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Unformatted text preview: A.K. Slone EG260 Dynamics (1) EG260 DYNAMICS I – Harmonic Excitation 2
EG260 DYNAMICS I – Harmonic Excitation 2 .......................................................... 1 1. Damped SDOF harmonically excited systems ................................................ 2 1. The particular solution .................................................................................... 3 2. The complete solution ..................................................................................... 7 2.1. Example 2.1 ..................................................................................................... 8 3. The transient and steady state responses. ...................................................... 11 3.1. Example 3.1 ................................................................................................... 16 4. Resonance in damped systems .................................................................. 18 5. Alternative solution methods ........................................................................ 21 5.1. The geometrical method. ............................................................................... 21 5.2. The frequency response method. ................................................................... 22 5.3. The transform method. .................................................................................. 26 6. Harmonic excitation of the base .................................................................... 28 6.1. Force transmitted to the mass ........................................................................ 35 6.2. Example 6.2 ................................................................................................... 36 7. Rotating imbalance ........................................................................................ 37 7.1. Force transmitted to the base ......................................................................... 41 7.2. Example 7.2 ................................................................................................... 42 7.3. Example 7.3 ................................................................................................... 45 8. Measurement. ................................................................................................ 49 8.1. Measurement devices .................................................................................... 49 ©a.k.slone 2010 1 of 1 A.K. Slone EG260 Dynamics (1) 1. Damped SDOF harmonically excited systems x(t)
k
F(t) m fk
fc F N c
Frictionfree surface mg
Free body diagram Figure 1 Viscously damped SDOF springmass system In this document the case of a viscously damped SDOF
system subject to harmonic excitation is considered.
Using Newton’s Second Law of motion the equation of
motion for the system shown in From Figure 1 is:
m&& + cx + kx = F (t )
x & (1) For harmonic excitation the forcing term is assumed to
be F (t ) = F0 cos (ωt ) thus the equation of motion is:
m&& + cx + kx = F0 cos(ωt )
x & (2) Equation (2) may be divided by m to give: && +
x ©a.k.slone 2010 c
k
&
x + x = f 0 cos(ωt )
m
m (3) 2 of 2 A.K. Slone where EG260 Dynamics (1) f0 = F0
m . From the formulae for the natural frequency and the
m
c
2
damping ratio k = ωn and m = 2ζω n . Hence equation
(3) may be expressed as :
2
&& + 2ζω n x + ωn x = f 0 cos (ωt )
&
x (4) As with any second order differential equation the
solution to the nonhomogenous equation (4) is formed
by obtaining a solution to the homogenous equation and
the particular solution and adding the two solutions
together.
1. The particular solution
The particular solution of equation (4) will be of a
similar form to the right hand side, i.e. the forcing term.
But here the frequency of the forced response will be of
the same frequency as the forcing term but with different
amplitude and phase. The phase shift is due to the effect
of the damping force.
Thus the particular solution is assumed to be of the form:
x p (t ) = X cos(ωt − θ ) (5) Successive differentiation of equation (5) gives: ©a.k.slone 2010 3 of 3 A.K. Slone EG260 Dynamics (1) &
x p (t ) = − Xω sin(ωt − θ ) &&p (t ) = − Xω 2 cos(ωt − θ )
x and (6)
(7) Substituting equations (5), (6) and (7) into equation (4)
gives:
2
− Xω 2 cos(ωt − θ ) − X 2ζω nω sin(ωt − θ ) + Xω n cos(ωt − θ ) = f 0 cos ωt (8) Collecting terms: {( ) } 2
X ωn − ω 2 cos(ωt − θ ) − 2ζω nω sin(ωt − θ ) (9) = f 0 cos ωt Using the trigonometrical relationships:
sin(ωt − θ ) = sin ωt cos θ − cos ωt sinθ
cos(ωt − θ ) = cos ωt cos θ + sin ωt sinθ Equation (9) may be expressed as: {( ) } 2
X ω n − ω 2 (cos ωt cos θ + sin ωt sin θ ) − 2ζω n ω (sin ωt cos θ − cos ωt sin θ ) = f 0 cos ωt (10)
Collecting cosine and sine terms equation (10) gives: [(
[( )
) 2
⎧ ω n − ω 2 cos θ + 2ζω n ω sin θ cos ωt + ⎫
⎪
⎪
X⎨ 2
⎬
2
⎪ ω n − ω sin θ − 2ζω n ω cos θ sin ωt ⎪
⎩
⎭
= f 0 cos ωt ©a.k.slone 2010 (11) 4 of 4 A.K. Slone EG260 Dynamics (1) Comparing righthand side and lefthand side terms
gives: [(
X [(ω )
− ω )sin θ − 2ζω ω cos θ ] = 0 2
X ω n − ω 2 cos θ + 2ζω n ω sin θ = f 0 2
n 2 n (12)
(13) Equations (12) and (13) may be expressed in matrix form
as: ( ) 2ζω n ω ⎤ ⎡ X cos θ ⎤ ⎡ f 0 ⎤
=
⎥
2
ωn − ω 2 ⎦ ⎢ X sin θ ⎥ ⎢ 0 ⎥
⎣
⎦ ⎣ ⎦ 2
⎡ ωn − ω 2
⎢
⎣ − 2ζω n ω ( ) (14) The matrix equation (14) may be solved using standard
matrix methods, i.e., the solution of the matrix equation
1
Ax = b is x = A −1b where A is the inverse of matrix A.
(note:
This is only a 2x2 matrix so the inverse is easy to
find and the equation may also be solved using
simultaneous equation methods, but for more complex
matrix equations packages such as MATLAB may be
used.)
