phase angle is 0 for 0 < r < 1 and 180 0 for r >1. The excitation and response are in phase for 0 < r < 1 and out of phase r >1. 2. For ζ > 0 and 0 < r < 1, phase angle is 0 < φ < 900, hence response lags excitation 3. For ζ > 0 and r > 1, phase angle is 90 0 < φ < 1800, hence response leads excitation 4. For ζ > 0 and r = 1, phase angle is φ = 900, hence phase difference between response and excitation is 900 5. For ζ > 0 and large value of r , phase angle is approaching 1800, hence response and excitation are out of phase
φ = tan−
ζ −
HARMONICALLY EXCITED VIBRATIONS Magnification Factor
• Total response: x(t) = xp (t) + xh (t) = X0
− cos (ωd t- φ0 )+ X cos (ωt- φ )
(23)
X and φ are given by equation (19) X0 and φ0 are determined from initial conditions.
• Initial conditions: x(t) = x 0 (displacement at t = 0)
ሶ (t) = ሶ 0
(velocity at t = 0)
By substituting initial conditions in equation (23)
φ0 + X cos φ ሶ 0 = ζ X0 cos φ0 + X ωd sin φ0 + ω X sin φ x 0 = X0 cos
(24)
Problem 1 Find the total response of a single degree of freedom system with m = 10kg, c =20 Ns/m, k =4000 N/m, x0 =0.01 m and ሶ 0 = 0 under the following conditions 1. An external force F(t) = F0 cos ωt acts on the system with F0= 100 N and ω = 10 rad/sec 2. Free vibration
Problem 2 A spring mass system with k = 5000 N/m, is subjected to a harmonic force of magnitude 30 N and frequency 20 Hz. The mass is found to vibrate with an amplitude of 0.2m. Assume initial condition to be zero, determine mass of the system
HARMONICALLY EXCITED VIBRATIONS M=
Quality Factor and bandwidth
• For small value of damping ( ζ <0.05), we can take X X ( )max = ( )at ω =ωn = = Quality factor, Q (25) δst δst • The points R1 and R2 where the amplification factor falls to
are called half power points because the
power absorbed (ΔW) by the damper is proportional to square of amplitude
ΔW = ω X 2
(26)
• The difference between frequencies associated with half power points R1 and R2 is called the bandwidth of the system.
− + ζ
HARMONICALLY EXCITED VIBRATIONS M=
Quality Factor and bandwidth
To find R1 and R2 , let
X δst =
in equation (20), = = ζ − + ζ
r 4 – r 2 (2-4 ζ2 )+ (1- 8 ζ2) = 0
(27)
Solution of (27)
= (1-2ζ2 ) - 2ζ 1 ζ , = (1-2ζ2 ) + 2ζ 1 ζ
(28)
For small values of ζ
= 1-2ζ and = 1+2ζ
(29)
− + ζ
HARMONICALLY EXCITED VIBRATIONS Quality Factor and bandwidth
= 1-2ζ and = 1+2ζ r 1 =
(29)
and r 2 =
By (29)
- = ( - ) (
+
) = ≅ 4ζ
(30)
Also
+ = 2 Using (30) and (31)
n
Bandwidth Δ = - = 2 ζn By (25) and (32), Q≅
n ≅ 2ζ −
(31) (32) (33)
Quality factor can be used for estimating the equivalent viscous damping in a mechanical system
Problem 3 A uniform slender bar of mass m , may be supported in one of the two ways as indicated in Figure. Determine the arrangement that result in a reduced steady-state response of the bar under a harmonic force , F 0 sin , applied at the middle of the bar, as shown in figure.
