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Introduction to Mechanical Vibrations and Structural Dynamics
The one semester schedule :
1. Vibration - classification. Free undamped single DOF vibration, equation of motion,
solution, integrational constants, initial conditions.
2. Free damped single DOF vibration, equation of motion, solution, integrational constants,
initial conditions.
3. The bending stiffness, the spring assemblies.
4. Forced vibration, vibration under constant force.
5. Forced vibration, harmonically varying external force, particular solution, harmonic
response, amplitude and phase characteristic.
6. Forced vibration due to the centrifugal force.
7. The rotational vibration, free and forced, torsional stiffness.
8. The two DOF free vibration, the mode shape concept, the mode shape solution, the mode
shape orthogonality.
9. The two DOF free vibration - complete solution including integrational constants.
10. The modal coordinates, the modal transformation, the free (unconstraint system), the
symmetrical system.
11. The modal analysis of the n-DOF system, matrix formulation.
12. The harmonic response of the n-DOF system.
13. Vibration of rectilinear beams.
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1. Vibration - classification. Free undamped single DOF vibration, equation of motion,
solution, integrational constants, initial conditions.
The vibration classification.
free (natural) vibration forced vibration
undamped vibration damped vibration
linear vibration non-linear vibration
vibration of discrete mass (masses) vibration of continuously distributed mass
vibration with 1 degree of freedom (DOF) vibration with more DOF
Free undamped single DOF vibration
vibration
the mass the spring the spring
The spring stiffness :
k
0
F
undeformed spring
the spring free length
deformed spring
the spring lengthening
The main spring parameters :
0 - the spring free length, the length of undeformed spring (not lengthened nor pressed)
[units are m, mm, ... in square brackets the units will be mentioned in the whole text],
- the spring lengthening (elongation) [m, mm],
F - the lengthening force [N],
Fk - the spring stiffness [N/m, N/mm].
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The example :
The stiffness of the spiral spring is :
3
4
Dn8
dGk
where : G - the elastic modulus (Young modulus) for shear deformation
[Pa N/m2, MPa N/mm
2],
d - the spring wire diameter [m, mm],
D - the spiral diameter [m, mm],
n - number of spring coils [-].
The spring potential (deformational) energy :
k
0
F
undeformed spring
the spring free length
deformed spring
the spring lengthening
2
21
0
2
21
00
P kxkdxxkdxFE
The action - reaction principle :
k
0
F
undeformed spring
the spring free length
deformed spring
FS
the spring lengthening
F - the external lengthening force, the action (black on figure),
FS - the spring force, the internal reaction force, caused by the spring deformation,
the spring resistance against deformation (red on figure).
kFFS
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The equation of motion
x
k
k
m
m
0
vx ax
FS = k·x
the
equ
ilibri
um
positio
n
undeformed spring
the spring free length
deformed spring
xkFFam S
i
i
0xkxm
0xkam
where m is the mass, xa is the acceleration, k is stiffness, x is displacement.
0xx2
0
where : m
k0
The solution is :
00t tCx sin displacement [m]
000 tCxv cos 1st derivative, velocity [m/s]
00
2
0 tCxva sin 2nd
derivative, acceleration [m/s2]
placed into the equation of motion :
0xkxm
0tCktCm 0000
2
0 sinsin
0tCktm
kCm 0000 sinsin
0tCktkC 0000 sinsin
00
5
In the solution :
00t tCx sin
are : t - time, independent variable [s],
m
k0 - the natural circular frequency [s
-1],
2f 0
0 - the natural frequency, number of cycles per second [s-1
, Hz],
00
0
2
f
1T
- the period, the time of one cycle [s],
C, 0 - integrational constants.
The solution time curve :
t
x
C
0
0
2T
0
0t
00t tCx sin
The integrational constants physical meaning :
C - the amplitude, the displacement maximum value, [m, mm]
0 - the phase shift, the phase “angle”, [rad]
divided by circular frequency gives the shift of the sinus curve to the left
on the time axis (in seconds).
The integrational constants solution
The initial conditions - the status in the beginning of the vibration, in time t = 0
t = 0 ... x = x0 initial displacement,
v = v0 initial velocity.
In the solution :
00t tCx sin
000 tCxv cos
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use the initial conditions :
0000t x0Cx sin
00000t v0Cv cos
simply :
00 xC sin
000 vC cos
the solution of the two equations with two unknowns :
2
0
2
02
0
vxC
0
00
0v
x arctan
The amplitude and phase shift are determined by the initial conditions.