Solving equation (14) for θ gives:
⎛ 2ζω n ω
2
2
⎝ ωn − ω θ = tan −1 ⎜
⎜ From equation (14) (ω ©a.k.slone 2010 2
n −ω2 ) 2 ( ) ⎞
⎟
⎟
⎠ sin 2 = 2ζω n ω cos 2 θ (15) 2 (16) 5 of 5 A.K. Slone EG260 Dynamics (1) 2
2
2
Adding (ωn − ω ) cos θ to both sides of equation (16)
gives:
2 (ω − ω ) sin + (ω − ω ) cos θ
= (2ζω ω ) cos θ + (ω − ω ) cos
2 2 2
n 2 2 2 2 2
n 2 2 2 2
n n 2 2 (17) θ Collecting terms gives: {(2ζω ω ) + (ω
2 2
n 2
n −ω2 ) }cos θ = (ω
2 2 2
n −ω2 ) (sin + cos θ )
2 2 2 (18)
Hence
cos θ = Similarly adding (2ζω n ω )
(16) gives: 2 ((ω 2
n −ω2 ) + (2ζω ω ) )sin
2 2 2 n (ω (ω
2
n sin 2 θ −ω2 2
n −ω2 ) ) + (2ζω ω )
2 2 (19) n to both sides of equation θ = (2ζω n ω ) (cos 2 θ + sin 2 θ )
2 (20) Hence:
sin θ = (ω 2ζω n ω
2
n − ω2 ) + (2ζω ω )
2 (21) 2 n Substituting equations (19) and (21) into equation (12)
gives:
⎡
X⎢
⎢
⎣ (ω 2
n −ω2 (2ζω ω ) + (ω ©a.k.slone 2010 2 n ) 2 2
n (2ζω ω )
(2ζω ω ) + (ω 2 −ω2 ) 2 + n 2 n 2
n −ω2 ) 2 ⎤
⎥= f
0
⎥
⎦ (22)
6 of 6 A.K. Slone EG260 Dynamics (1) Thus (ω X ) ( − ω 2 + 2ζωnω
2 2
n ) 2 = f0 (23) which on rearrangement gives the forcing term amplitude
as:
X= (ω 2
n ) f0 − ω 2 + (2ζω nω )
2 2 (24) Substitution into equation (5) gives the particular
solution as:
x p (t ) = (ω f0 2
n −ω ) + (2ζω ω ) 2 2 2 cos(ωt − θ ) (25) n From equations (19) and (21) the forcing term phase
angle is:
⎛ 2ζω nω ⎞
⎟
2
ωn − ω 2 ⎟
⎝
⎠ θ = tan −1 ⎜
⎜ (26) 2. The complete solution For the underdamped case (0< ζ <1), the solution to the
complementary function, i.e. the case of free vibration,
is: x(t ) = Ae−ζω t sin(ωd t + φ )
n ©a.k.slone 2010 (27) 7 of 7 A.K. Slone EG260 Dynamics (1) Thus the complete solution is:
x(t ) = Ae −ζωnt sin(ω d t + φ ) + X cos (ωt − θ ) (28) where X and θ are given by equations (25) and (26)
The solution of equation (28) requires the evaluation of
the constants A and φ, which are determined from initial
conditions.
Successive differentiation of equation (28) gives:
x(0 ) = A sin(φ ) + X cos(θ ) &
x(0 ) = −ζω n A sin (φ ) + ωd A cos (φ ) + ωX sin (θ ) (29) (30) From which A and φ, may be determined from initial
conditions.
2.1. Example 2.1 Compute the constant of integration for an undamped
system subject to a harmonic excitation 1000 cos(5t ) and
compare these values for the unforced case, given the
parameters ω n = 10 rad/s, ζ = 0.01, m = 100 kg and the
initial conditions xo = 0.05 m and v0 = 0 m/s.
Solution
f0 = 1000
= 10 ζ = 0.01 and ω = 5 rad/s
100 ©a.k.slone 2010 8 of 8 A.K. Slone EG260 Dynamics (1) From equation (25)
X =
= (ω f0 2
n ) − ω 2 + (2ζω nω )
10
2 2 + (100 − 25) + (2 × 0.01 × 10 × 5)
X = 0.133 m
2 2 (31) From equation (26) the forcing term phase angle is:
⎛ 2ζω nω ⎞
⎟
2
ωn − ω 2 ⎟
⎝
⎠ θ = tan −1 ⎜
⎜ ⎛ 2 × 0.01 × 10 × 5 ⎞
= tan −1 ⎜
⎟
⎝ 100 − 25 ⎠ (32) θ = tan −1 (0.01333) Thus: θ = 0.013 rad (33) The damped natural frequency is:
ωd = ω n 1 − ζ 2 = 10 1 − 0.012 = 9.999 (34) Substituting into equation (28) gives:
x(t ) = Ae −0.1t sin(9.999t + φ ) + 0.133 cos(5t − 0.013) (35) Differentiating gives: &
x(t ) = −0.1Ae −0.1t sin(9.999t + φ )
+ 9.999 Ae −0.1t cos(9.999t + φ )
− 0.665 sin(5t − 0.013) ©a.k.slone 2010 (36) 9 of 9 A.K. Slone EG260 Dynamics (1) Setting t = 0 in equation (35) gives:
x(0 ) = 0.05 = A sin(φ ) + 0.133 cos(0.013) Hence:
A= (37) 0.05 − 0.133 cos (0.013) − 0.083
=
sin(φ )
sin(φ ) (38) Setting t = 0 in equation (36) gives:
&
x(0) = −0.1A sin(φ ) + 9.999 A cos(φ ) (39) − 0.665 sin(− 0.013) Substituting for A from equation (38) gives:
&
x(0) = 0 = −0.1(− 0.083) + 9.999(− 0.083) cot (φ ) (40) + 0.665 sin(0.013) From equation (40) φ = 1.55 rad and so A = −0.083 m
For the free vibration case F0 =0, the response for the
underdamped case is:
x(t ) = Ae −0.1t sin(9.999t + φ ) (41) Setting t =0 in equation (41) gives:
x(0 ) = 0.05 = A sin(φ ) (42) Differentiating equation (41) gives; ©a.k.slone 2010 10 of 10 A.K. Slone EG260 Dynamics (1) &
x(t ) = −0.1Ae −0.1t sin(9.999t + φ ) + 9.99 Ae −0.1t cos(9.999t + φ ) (43)
Setting t = 0 gives:
&
x(0 ) = 0 = −0.1A sin(φ ) + 9.99 A cos(φ ) (44) 9.99
= 0.999
0.1
φ = 1.56 rad (45) So
tan(φ ) = From equation (42)
0.05
sin(φ )
A = 0.05 m
A= (46) Comparing the values of A and φ in equations (45) and
(46) for the case of free vibration with those when the
system is subject to a driving force where φ = 1.55 rad
and so A = −0.083 m, the effect of the driving force has
been that the amplitude is greatly reduced and the phase
angle is increased.