HARMONICALLY EXCITED VIBRATIONS Base Excitation
Base / support of a spring mass damper system undergoes harmonic motion y(t) – displacement of base x(t) – displacement of mass from its static equilibrium position Equation of Motion is mሷ + c( ሶ - ሶ ) + k(x-y) = 0
(34)
, equation (34) becomes mሷ + c ሶ + kx = ky + c ሶ = kYsin + cYcos = A sin (- ) (35)
If y (t) = Y sin
HARMONICALLY EXCITED VIBRATIONS Base Excitation − (36) This shows that giving excitation to base is equivalent to applying a harmonic force of magnitude A to the mass
A = Y
+(c2 ) and = tan−
By using the solution indicated by (19), the steady state response of the mass, xp (t) can be expressed as xp (t) =
[(k−
+(c ) m2 )2 + c 2) ]1/2 sin (t-φ1 - )
(37)
Where
c φ1 = tan− k− m 2
(38)
HARMONICALLY EXCITED VIBRATIONS Base Excitation
Using trigonometric identities, equation (37, 38) can be rewritten in more convenient form. As xp (t) = X sin (t-φ) =
+ ζ) ζ) + −
Τ
,φ
= tan−
ζ + ζ −
(39)
• The ratio of response x p (t) to that of the base motion y(t) , i.e. is called displacement transmissibility
HARMONICALLY EXCITED VIBRATIONS Base Excitation
Transmissibility =
+ ζ) ζ) + −
Τ
φ = tan−
ζ + ζ −
HARMONICALLY EXCITED VIBRATIONS Base Excitation
Transmissibility
Some aspects of displacement transmissibility , Td: • Value of Td = 1 when r = 0 and close to unity when r is small. • For undamped system, Td → ∞, at resonance • < 1, for values of r > 2 for any amount of damping • = 1, for values of r = for any amount of damping • For r < 2, smaller damping ratios leads to larger values of • For r > 2, smaller damping ratios leads to smaller values of (soft spring) • attains a maximum for 0 < ζ < 1, at r = < 1 given by
=
ζ
1 8ζ 1
Τ
(40)
=
Τ
ζ)
+ ζ) + −
HARMONICALLY EXCITED VIBRATIONS Base Excitation Force Transmitted: Force transmitted to the base / support due to reactions from spring and dashpot
= (ሶ - ሶ )= - m ሷ Assume the solution as x p (t) = X cos
(41)
(- )
(42)
By (41 & 42), F= m 2 X sin
(- ) = FT sin (- )
(43)
FT is the amplitude / maximum value of the force transmitted to the base is given by
FT
=
+ ζ) ζ) + −
Τ
(44)
Force Transmissibility
The transmitted force is in phase with the motion of the mass x(t).
HARMONICALLY EXCITED VIBRATIONS Base Excitation FT =
+ ζ) ζ) + −
Τ
HARMONICALLY EXCITED VIBRATIONS Base Excitation
Relative Motion: If z = x – y ( motion of mass relative to the base / rattle space), equation of motion (34) becomes m ሷ + c ሶ + k z = -m ሷ = m 2 Ysin
(45)
Steady state solution of (45) is given by, =
(46)
√( ζ) + − )
, φ1 = tan−
ζ −
HARMONICALLY EXCITED VIBRATIONS Base Excitation
φ1 = tan−
ζ −
Problem 1 (a) 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 400kN /m and a damping ratio of ζ = 0.5. If the vehicle speed is 20 km/hr, determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude of Y=0.05 m and wavelength of 6m. (b) If the vehicle speed increases to 100 km/hr, what will be the displacement amplitude of the vehicle? (c ) Analyse the system under the condition of ζ = 0.05 and speed of 20 km/hr and 100 km/hr. Comment .
Problem 2 An automobile is modeled as a single degree of freedom system vibrating in the vertical direction. It is driven along a road whose elevation varies sinusoid ally. The distance from peak to trough is 0.2 m and the distance along the road between the peaks is 35 m. If the natural frequency of automobile is 2 Hz and the damping ratio of shock absorber is 0.15, determine the amplitude of vibration of the automobile at a speed of 60 km/hr. If the speed of the automobile is varied, find the most unfavourable speed for passenger
Comparison of objectives
r=√2 Displacement of mass
Rattle space
Force Transmitted to ground
Conflicting Objectives in damper design The traditional engineering practice of designing a spring and a damper, are two separate functions that has been a compromise from its very inception in the early 1900’s. Passive suspension design is a compromise between ride comfort and vehicle handling, as shown in Figure . In general, only a compromise between these two conflicting criteria can be obtained if the suspension is developed by using passive springs and dampers.. This also applies to modern wheel suspensions and therefore a break-through to build a safer and more comfortable car out of passive components is below expectation. The answer to this problem seems to be found only in the development of an active suspension system
ACCELERATION
RATTLE SPACE
P. Kousik, Pravin M Singru and Narayan Manjrekar,” Comparing PID and H -∞ Controllers on a 2-DoF Nonlinear Quarter Car Suspension System” Journal of Vibro-Engineering (paper No. 15281) available online from 19.11.2014. (IF=0.66)
Rotating Unbalance • Unbalance is main cause of vibration in many machines
• M- total mass of machine • m/2 – two eccentric masses rotating in opposite directions with constant angular velocity
• Centrifugal force (me2)/2 due to each mass will cause excitation of two masses
• As masses are rotating in opposite direction, horizontal components cancel each other
• Vertical component along AA causes excitation
Rotating Unbalance =
• Angular position of the masses is measured from a horizontal position, total vertical component of the excitation F(t) = me 2 sin t
• Equation of motion is Mሷ + c ሶ + kx = me2 sin t
(47)
• Solution is xp (t) = X sin (t-φ) =
− +
ζ
,
φ = tan−
ζ −
(48)
− + ζ
Rotating Unbalance
φ = tan−
ζ −
Rotating Unbalance • All curves begin at zero amplitude. The amplitude near resonance is markedly affected by damping. So if machine is to be run near resonance, damping should be introduced purposefully to avoid dangerous amplitudes.