The alternative expression of the displacement solution :
tBtAxv
tBtAtCx
0000t
0000t
cossin
sincossin
where :
0
0
CB
CA
cos
sin
are integrational constants (in opposite to C and 0, A and B have no direct physical meaning)
Integrational constants solution from initial conditions :
0000000t
0000t
vB0B0Av
xA0B0Ax
cossin
sincos
The integrational constants are then :
0
0
0
vB
xA
and :
2
0
2
02
0
22 vxBAC
0
00
0v
x
B
A arctanarctan
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The note to the numerical calculation of the arctg function :
The function arctg has always 2 roots.
For example : arctg 0.5 = 26.6°, arctg 0,5 = 206,6°. Because both tg 26.6° = tg 206,6° = 0.5
A C
B
0
I quadrant II quadrant
III quadrant IV quadrant
B
A0 arctan
The common calculator always returns the root in the interval 0 -90°, 90°.
But if B<0, the proper root is shifted 180º or rad.
B
A0 arctan
B < 0 B > 0
A > 0 0 90°, 180° (II. quadrant)
41532
1.arctan
while calculator returns -26.6° !
0 0, 90° (I. quadrant)
6262
1.arctan
A < 0 0 -180°, -90° (III. quadrant)
6206
2
1.arctan
or
4153
2
1.arctan
while calculator returns 26.6° !
0 -90°, 0 (IV. quadrant)
6262
1.arctan
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2. Free damped single DOF vibration
The damping is a physical phenomenon, which causes the decrease of vibration until it
completely disappears.
The causes of damping :
- the environmental resistance against motion (in liquid, in air),
- the internal friction inside material structure - the material damping,
- technical device - damper.
the symbol of damping
b
Fb F
The force resistance against motion - the damping force :
vbFb
here : b - coefficient of damping [N·s·m-1
],
v - velocity [m/s].
The equation of motion :
x
k m
vx ax
b
Fs
Fb
vbxkFFFam bS
i
i
0xkxbxm
0xkvbam
0xx2x2
0
where : m2
b
is the decay constant
the solution is :
0
t
t teCx sin
9
where : 22
0 - natural circular frequency of damped vibration [s-1
],
2f - the natural frequency (the num. of cycles per sec.) [Hz],
2
f
1T - the period (the time of one cycle) [s],
C, 0 - integrational constants.
The solution time curve :
0 1·T 2·T 3·T 4·T 5·T 6·T 7·T
C
t
x
0
t
t teCx sin
teC
x(t
)
x(t
+T
)
T
The integrational constants solution
The initial conditions - the status in the beginning of the vibration, in time t = 0
t = 0 ... x = x0 initial displacement,
v = v0 initial velocity.
used in the solution, (alternative expression) :
tBtAeteCx t
0
t sincossin
tBAtABexv t sincos
AB0BA0ABev
A0B1A10B0Aex
0
0
0
0
sincos
sincos
0xA
000 xvAv
B
2
2
002
0
22 xvxBAC
00
0
0xv
x
B
A
arctanarctan
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Further :
0
is the damping ratio,
222
0Tt
t
1
22T
x
x
ln is the logarithmic decrement.
22 4
The exponential function
teCx
determines the time of the vibration disappearing :
0
C
t [s]
x [m]
t
tmax eCx
1time constant
xmax = 37 % C
the tangential line
Let us define so called time constant :
1
The function is then :
t
t eCeCx
The numerical values of the function can by calculated as follows :
t = C370eCeCeCx 1
tmax_
, xmax = 37% C
t = 2· C140eCeCeCx 2
22
2tmax_
, xmax = 14% C
t = 3· C050eCeCeCx 3
33
3tmax_
, xmax = 5% C
t = 4· C020eCeCeCx 4
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4tmax_
, xmax = 2% C
t = 5· C0070eCeCeCx 5
55
5tmax_
, xmax = 0,7% C
That can be seen that in the time t = 5· the rest of the vibration amplitude is less than 1% of
the initial value C. This time can be understood as the vibration disappearing time.
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3. The bending stiffness, the bending vibration, the spring assemblies
The bending stiffness
Suppose the mass particle on the free end of the flexible beam (of the negligible mass).