3. The transient and steady state responses. Equation (28) can be considered as a combination of a
transient response and a steady state response, as shown
below:
©a.k.slone 2010 11 of 11 A.K. Slone EG260 Dynamics (1) x(t ) = Ae −ζωnt sin(ω d t + φ ) +
X cos (ωt − θ )
14 244
4
3
144 2444
4
3
transient response steady state response (47) If the system has relatively large damping, the term e −ζωnt
causes the transient response to quickly die out, in some
cases within a fraction of a second and so need not be
considered. However, if the system is lightly damped, i.e.
ζ very small, the transient term in the solution may last a
sufficiently long time so that it could be significant.
In a number of applications, the damping is sufficiently
large that the transient term may be ignored. But there
are a number of applications, such as earthquake analysis
and satellite analysis where the transient term is
significant. The transient response may also be
significant if the amplitude is large. The usual practice in
engineering is to design and analyse the system based on
the steady state response alone. But in all cases the
transient response must be checked to see if it is
reasonable to ignore it.
Assuming that the transient term may be ignored, the
frequency ratio r is defined as r = ω
, which is a
ωn dimensionless quantity and may be used in equations
(26) and (to give:
⎛
ω ⎞
⎜ 2ζ n ω ⎟
2
⎛ 2ζω ω ⎞
θ = tan −1 ⎜ 2 n 2 ⎟ = tan −1 ⎜ 2 ω 2 ⎟
(48)
⎜ω −ω ⎟
⎜ ωn ω ⎟
⎝ n
⎠
⎜ 2− 2⎟
⎝ω
ω ⎠ ©a.k.slone 2010 12 of 12 A.K. Slone EG260 Dynamics (1) ⎛ 2ζr ⎞
2 ⎟
⎝1− r ⎠ θ = tan −1 ⎜ (49) and the forcing term amplitude is:
X = (ω f0 2
n − ω2 ) + (2ζω ω )
2 2 (50) n 2
Dividing top and bottom by ω n X = 2
f 0 ωn (51) (1 − r ) + (2ζr )
2 2 2 Thus
2
Xω n
Xk
=
=
fo
F0 1 (1 − r ) + (2ζr )
2 2 2 (52) 2
Xω n
Xk
The term
or
in equation (52) is called the
F0
f0 normalised amplitude and is plotted against the
frequency ratio in Figure 2 for several values of the
damping ratio ζ; similarly the phase of the steady state
response is plotted against the frequency ratio in Figure 3
As the driving frequency ω approaches the undamped
natural frequency ωn, i.e. r → 1 , the magnitude of the
amplitude reaches a maximum for light values of
damping, ( ζ ≤ 0.1) ©a.k.slone 2010 13 of 13 A.K. Slone EG260 Dynamics (1) Figure 2 Normalised Amplitude  steady state damped system In Figure 3 as the driving frequency approaches the
undamped natural frequency, i.e. the definition of
resonance for the damped case, the phase shift crosses
through π 2
• For ζ > 0 and 0 < r < 1,
π the phase angle is given by 0 < θ < , implying the
2 response lags the excitation.
• For ζ > 0 and r > 1,
π the phase angle is given by < θ < π , implying the
2 response leads the excitation.
• For ζ > 0 and r = 1, ©a.k.slone 2010 14 of 14 A.K. Slone EG260 Dynamics (1) π the phase angle is θ = , implying the phase
2 difference between the excitation and the response
is π 2 • For ζ > 0 and large value of r,
the phase angle approaches π, implying that the
excitation and the response are out of phase. Figure 3 Phase of steady state response for damped system It is important in a design phase to examine how the
amplitude of the steady state vibration is affected by
changing the damping ratio.
Figure 4 shows the same information as in Figure 2
except that it is plotted on a log scale so that curves for
©a.k.slone 2010 15 of 15 A.K. Slone EG260 Dynamics (1) both small and large values of damping may be shown on
the same graph.
Figure 4 shows that as the damping ratio ζ increases the
peak in amplitude decreases until it disappears.
Conversely as the damping ratio decreases the peak value
increase and become much sharper. In the limit as ζ → 0 ,
the peak reaches an infinite value, which is consistent
with the undamped response at resonance. Figure 4 Log scale magnitude of steady state response. 3.1. Example 3.1 A small spring 30 mm long is welded to a stationary
table so that it is fixed at the point of contact, with a 12
©a.k.slone 2010 16 of 16 A.K. Slone EG260 Dynamics (1) mm bolt welded to the other end, which s free to
move. The mass of the system is 49.2x103 kg, the
spring constant is measured to be 857.8 N/m and the
damping rate of the spring is measured to be 0.11 kg/s.
The spring is initially deflected 10mm.
