mechanical vibrations - philadelphia university...mechanical vibrations v i b r a t i o n s. v...
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Mechanical Vibrations
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Introduction v
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1
2
Examplesv
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Examplesv
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Equation of motion v
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Equation of motion
Newton's 2nd
law of motion
Other methods
D’Alembert’sprinciple
Virtual displacement
Energy conservation
Newton's 2nd law of motionv
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Select a suitable coordinate
Determine the static equilibrium configuration of the system
Draw the free-body diagram
Apply Newton s second law of motion
The rate of change of momentum of a
mass is equal to the force acting on it.
Newton's 2nd law of motion
Energy conservation
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0
tan
UTdt
d
tconsUT
Potential energy
2
2
1kxU
Kinetic energy 2.
2
1xmT
0..
kxxm
0....
..
2
2
kxxmxmkxtF
xmdt
txdm
dt
txdm
dt
dtF
Vertical system v
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0..
; ..
kxxm
kmgmgxkxm stst
Mathematical reviewv
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tFtxtAtxtAtxtA 321
...
F(t) = 0 → Homogenous F(t) ≠ 0 → none homogenous
x(t) = C1x1 + C2x2 x(t) = C1x1 + C2x2 +xp
If A1,A2 and A3 are constants:
Characteristic equation:
A1s2+A2s+A3=0
1
31
2
22
2
4
A
AAAAs
xp form is the same type
asF(t)
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Only one real root:
s1,2
x(t) = C1est + C2t e
st
Two unequal real root: s1,2
x(t) = C1es1t + C2 es2t
Complex root:
s1,2 = α±βi
x(t) = eαt {C1cos(βt) + C1sin(βt)}
1
31
2
22
2
4
A
AAAAs
Solution
is 2nd order homogenous differential equation
where: A1 = m, A2=0 and A3=k
the solution : x(t) = eαt {C1cos(βt) + C1sin(βt)}
where α = 0 and β = ωn
x(t) = C1cos(ωnt) + C2sin(ωnt)
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0..
kxxm
Natural frequency
nim
ki
m
mks
2
4
Solution
C1 and C2 can be determined from initial conditions (I.Cs). For
this case we need two I.Cs.
The I.Cs for this case would be:
So:
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on
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xCtx
xCtx
.0
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0
2
1
1.sin
.
cos Eqtx
txtx n
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no
Harmonic motionIntroduce Eq.2 into Eq.1: x(t ) = A cos(ωnt – ϕ ) = Ao sin(ωnt + ϕo)
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Harmonic motion v
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tx
txtx n
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sin
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cos
tCtCtx nn sincos 21
Harmonic functions in time.
Mass spring system is called
harmonic oscillator
Assume:
C1 = A cos(ϕ) --- Eq.2(a)
C2 = A sin(ϕ) --- Eq.2(b)
AA
Amplitudex
xCCA
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2
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2
2
1
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no
no
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x
x
x
x
C
C
.tan
angle Phase
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tantan
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1
21
Harmonic motionv
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Natural Frequency (N.F)
A system property (i.e. depends of system parameters m
and k )
Unit: rad/sec
It is related to the periodic time (τ ) : τ = 2π/ωn
Periodic time is the time taken to complete one cycle
(i.e. 4 strokes)
The relation between the ωn and τ is inverse relation
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Example 2.1The column of the water tank shown in Fig
is 90m high and is made of reinforced
concrete with a tubular cross section of
inner diameter 2.4m and outer diameter
3m. The tank mass equal 3 x 105 kg when
filled with water. By neglecting the mass of
the column and assuming the Young’s
modulus of reinforced concrete as 30 Gpa.
determine the following:
the natural frequency and the natural
time period of transverse vibration of the
water tank
the vibration response of the water tank
due to an initial transverse displacement of
0.3m.
the maximum values of the velocity and
acceleration experienced by the tank.