• At very large speeds r>>1, is almost unity and effect of damping is negligible
=0
• For 0 < ζ < √, maximum of( ) occurs when Solution of (49) gives Maximum of (
• For ζ > √,
,r =
)= −
(49)
>1, −
(50)
does not attain a maximum. Its value grows from
0 at r =0 to 1 at r → ∞
Peak occurs to the right of the resonance
Problem 3 When an exhaust fan of mass 380 kg is supported on springs with negligible damping, the resulting static deflection is found to be 45 mm. If the fan has rotating unbalance of 0.15 kg-m, find (a) the amplitude of vibration at 1750 rpm. (b) the force transmitted to the ground at this speed.
Problem 4 One of the tail rotor blades of a helicopter has an unbalanced mass of m = 0.5 kg at a distance of e = 0.15 m from the axis of rotation. The tail section has a length of 4m, mass of 240 kg, flextural stiffness (EI) of 2.5 MNm2 and a damping ratio of 0.15. The mass of the tail rotor blades, including their drive system, is 20 kg. Determine the forced response of the tail section when blades rotate at 1500 rpm.
SOLUTION
Whirling of Shaft • In practical applications like turbines, compressors, motors, a heavy rotor is mounted on a lightweight, flexible shaft supported on bearings.
• Unbalance in rotors due to manufacturing considerations.
• Unbalance effects, stiffness & damping of shaft, gyroscopic effects and fluid friction in bearings will bend the shaft at certain rotational speeds, called whirling / Definition: Rotation of the plane whipping / critical speeds made by the line of centers of the bearings and bent shaft
Whirling of Shaft Equation of Motion: Assumption : Rotor is subjected to steady state excitation due to mass unbalance Forces acting on rotor
• Inertia force • Spring force due to shaft elasticity • External and internal damping O is equilibrium position of shaft when balanced perfectly
Whirling of Shaft
C- geometric geometric center center G – CENTER OF MASS
• Shaft (line CG) is assumed to rotate with a constant angular velocity,
• In steady state, rotor deflects radially by a distance A =OC (x,y) eccentric ricity ity = CG • a- eccent • Angular velocity of the line OC = ሶ (Whirling speed,)
EOM
• Inertia force = elastic force + internal damping force + external damping force
(51)
• Radius vector of CG of disc, OG , R= (x+a (x+a cos cos t) Ԧ + (y + a sin
t) Ԧ
(52)
Whirling of Shaft ሷ Inertia force Fi = m
(52)
∴ Fi =m[(ሷ a2 cost) Ԧ + (ሷ a 2 sint) Ԧ ] Elastic force Fe = - k (x Ԧ + y Ԧ)
(53) (54)
k – shaft stiffness Internal damping force Fdi = -ci [(ሶ
y) Ԧ + (ሶ + x) Ԧ
(55)
- internal / rotary damping coefficient Fde = -c (ሶ Ԧ + ሶ Ԧ)
(56)
c- external damping coefficient. Substituting (52) to (56) in (51), EOM in scalar form mሷ + ( c+ ci) ሶ + kx - ci y = m a 2 cost
(57)
m ሷ + ( c+ c ) ሶ + ky - c x = m a 2 sin
(58)
t
EOM of lateral vibration of shaft, Coupled ,Dependent on
Whirling of Shaft Define a complex quantity w = x + iy Adding (57) + i (58), EOM mሷ + ( c+ ci) ሶ + kw - i ci w = m a 2
(59) (60)
Critical speed - Shaft is said to be revolving at critical speed when frequency of rotation = one of the natural frequency - When shaft is revolving at critical c ritical speed , there are large deflections , causing bearing failure - A rapid transition through critical speed is expected to limit whirl amplitude.