Calculate the steady state response given ω = 132
rad/s, f0 = 10 N/kg. Calculate the change in amplitude
if ω = 125 rad/s
Solution Using the data given the undamped natural frequency
and damping ratio are:
k = 857.8 m = 49.2 × 10 −3
∴ ωn = 857.8
= 132.04 rad/s
49.2 × 10 −3 c
0.11
=
2 mk 2 857.8 × 49.2 × 10 −3
ζ = 0.0085 c = 0.11 ζ = The magnitude of the particular solution is x p (t ) = X , the
amplitude of the steady state response, which from
equation ( 25) is: ©a.k.slone 2010 17 of 17 A.K. Slone X= EG260 Dynamics (1) f0 (ω −ω ) + (2ζω ω)
2
n = 2 2 2 n 10
2 −132 ⎞2 + ⎛ 2× 0.0085 13204×132⎞2
2⎟ ⎜
⎟
13204
.
× .
⎟
⎝
⎠
⎟ ⎛
⎜
⎜
⎜
⎝ (53) ⎠ X = 0.0337m If the driving frequency is changed to 125 rad/s, the
amplitude is:
X = (132.04 2 ) 10 − 1252 + (2 × 0.085 × 132.04 × 125)
2 2 (54) X = 0.005 m Comparing the two values of the amplitude of the steady
state response, given by equations (53) and (54), shows
that a slight change in the driving frequency, from one
near resonance at 132 rad/s to driving frequency of 125
rad/s, causes an order of magnitude change in the
amplitude of the steady state response.
4. Resonance in damped systems Resonance is defined to occur if ω = ωn , i.e. when the
driving frequency is equal to the undamped natural
frequency, which corresponds to a phase shift of π rads. 2
This definition, which is equivalent to r = 1 or ω = ωn is frequently applied to damped systems, but Figure 4 ©a.k.slone 2010 18 of 18 A.K. Slone EG260 Dynamics (1) shows that the maximum value of Xk
occurs at different
F0 values of r depending on the value of the damping ratio
ζ.
• If ζ > 1
2 Maximum value of Xk
at r = 0
F0 1 • If 0 ≤ ζ ≤ 2 = 0.707 Maximum value of Xk
at r = 1 − 2ζ 2
F0 The value of driving frequency corresponding to the
maximum value of Xk
is called the driving frequency,
F0 ω p where: ω p = ωn 1 − 2ζ 2 Hence as damping decreases ω p → ωn , this gives the
undamped resonance condition.
As ζ increases, i.e. as the damping increases, the curves
in Figure 4 have peaks that are increasingly further away
from the line r = 1. Eventually the damping increases so
that ζ > 1
and the maximum value of
2 Xk
occurs at
F0 r = 0. ©a.k.slone 2010 19 of 19 A.K. Slone EG260 Dynamics (1) In many practical situations there is only light damping
so that ζ is small such that 1 − 2ζ 2 → 1 and so
ω p = ωn 1 − 2ζ 2 → ωn , i.e. r = 1. So that, in lightly damped cases, the undamped resonance condition ω = ωn
is often used.
Figure 5 shows the change in peak frequency with the
damping ratio. As the damping increases in the range
0≥ζ ≤ 1
= 0.707 , the size of the peak frequency ratio
2 decreases more sharply as ζ approaches 0.707. 1
Figure 5 Peak Amplitude for ζ ≤ 2 Figure 6 shows on a log scale shows how the peak
response given by equation (52) decreases as damping
increases, where it may be seen that the peak value
changes in orders of magnitude as the damping increases ©a.k.slone 2010 20 of 20 A.K. Slone EG260 Dynamics (1) Figure 6 Normalise Amplitude versus ζ 5. Alternative solution methods So far we have only considered one method for solving
the equations of motion for a harmonically excited SDOF
springmassdamper system, the method of undetermined
coefficients. Alternative methods, such as a geometric
approach, a frequency approach and a transform
approach are described below.
5.1. The geometrical method. The geometrical approach uses the fact that position,
velocity and acceleration will be out of phase with each
other by π/2 radians. Therefore express displacement,
velocity and acceleration as a vector. Compute X in terms
of F0 via vector addition
This method is of mathematical rather than practical
interest to the practicing engineer. If you are interested in
further explanation see the section on complex vector
representation of harmonic motion in Rao1
1 Mechanical Vibrations, S. S. Rao, ISBN 0131967517 ©a.k.slone 2010 21 of 21 A.K. Slone EG260 Dynamics (1) 5.2. The frequency response method. If the forcing function F (t )for a harmonically excited
system is expressed in complex form as: F (t ) = F0 e jωt (55) Then the equation of motion for the springmassdamper
system in Figure 7 may be expressed as: m&& + cx + kx = F0 e jωt
x &
x(t)
k (56) fk
mg Mass, m c
Frictionfree surface N Figure 7 Springmassdamper SDOF system But the excitation is real, so the response will also be the
real part of x(t), where x(t) is a complex function that by
definition must satisfy the equation of motion as given by
equation (56)
The particular solution must also be a complex number
as it is of the same form as the right hand side of
equation (56), thus:
©a.k.slone 2010 22 of 22 A.K. Slone EG260 Dynamics (1) x p (t ) = Xe jωt (57) Differentiating gives: &
x p (t ) = Xjωe jωt (58) && p (t ) = − Xω 2 e jωt
x (59) Substituting equations (57) (58) and (59) into equation
(56) gives:
− mω 2 Xe jωt + jcωXe jωt + kXe jωt = F0 e jωt (60) Collecting terms and simplifying gives: {( ) } X k − mω 2 + jcω = F0 (61) Hence
X= F0 (k − mω ) + jcω
2 (62) X = H ( jω )F0 (63) and where the frequency response function, H(jω) is:
1
H ( jω ) =
(k − mω 2 ) + jcω ©a.k.slone 2010 (64) 23 of 23 A.K. Slone EG260 Dynamics (1) Using standard methods for complex number
manipulation, multiply the numerator and denominator of
the right hand side (RHS) of equation (62) by the
complex conjugate of the RHS denominator gives: (
( )
(
)
)
F0 ((k − mω 2 ) − cjω )
X=
2 2
(k − mω ) + (cω )2 ⎛
⎞⎛ k − mω 2 − cjω ⎞
F0
⎟⎜
⎟