The particle can vibrate.
kb
m
y
m
kb
m
Let us express so called bending stiffness of the fixed beam.
y
E·J
F
The bending of the fixed beam under the force can be expressed as :
JE3
Fy
3
where F - the bending force [N],
- the beam length [m],
E - the beam material Young modulus [Pa N/m2],
J - the beam cross section quadratic moment of inertia [m4].
The bending stiffness can be expressed as :
3b
JE3
y
Fk
in terms [N/m]
The beam cross section quadratic moment of inertia can be calculated as :
d
full circle
b
t
rectangle 4
641 dJ 3
121 tbJ
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All cases of vibration can be subsequently solved as the mass-spring system.
Of course the beam-support system can be different :
/2 /2
m E·J
kb
m
3b
JE48k
b a
m E·J
kb
m
22bba
JE3k
b a
m E·J
kb
m
2bb
JE3k
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The stiffness of the hydraulic circuit
Suppose the hydraulic cylinder, exposed to the force.
F
A
y
V0
V
the hydraulic cylinder
the hydraulic liquid
the piston
the force
F - the force, pressing the hydraulic liquid [N],
V0 - the initial volume of the liquid [m3],
V - the volume change (decrease), the compressed volume [m3],
K - the module of the liquid volume compressibility [Pa] (approx. K = 1÷2 GPa),
p - the hydraulic pressure [Pa],
A - the piston cross section area [m2],
y - the piston displacement [m].
The pressure - relative compression formula (analogous to the Hook’s law for solid materials)
0V
VKp
here 0V
V represents the relative volume change (decrease).
Pressure can be expressed as :
A
Fp
The compressed volume can be expressed as :
yAV
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Then :
0V
VKp
is :
0V
yAK
A
F
and finally :
0
2
hydV
AK
y
Fk
can be understood as hydraulic stiffness in terms [N/m].
The following solution is equal to the mass - spring system.
S
y
khyd V0
V
m m
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The hydraulic circuit can contain more than only the hydraulic cylinder.
S
y
V0
V
m
In this case V0 is not only the volume under the cylinder piston, but volume of the hydraulic
liquid in the whole circuit.
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The spring assemblies
The two springs can be combined in two ways.
The parallel assembly :
The two springs with stiffnesses k1 and k2 are combined side by side as shown on figure.
k1
F
k2
k1
k2
Fs1
Fs2
The external force F drags both springs through common base. There are two important keys :
1. the deformation of both springs is the same,
2. the spring forces Fs1 and Fs2 in both springs (red on figure) are summarized and in
equilibrium with the external force F (black on figure).
22s
11s
kF
kF
Fkk
Fkk
FFF
21
21
2s1s
The sum of stiffneses (k1+k2) can be interpreted as so called total stiffness :
21total kkk
and the system behaves as with one spring
k1
k2
ktotal
m
m
real system
substitutional system
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The serial assembly :
The two springs with stiffneses k1 and k2 are combined in one line as shown on figure.
total = F/ktotal
1 = Fs1/k1
k1
F
k2
Fs1 Fs2
1 = Fs1/k1
Fs2 A B
The external force F drags both springs in point B. There are two important keys :
1. the total deformation total is the sum of both partial deformations of both springs,
21total
2. the spring forces Fs1 and Fs2 (red on figure) are in equilibrium in point A and are equal,
the spring force Fs2 and the external force F (black on figure) are in equilibrium in point B
and are equal.
FFF 2s1s
The partial deformations can be expressed as :
1
1s
1k
F
2
2s
2k
F
likewise :
total
totalk
F
2
2s
1
1s
total
21total
k
F
k
F
k
F
In the nominator F = Fs1 = Fs2 can be canceled and finally :
21total k
1
k
1
k
1
21
total
k
1
k
1
1k
21
21total
kk
kkk
k1 k2
B
ktotal
m
m
real system
substitutional system
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The classification of the spring assembly can be confusing due to the visual form.
k1
m
k2
The assembly sketch on fig. looks similar to the assembly sketch of the serial assembly.
But only “looks similar”.
But ...
1
F
Fs1 Fs2
2
x
1. The body displacement x represents deformation of both springs 1 and 2, then :
x21
2. The sum of spring forces must be in the equilibrium with external force :
FFF 2s1s
That is clear, that the example represents the parallel assembly and total stiffness is :
21total kkk
k1
m
k2
ktotal
m
real system
substitutional system
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