Whirling of Shaft Response of the system
• Assume that the excitation to be harmonic force due to the unbalance of the rotor and internal damping (c i ) is negligible
• Solving (60) mሷ + c ሶ + kw kw = m a 2
(61)
• Assume a solution w(t) = C −(+) + A (−) Homogeneous solution
C, , A, are constants
Particular solution
(62) Whirl
Whirling of Shaft Substituting steady state part of (62) in (61) m a 2 X= = [(k− m2 )2 + c 2) ]1/2
φ = tan−
− + ζ
(63)
c k− m 2
• Differentiating (63) w.r.t. and setting result to zero, we get maximum whirling / critical speed
=
−
. 5
• Critical speed is same as only when c = 0
(64)
Whirling of Shaft • At low speeds, amplitude is determined by spring constant, k since the two terms m2 and c2 2 are small. Also the value of phase angle is zero. • As speed increases, amplitude reaches a peak (at resonance) • At resonance, response is limited by damping term and phase lag is 900 • As speed increases, amplitude is dominated by the mass term m24 . This term is 1800 out of phase with the unbalanced force, the shaft rotates in a direction opposite to that of the unbalance force and hence response is limited.
m a 2 X= [(k− m2 )2 + c 2) ]1/2 c φ = tan− k− m 2
Whirling of Shaft Bearing Reactions: Find deflection of mass center of the disc from the bearing axis, R R= (x + a cos t) Ԧ + (y + a sin
t) Ԧ
= +2Aa cos φ
(52) (65)
Using (63), (65) can be written as
=
+ ζ) ζ) + −
Τ
(66)
The bearing reactions can then be determined from centrifugal force m 2 R.
Problem 5 A steel shaft of diameter 25 mm and length 1 m is supported at the two ends in bearings. It carries a turbine disc, of mass 20 kg and eccentricity 0.005 m, at the middle and operates at 6000 rpm. The damping in the system is equivalent viscous damping with ζ =0.01. (i) Determine the whirl amplitude of the disc at (a) operating speed (b) critical speed, © 1.5 times the critical speed. (ii) Find the bearing reactions and the maximum bending stress induced in the shaft at (a) operating speed (b) critical speed, © 1.5 times the critical speed.
TUTORIAL TEST A centrifugal pump, weighing 600 N and operating at 1000 rpm, is mounted on six springs 6000 N/m each. Find the maximum permissible unbalance in order to limit the steady state deflection to 5 mm peak to peak
Problem 6 A shaft , having a stiffness of 3.75 MN/m, rotates at 3600 rpm. A rotor, having a mass of 60 kg and eccentricity of 2000 micron, is mounted on the shaft. Determine (a) the steady state whirl amplitude of the rotor (b) the maximum whirl amplitude of the rotor (b) the maximum whirl amplitude of the rotor during startup and stopping conditions. Assume the damping ratio = 0.05
Experimental Vibration Analysis Why we should measure vibrations? 1. Operating speeds have increased and weights of structures have reduced due to increasing demand for higher productivity and economical designs. This makes occurrence of resonance more frequent and reduce reliability. So periodic measurements of vibration characteristics is essential. 2. Identifying natural frequency of system experimentally leads to selecting useful operating speed. 3. Assumed behavior may be different forcing us to characterize the actual system 4. Survivability of equipment under certain conditions need to be ascertained. 5. System identification in terms of its mass, stiffness and damping
Experimental Vibration Analysis Basic vibration measuring scheme:
Choice of equipment depends on: 1. Expected range of frequencies and amplitudes 2. Sizes of the machines / structures involved 3. Conditions of operation of machines / equipment / structure
Experimental Vibration Analysis Transducers: a device that transforms values of physical variables into equivalent electrical signals. Examples: 1. Variable resistance transducers- strain gage 2. Piezoelectric transducer 3. Electrodynamic transducer 4. Linear variable differential transformer transducer
Experimental Vibration Analysis Variable resistance transducers- strain gage
K- Gage factor
∆ = 0 r0 =
+
(65)
Δ - STRAIN MEASURED =
Experimental Vibration Analysis Piezoelectric transducer:
• Quartz, tourmaline, lithium sulfate, Rochelle salt generate electric charge when subjected to a deformation / mechanical stress. Charge disappears on removal of load E = v t px
(66)
• v- voltage sensitivity • t – thickness of crystal
• When base vibrates, load acts on piezoelectric •
crystal, output voltage generated by crystal will be proportional to the acceleration Compact, rugged, high sensitivity and high frequency range.