X =⎜
2
2
⎜ k − mω + cjω ⎟⎜ k − mω − cjω ⎟
⎝
⎠⎝
⎠ (65) (66) Separating the real and imaginary parts gives:
⎡
⎤
k − mω 2
cω
−i
X = F0 ⎢
⎥
2
2
2
2
k − mω 2 + (cω ) ⎥
⎢ k − mω 2 + (cω )
⎣
⎦ ( ) ( ) (67) Using the relationship: α − βj = Ae− jθ (68) where
A= α2 + β2
tan(θ ) = β
α (69) Let
α= k − mω 2 (k − mω ) 2 2 + (cω ) 2 β= cω (k − mω ) 2 2 + (cω ) 2 (70) Then
©a.k.slone 2010 24 of 24 A.K. Slone EG260 Dynamics (1) A= (k − mω ) + (cω )
{(k − mω ) + (cω ) }
2 2 2 2 2 2 (71) 2 Thus
A= 1 (k − mω ) 2 2 + (cω ) 2 (72) Hence the complex number: (k − mω 2 ) − j
(cω )
1
= e − jθ
(k − mω 2 )2 + (cω )2 (k − mω 2 )2 + (cω )2
(k − mω 2 )2 + (cω )2
(73)
Substituting equation (73) into equation (67) gives:
F0 X= (k − mω 2 )2 + (cω )2 e − jθ (74) Substituting into equation (57) gives the particular
solution:
⎡
⎤
F0
⎢
⎥
x p (t ) = ⎢
e − jθ ⎥ e jωt
2
⎢ k − mω 2 + (cω )2
⎥
⎣
⎦
F0
x p (t ) =
e j (ωt −θ ) ( ) (75) (k − mω 2 )2 + (cω )2 ©a.k.slone 2010 25 of 25 A.K. Slone EG260 Dynamics (1) Using the Euler relationship:
eαj = cos(α ) + j sin(α ) (76) Equation (75) may be expressed as
x p (t ) = F0 (k − mω ) 2 2 + (cω )2 (cos(ωt − θ ) + j sin(ωt − θ )) (77)
Comparing equation (77) with equation (25) it can be
seen that considering only the real part of the solution as
given by equation (77), both methods yield the same
result.
The frequency response approach is very similar to the
eigenvalue approach that is commonly used in
aerodynamics.
5.3. The transform method. Consider the equation for a springmass damper system
subject to harmonic excitation. &
m&& + cx + kx = F0 cos ωt
x
So far we have only considered solutions for SDOF
systems. An alternative method is the Laplace transform
method, which is also applicable to multidegreepf
freedom, (MDOF) systems. ©a.k.slone 2010 26 of 26 A.K. Slone EG260 Dynamics (1) The equation of motion may be transformed using the
Laplace transform:
∞ X ( s ) = L( x( t )) = ∫ x( t )e − st dt (78) 0 To give: ( ms 2 + cs + k ) X ( s ) = F0 s
s2 + ω 2 (79) Equation (79) is an algebraic equation in s, which may be
solved for X(s) to give: X( s ) = F0 s
( ms 2 + cs + k )( s 2 + ω 2 ) (80) Then an inverse transformation must be performed to
obtain the solution in the time domain.
For zero initial conditions the transformation gives: ( ms 2 + cs + k ) X ( s ) = F ( s ) X( s )
1
= 2
= H (s )
F ( s ) ms + cs + k (81) (82) 1 where H ( s ) = ms 2 + cs + k
©a.k.slone 2010 27 of 27 A.K. Slone EG260 Dynamics (1) But this is directly comparable to the frequency response
function, as given in equation (64) with (jω) = s
6. Harmonic excitation of the base Figure 8 shows a spring massdamper SDOF system
undergoing harmonic excitation of the base. The
displacement of the base is denoted by y(t) and the
displacement of the mass from its static equilibrium
position at time t is denoted by x(t). +x m k
c +y
y(t)= Ysin(ωt) Figure 8 Base excitation springmassdamper SDOF system The net elongation from the equilibrium position is (x –y)
and the relative velocity between the two ends of the
& &
damper is ( x − y ) . ©a.k.slone 2010 28 of 28 A.K. Slone EG260 Dynamics (1) From Figure 9 the equation of motion is: & &
m&& + c( x − y ) + k (x − y ) = 0
x (83) + &&
x +x
M k (x − y ) & &
c( x − y ) Figure 9 Free body diagram If the base is moving under harmonic motion such that
y(t ) = Y sin (ω t ) then equation (83) becomes: &
&
m&& + cx + kx = ky + cy
x
⇒
&
m&& + cx + kx = kY sin (ωb t ) + cωbY cos (ωb t )
x
⇒
2
2
&& + 2ζωn x + ωn x = ωn Y sin (ωb t ) + 2ζωn ωbY cos (ωb t )
&
x (84)
One interpretation of equation (84) is that system is acted
upon by two harmonic forces kY sin (ωt ) and cω cos(ωt ) ©a.k.slone 2010 29 of 29 A.K. Slone EG260 Dynamics (1) From Equation (24) the magnitude of the particular
solution for a cosine excitation is:
X= (ω 2
n −ω2 ) f0 2 + (2ζω nω )2 (24) Substituting f 0 = 2ζω n ω b in equation (24) gives the
particular solution for the cosine forcing function:
2ζω nωbY
x1 (t ) =
cos(ωbt − θ1 )
p
(85)
2
2
2 2
ωn − ωb + (2ζω nωb ) ( ) ⎛ 2ζω n ω b
⎜ ω 2 −ω 2
b
⎝ n θ 1 = tan −1 ⎜ ⎞
⎟
⎟
⎠ To calculate the particular response to the sine forcing
function, the same method of undetermined coefficients
is used. Applying the procedures used to calculate
2
equation (24) for the harmonic input ωn Y sin (ωbt ) gives:
x2
p (t ) = ( 2
ωn Y ) + (2ζω ω ) 2
2 2
ωn − ωb n b 2 sin (ωbt − θ1 ) (86) Here the particular solution is assumed to be of the form:
x 2 (t ) = X sin (ωbt − θ1 ) note that the phase angle is the
p
same as in equation (85). ©a.k.slone 2010 30 of 30 A.K. Slone EG260 Dynamics (1) By linear superposition the two particular solutions may
1
2
be added i.e. x p (t ) = x p (t ) + x p (t )
solution: x p (t ) = ω n Y ( 2
ω n + (2ζω n ω b )2 2
ωn ) 2 2
− ωb + (2ζω n ω b ) to give the particular cos(ω b t − θ 1 − θ 2 )
2 ⎛ ω
n
θ 2 = tan
⎜ 2ζω
b
⎝
−1 ⎜ Where (87) ⎞
⎟
⎟
⎠ f = F0 was the amplitude of the driving force of the
syste m, but here the amplitude of the harmonic
excitation is A, by equation (Error! Reference source
not found., thus the magnitude of the harmonic
excitation is: Y k 2 + (cω ) 2 X1 = (k − mω ) + (cω )
2 2 2 (88) Equation (15) gives the corresponding phase angle φ1. ⎛ cω ⎞
⎟
⎝ k − mω 2 ⎠ φ1 =tan −1 ⎜ (89) Thus the particular solution for base excitation is ©a.k.slone 2010 31 of 31 A.K. Slone EG260 Dynamics (1) x p (t ) = Y k 2 + (cω ) 2 sin(ωt − φ1 − α ) (k − mω ) + (cω )
2 2 2 (90) where φ1 and α are given by equations (89) and (Error!