Experimental Vibration Analysis Electrodynamic transducer:
• When a conductor coil moves in a magnetic field, voltage E is generated = DL v • D- magnetic flux density (Tesla) • L- length of conductor • v-velocity of conductor w.r.t. magnetic field. • Velocity pickups
• Basis of electrodynamic shaker
(67)
Experimental Vibration Analysis Linear variable differential transformer transducer
Vibration Measurements Vibration pickups: • A transducer is used in conjunction with another device to measure vibrations. • Seismic instrument • Instrument is fastened to the body • Bottom ends of spring and damper will have same motion as the cage, y and their vibrations excites the mass by x • Displacement of mass relative to cage , z = x-y mሷ + c( ሶ - ሶ ) + k(x-y) = 0 (68) y (t) = Y sin (69)
Vibration Measurements Vibration pickups:
m ሷ + c ሶ + k z = -m ሷ = m 2 Ysin (70) Steady state solution of (70) is given by, z(t) = Z sin (t- φ) =
(71A)
, (71) √( ζ) + − )
φ = tan−
ζ −
(72)
Vibration Measurements Seismometer / Vibrometer • It is observed that Τ ≈ 1, when Τ ≥ 3 • Relative displacement between mass and the base (sensed by the transducer) is essentially same as the displacement of the base. Consider (71) z(t) ≈ Y sin (t- φ) (73) if (74) ≈1 √( ζ) + − ) • Comparison of (73) with y(t) = Y sin t shows that z(t) gives directly the motion y(t) except for phase lag φ • Phase lag φ ≈ 1800 for ζ = 0 • Recorded displacement z(t) lags behind the displacement being measured y(t) by time t’ = φൗ and this time lag is not important if y(t) consists of a single harmonic component .
Vibration Measurements • Since r = Τ ≥ 3 is large and the value of
is fixed, the natural frequency nmust be low. • Mass must be large and spring must have low stiffness, so instrument becomes bulky (undesirable) • In practice, r is not very large and hence Z ≠ Y exactly, but compensation technique is used to make it exact.
Problem 7 A vibro-meter having natural frequency of 4 rad/s and ζ =0.2 is attached to a structure that performs a harmonic motion. If the difference between the maximum and minimum recorded values is 8 mm, find the amplitude of motion of the vibrating structure when its frequency is 40 rad/s
Vibration Measurements Accelerometers • Measures acceleration of vibrating body • By (71A) and (71) -
=
[
ζ)
+
− ]
{- Y sin (
This shows that if [(
ζ)
Equation (75) becomes -
+
− ]
=
)}
(75)
≈1 {- Y sin (
(76)
)}
(77)
• Comparing (77) with ሷ (t) = - Y sin , the term - gives acceleration of the base , except for phase lag
.
• Time lag by which recording lags acceleration, t’ = φൗ • This time lag is not important if y(t) consists of a single harmonic component.
Vibration Measurements Accelerometers
Vibration Measurements Accelerometers
1 ≈1 [( 2ζ) 1 2]
Refer equation LHS of (76) is plotted.
• LHS lies between 0.98 to 1.02 for 0 ≤ ≤ 0.6, if ζ lies between 0.65 to 0.7
• Since r is small, of the instrument has to be large compared to the frequency of vibration to be measured.
• Mass needs to be small and spring need to be have large value of k (short spring), so the instrument is small. ACCELEROMETER OPERATING
Vibration Measurements Accelerometers
Sensitivity 101.3 mV/g
Sensitivity 102.2 mV/g.