Reference source not found. respectively.
Using trigonometrical identities Equation (90) may be
written in a more simple form as: x p (t ) = X sin(ωt − φ ) (91) where k 2 + (cω ) 2 X
=
Y (k − mω ) + (cω )
2 2 (92) 2 and
−1 ⎛
⎜ ⎞ mcω 3 ⎟
φ = tan ⎜
2 ⎟
2
⎝ k k − mω + (cω ) ⎠
Using ζ = c
2 mω n ( and ) r= (93) ω
ω n equations (92) and (93) may be written as: X =Y 1 + (2ζr ) 2 (1 − r ) + (2ζr )
2 2 2 (94) and ©a.k.slone 2010 32 of 32 A.K. Slone EG260 Dynamics (1) ⎛
2ζr 3
φ = tan ⎜
⎜ 1 + 4ζ 2 − 1 r 2
⎝
−1 ( ) ⎞
⎟
⎟
⎠ (95) Equation (94) may be expressed as the ratio of the
maximum response amplitude and the input displacement
amplitude: X
=
Y 1 + (2ζr ) 2 (1 − r ) + (2ζr )
2 2 2 (96) X Y = Td is called the displacement transmissibility and describes how motion is transmitted from the base to the
mass of the system as function of the frequency ratio
, r = ω ω n where ω is the frequency of the base. A plot of
Td against the frequency ratio, r, is shown in
Comparing Figure 10 with the plot of normalised
magnitude for harmonic excitation given in Figure 4 it
may be seen that the frequency ratio r in both is
independent of the damping ratio. However, the shape of
the two plots varies slightly due to the effect of the
numerator having an extra term of 2ζr for base
excitation, which is due to the load transmitted to the
damper in base excitation. ©a.k.slone 2010 33 of 33 A.K. Slone EG260 Dynamics (1) Figure 10 Displacement transmissibility as a function of frequency ratio From Figure 10 the following further observations may
be made:
For r = 0 Td ≡ 1 For small values of r, Td is close to 1
For r > 2 Td < 1 for all values of ζ For r = 2 Td = 1 for all values of ζ For r < 2 smaller values of ζ lead to larger values of Td
©a.k.slone 2010 34 of 34 A.K. Slone EG260 Dynamics (1) For r > 2 smaller values of ζ lead to smaller values of Td
For an undamped system ζ = 0 and r = 1, i.e. resonance, Td → ∞ then 6.1. Force transmitted to the mass For the system shown in Figure 8 a force is transmitted to
the mass due to the reaction from the spring and dashpot.
This force, F(t) is given by: & &
F (t ) = k ( x − y ) + c( x − y ) (97) By Newton’s second law, this force must balance the
inertial force, hence: & &
F (t ) = k ( x − y ) + c(x − y ) = −m&&
x (98) The steady state solution is: Fp (t ) = FT sin(ωt − φ ) (99) Where FT is the amplitude or maximum value of the force
transmitted and is given by: FT
= r2
kY
©a.k.slone 2010 1 + (2ζr ) 2 (1 − r ) + (2ζr )
2 2 2 (100) 35 of 35 A.K. Slone EG260 Dynamics (1) The ratio FT kY is the force transmissibility. From
equation (99) is can be seen that the force transmitted is
in phase with the motion of the mass x(t).
Figure 11 shows the normalised force transmitted to the
mass F versus the frequency ratio for differing damping
T kY ratios showing how the normalised force varies with
frequency ratio. Figure 11 Log Plot of force transmitted to the base 6.2. Example 6.2 A floor deflects 5 cm under the weight of a 2000N
machine resting on rubber mounts with ζ=0.01. If the
©a.k.slone 2010 36 of 36 A.K. Slone EG260 Dynamics (1) floor is vibrating with amplitude of 0.2 cm very near this
machine’s resonance, calculate the machine’s
displacement amplitude and transmitted force. Solution Y = 2cm = 0.002m and the stiffness of the floor is:
k= 2000 N
= 40kN/m
0.05m Assume r ≈ 1
FT = kYr 2 2
1 + 4ζ 2
1 + (2ζ r )
≈ kY
2
2ζ
( 1 − r 2 )2 + (2ζ r ) FT ≈ (40000 N/m )(0.002 m ) 1 + 4(0.01)
= 4001 N
2(0.01)
2 7. Rotating imbalance Rotating imbalance is present in a variety of engineering
applications having rotating parts such as gyros, cryocoolers, tyres and washing machines. The imbalance is
due to irregularities in the distribution of the rotating
component, which may cause substantial vibrations. A
schematic of rotating imbalance of mass m0 a distance e
, eccentricity, from the centre of rotation is shown in
Figure 12 ©a.k.slone 2010 37 of 37 A.K. Slone EG260 Dynamics (1) Guide Guide x(t) Machine of total mass m m0
e ωt
r c k
Rubber floor mounting
modelled as a spring
and damper Friction free
surface Figure 12 Schematic of a machine with rotating imbalance A free body diagram is shown in Figure 13, where ωr is
the frequency of the rotating machinery and m is the total
mass including both machine and imbalance mass. ©a.k.slone 2010 38 of 38 A.K. Slone EG260 Dynamics (1) a b Figure 13 Free body diagram of the imbalance (a),
and machine (b) In this example the vertical direction is denoted by x, and
summing the forces in the x direction for the rotating
imbalance in Figure 13a gives: m0 (&& + &&r ) = − Fr
x x (101) Summing the forces for the machine in Figure 13b gives: &
(m − m0 )&& = Fr − cx − kx
x (102) Adding equations (101) and (102) and rearranging gives: &
m&& + m0 &&r + cx + kx = 0
x
x (103) It is assumed that there is no motion in the horizontal
direction due to the presence of the guides. ©a.k.slone 2010 39 of 39 A.K. Slone EG260 Dynamics (1) x m0 ωrt e sin(ωrt)
e θ y e cos(ωrt)
Figure 14 Rotating of mass, m0 From Figure 14, the vertical, i.e. x, component of the
motion of the mass is xr = e sin(ω r t ) (104) So that the acceleration is &&r = −eω r 2 sin(ω r t )
x (105) Substituting in equation (103) and rearranging gives:
2
m&& + cx + kx = m0 eω r sin(ω r t )
x & (106) Dividing through by total mass, m, gives: m eω
&& + 2ζω n x + ω x = 0 r sin(ω r t )
&
x
m
2 2
n ©a.k.slone 2010 (107) 40 of 40 A.K. Slone EG260 Dynamics (1) From which it is apparent that both the mass of the
imbalance and the total mass influence the forcing
function.
Equation (107) represents a SDOF oscillator with a
harmonic forcing function. Thus we can use the methods
and solutions found in the previous section, where: x p ( t ) = X sin( ω r t − φ ) (108) ω r
Setting r = ω gives the magnitude and the phase of the
n
steady state motion of the mass m due to the rotating
imbalance of mass m0 moe
r2
X=
m (1 − r 2 ) 2 + (2ζr )2 and r⎞
⎛ 2ζ
⎟
2
⎝ 1− r
⎠ φ = tan −1⎜ (109) Thus the steady state solution is:
mo e
r2
r⎞
⎛ 2ζ
x p (t ) =
sin(ωr t − φ ) where φ = tan −1 ⎜
⎟
m (1 − r 2 ) 2 + (2ζr )2
⎝ 1− r 2 ⎠ (110)
7.1. Force transmitted to the base The force transmitted to the base of the machine is:
&
F (t ) = kx + cx ©a.k.slone 2010 41 of 41 A.K. Slone EG260 Dynamics (1) where the steady state displacement is given by equation
(110).
Differentiating equation (110) for velocity gives the force
transmitted to base of machine as:
me
F (t ) = 0
m kr 2 (1 − r ) 2 2 + (2ζr ) 2 me
sin (ωt ) + 0
m cω 2 r 2 (1 − r ) 2 2 + (2ζr ) cos(ωt ) 2 (111)
Thus the magnitude of the force transmitted to the base
is:
r 2 k 2 + c 2ω 2 me
F0 = 0
m (1 − r ) 2 2 + (2ζr )2 (112) which may be simplified as:
F0 = m0eω 7.2. 2 1 + (2ζr )2 (1 − r ) + (2ζr )
2 2 2 (113) Example 7.2 A machine with rotating balance, as shown in Figure 12
has a maximum measured deflection at resonance of 0.1
m. The damping ratio is estimated to be ζ = 0.05. From
the manufacturing process the out of balance mass is
estimated to be 10%. Estimate the eccentricity e and
hence the approximate location of the imbalance mass.
Also find what additional mass should be added
©a.k.slone 2010 42 of 42 A.K. Slone EG260 Dynamics (1) uniformly to the system to reduce the deflection at
resonance to 0.01 m ©a.k.slone 2010 43 of 43 A.K. Slone EG260 Dynamics (1) Solution Let total mass be M kg then rotating imbalance mass is
0.1M kg. ζ = 0.05, at resonance X = 0.1 m
At resonance r = 1 and using equation (109)
me 1
mo e
r2
= o
X =
m 2ζ
m ( 1 − r 2 )2 + (2ζ r )2
0.1M
1
e
M
2 × 0.05
e = 0. 1 m
0. 1 = (114) Now it is required to reduce the maximum amplitude to
0.01 m by means of an added mass say ∆m, again at
resonance: mo e 1
mo e
r2
=
X =
m 2ζ
m ( 1 − r 2 )2 + (2ζ r )2
⇒ 0.01 = ©a.k.slone 2010 0.1M
1
× 0.1 ×
M + ΔM
2 × 0.05 (115) 44 of 44 A.K. Slone EG260 Dynamics (1) Thus rearranging gives: 0.1M
M + ΔM
M + ΔM = 10 M
ΔM = 9 M
0.01 = (116) Hence, the total mass must be increased by a factor of 9
to reduce the maximum deflection to 0.1 m or 1 cm. Rotating imbalance is also important in helicopters and
prop aeroplanes. The tail rotor of a helicopter is used to
provide yaw control and torque balance. The main rotor
provides lift and horizontal motion and in doing so
introduces a clockwise thrust, which is overcome by the
counter clockwise thrust of the tail motor. An out of
balance rotor may cause damaging vibrations and limit
the performance of the helicopter.