Problem 8 An accelerometer has a suspended mass of 0.01 kg with a damped natural frequency of vibration of 150 Hz. When mounted on an engine undergoing an acceleration of 1 g at an operating speed of 6000 rpm, the acceleration is recorded at 9.5 m/s 2 by the instrument. Find the damping constant and the spring stiffness of the accerometer.
Vibration Measurements Vibration Exciter / Shaker
• Used in determination of dynamic characteristics of machines and structures • Types • • • •
Mechanical Electromechanical Electrodynamic Hydraulic
Vibration Measurements Vibration Exciter / Shaker
Electrodynamic Shaker: Reverse of electrodynamic transducer. When current passes through a coil placed in a magnetic field, force ,F proportional to the current, I and magnetic flux density D is produced which accelerates the component placed on shaker table (78) F= D I L Magnitude of acceleration depends on the maximum current and the masses of the component and moving elements of the shaker If current is harmonic , force produced is harmonic
Vibration Measurements Vibration Exciter / Shaker
• Coils and moving element should have a linear motion, so they are suspended from a flexible support (small stiffness)
• Two natural frequencies (i) flexible support (small) (ii) moving element (large)
• Operating range of exciter lies between these two resonant frequencies
Vibration Measurements Vibration Exciter / Shaker
• Modal Sho p 2110E electrodyn amic s haker / exciter • Up to 110 lbf pk sine force. • A large armature (3.25 in / 8.3 cm diameter platform table supporting payloads up to 10 lb / 4.5 kg)
• Designed with a through hole armature and includes a chuck and collet attachment, providing simple set-up with stingers for experimental modal analysis applications.
• When used in this configuration, these stingers greatly simplify test setup with an easy connection to the force sensor and test structure, and help decouple cross-axis force inputs, minimizing force measurement errors.
Modal Shop 2110E
electrodynamic shaker
Vibration Measurements Vibration Exciter / Shaker http://www.modalshop.com/filelibrary/110lbf-Vibration-Shaker-Datasheet%28DS-0078%29.pdf
Vibration Measurements Signal Analysis:
Response of a system under known excitation presented in convenient form
Time response doesn’t give useful information
Frequency response gives one or more discrete frequencies around which energy is concentrated
Vibration Measurements Spectrum analyser
• Analyses a signal in the frequency domain by separating the energy of the signal into various frequency bands
• The separation of signal energy into frequency bands is accomplished through a set of filters (ex. Octave band)
• Real time analyzers useful in machinery health monitoring • Digital filtering • Fast Fourier transform method (FFT)
Crystal Spider 81 Analyser
Vibration Measurements Spectrum analyser
Bandpass Filter:
• Circuit that permits the passage of frequency components of a signal over a frequency band and rejects all other frequency components of the signal
• Practical filter will have deviation • Good bandpass filter will have minimum ripples and the slopes of filter skirts will be steep to maintain the actual bandwidth close to the ideal value
• Cuttoff fr equencies: the frequencies f i and f u at which the response is 3 dB below its mean bandpass response
Vibration Measurements Spectrum analyser
Bandpass Filter: 1. Constant percent bandwidth filters • Ratio of bandwidth to constant tuned frequency (
) is constant
• Octave, one half octave and one-third octave band filters 2. Constant bandwidth filters • Bandwidth is constant • Difference lies in details provided by various bandwidths
Vibration Measurements Filters Constant percent bandwidth filters:
• Ratio of bandwidth to center (tuned) frequency • • • • •
− is
constant Octave, one half octave and one-third octave Octave band filters: Gives less detailed (too coarse) analysis for practical vibration / noise measurement one half octave filters provide twice the information but requires twice the amount of time to obtain the data. Generally octave and one-third octave filters are used in spectrum analyzers. Each filter is tuned to different central frequency to cover entire frequency range. The lower cut-off frequency of a filter is equal to upper cut-off frequency of previous filter.
Vibration Measurements Filters Constant bandwidth filters:
• The bandwidth is constant • Used to obtain a more detailed analysis • High frequency analysis
Problem 9 Determine the maximum percent error of an accelerometer in the frequency ratio range 0 < r < 0.65 with (a) damping ratio ζ = 0, (b) damping ratio ζ = 0.75