7.3. Example 7.3 Figure 15 shows a schematic of a helicopter tail section,
where the stiffness is provided by tail section in the
vertical direction, where k = 1 x 105 N/m and the
corresponding mass is 60 kg. The tail rotor has mass of
20 kg. A 500 g mass is attached to one of the blades at a
distance of 15 cm from the axis of rotation.
Calculate the magnitude of the deflection of the tail
section of the helicopter as the tail rotates at 1500 rpm,
assuming a damping ratio of 0.01. At what rotor speed is
©a.k.slone 2010 45 of 45 A.K. Slone EG260 Dynamics (1) deflection a
deflection. maximum? Calculate the maximum Figure 15 Schematic of helicopter tail section Solution This problem only models the vertical vibration of the
tail section. The rotor system may be modelled as a
machine of total mass 20.5 kg attached to a spring of
mass 60 kg, stiffness 1 x 105 N/m, as shown in Figure 16.
m0 m m0 ωrt m
X ms , k ms , k Figure 16 Vertical vibration of helicopter tail section
©a.k.slone 2010 46 of 46 A.K. Slone EG260 Dynamics (1) From the problem statement the mass of the spring is 60
kg and the imbalance mass, m0 is 0.5 kg. The total mass
of the tail rotor including the added mass is: m = 20 + 0.5 = 20.5 kg
The damping ratio ζ is 0.01 and the rotator rotates at 157
rpm.
In the example of significant spring mass, (see modelling
and energy methods on blackboard), the natural
frequency of the springmass system was found to be: ωn = m+ k
ms 105
=
20.5 + 60 3 3
ωn = 46.69 rad/s (117) The frequency of the rotation is found from the rpm. 1500 × 2π
rad/s
60
ωr = 157.6 rad/s ωr = (118) Thus the frequency ratio is: r= ωr 157.6
=
ωn 46.69 r = 3.16 ©a.k.slone 2010 (119) 47 of 47 A.K. Slone EG260 Dynamics (1) The magnitude of the oscillation is given by equation
(109) as:
mo e
r2
X=
m ( 1 − r 2 )2 + (2ζ r )2
0.5 × 0.15
3.16 2
=
2
20.5
( 1 − 3.16 2 )2 + (2 × 0.01× 3.16 ) (120) = 0.004 m The maximum deflection will occur at resonance, i.e.
r~1, so that: ωr = ωn = 49.69 rad/s
60
rpm
2π
= 474.5 rpm ωr = 49.69 × (121) From equation (109) the magnitude of the oscillation for
r=1
moe 1
m 2ζ
0.5 × 0.15
1
=
20.5 2 × 0.01
= 0.183m = 18.3 cm
X= (122) 18.3 cm represents an unacceptably large deflection of
the rotor; hence the tail rotor should not be allowed to
rotate at 474.5 rpm
©a.k.slone 2010 48 of 48 A.K. Slone EG260 Dynamics (1) 8. Measurement. Dynamics testing is a valuable complement to dynamics
analysis, this is achieved through either:
• Modal testing, in terms of
 Natural
frequencies, useful for selecting
operational speeds of nearby machinery to
avoid resonant conditions.
 Damping
 Mode shapes
• Operational testing
 Inservice loads (often random)
8.1. Measurement devices Transducers are used to measure vibration. A transducer
is device which converts mechanical vibration into an
electric signal such as voltage proportional to
acceleration. The size of the transducer is important and
should not influence the structure’s dynamics through
added mass or stiffness. In analytic models the mass of
the transducer is included in the analysis.
There are a number of different types of transducers:
• Response Transducers.
Two broad types based on either velocity or on
displacement. Response transducers based on
displacement may be further categorised as linear
variable differential transformers (LVDT), rotary
variable differential transformers (RVDT) and
optical sensors. ©a.k.slone 2010 49 of 49 A.K. Slone EG260 Dynamics (1) LVDT are based on a system of coils and
magnets. They are often noisy and are best for lowfrequency, highdisplacement application.
Optical sensors are applicable where nocontact is required. Accelerometers.
As the name implies they convert the acceleration
into an electric signal. Accelerometers are based on
piezoelectric elements and are springmassdamper
systems; therefore internal resonance is a potential
problem and has to be considered when constructing
such a measurement device. A simple accelerometer is
shown in Figure 17a • a) b)
casing k + + m&&
x Voltage
m m
y(t c k(xy) & &
c(x − y ) Figure 17 Schematic of accelerometer mounted on structure Figure 17b shows the forces acting on the mass m.
Both the damper and the spring, by means of the
casing, are connected to the ground hence the force
©a.k.slone 2010 50 of 50 A.K. Slone EG260 Dynamics (1) k(x –y) is parallel to the damping force. Balance of forces
gives: & &
m&& = −k ( x − y ) − c(x − y )
x (123) Assuming the base is undergoing SHM such that: y (t ) = Y cos ωt (124) The relative motion of the accelerometer mass is defined
as: z (t ) = x(t ) − y (t ) (125) Equation (123) may be rewritten in terms of z to give: &
m&& + cz + kz = mω 2Y cos (ωt )
z (126) which is of the same form as equation (2), the only
difference is that equation (126) is in terms of the relative
displacement whereas equation (2) is in terms of the
absolute displacement. Bearing this in mind, the solution
to equation (126) will follow that for the preceding
paragraphs. ©a.k.slone 2010 51 of 51 ...
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This note was uploaded on 05/18/2010 for the course ENGINEERIN EG 260 taught by Professor Stone during the Spring '10 term at Swansea UK.
 Spring '10
 Stone
 Dynamics

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