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Example 2.1 solution:
Initial assumptions:
1. the water tank is a point mass
2. the column has a uniform cross section
3. the mass of the column is negligible
4. the initial velocity of the water tank equal zero
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Example 2.1 solution:
a. Calculation of natural frequency:
1. Stiffness: But:
So:
2. Natural frequency :
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3
3
l
EIk 44444 3475.24.23
6464mddI io
mNxxx
k / 812,28990
3475.2103033
9
srxm
kn /ad 9829.0
103
812,2895
Example 2.1 solution:
b. Finding the response:
1. x(t ) = A sin(ωnt + ϕ )
So, x(t ) = 0.3 sin (0.9829t + 0.5π )
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mxx
xA o
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. 2
2
20
tan.tan 11
no
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no x
x
x
Example 2.1 solution:
c. Finding the max velocity:
Finding the max acceleration :
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29829.0cos9829.03.0
. ttx smx /2949.09829.03.0
.max
29829.0sin9829.03.0
..2
ttx 22max /2898.09829.03.0
..smx
Example2 : Q2.13
Find the natural frequency of
the pulley system shown in Fig.
by neglecting the friction and
the masses of the pulleys.
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Example2 : Q2.13
Solution:
1. Free body diagram
2. x = 2x1 +2x2 ---- Eq.1
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x1x2
x
PP
PP
Example2 : Q2.13
Solution:
3. Equilibrium for pulley_1 : 2P = k1 x1 = 2k x1 ---- Eq.2
4. Equilibrium for pulley_2 : 2P = k2 x2 = 2k x2 ---- Eq.3
5. Substitute Eqs 2 and 3 in Eq.1:
6.Let keq is the equivalent spring constant for the system:
7. Mathematical model:
8. Natural frequency:
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k
P
kkP
k
P
k
Px
4
2
1
2
14
22
22
21
4
k
x
Pkeq
0..
kxxm
m
k
m
keq
n4
Rotational system
Governing equation:
Assume θ is very small
Natural frequency (ωn)
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.. OOO JJM
0sin..
mglJO
sin
0..
mglJO
mgl
J
J
mgl O
O
n 2
Torsional system
Governing equation:
Natural frequency (ωn)
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.. OOO JJM
0..
TO kJ
T
O
O
Tn
k
J
J
k 2
g
WDDhJO
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Solutionv
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2
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ttt n
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sin
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cos
tAtAt nn sincos 21
Example 2.3
Any rigid body pivoted at a
point other than its center of
mass will oscillate about the
pivot point under its own
gravitational force. Such a
system is known as a compound
pendulum (see the Fig). Find
the natural frequency of such a
system.
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Solution
the governing equation is found as:
Assume small angle of vibration:
So:
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0sin..
WdJO
0..
WdJO
OO
nJ
mgd
J
Wd
Example 2.4: Q2.12
Find the natural frequency of the system shown in Fig. with thesprings k1 and k2 in the end of the elastic beam.
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Example 2.4: Q2.12
Solution: F.B,D
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Example 2.4: Q2.12
Solution:
keq is equivalent stiffness for the combination of k1, k2 and kbeam
k1 and k2 equivalent: apply energy concept
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3
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EIkbeam
2
22
2
11
2
22
2
11
2
2,1,2
1
2
1
2
1
x
xk
x
xkkxkxkxk eqeq
Example 2.4: Q2.12
Solution:
Finding keq
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beameq
beameq
eq
beameqeq kk
kkk
kkk
2,1,
2,1,
2,1,
111
Example 2.4: Q2.12
Solution:
Finding natural frequency
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beam
beam
n
beameq
beameqeq
n
kx
xk
x
xkm
kx
xk
x
xk
kkm
kk
m
k
2
22
2
11
2
22
2
11
2,1,
2,1,
x1, x2 and x can
be found from
strength relation
Example 2.5: Q2.7
Three springs and a mass are attached to a rigid, weightless bar PQas shown in Fig. Find the natural frequency of vibration of thesystem.
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Example 2.5: Q2.7
Solution :Assume small angular motion
Let keq is the equivalent stiffness for the whole system
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2
3
2
22
2
112,1,
2
22
2
11
2
32,1,2
1
2
1
2
1
l
lklkklklklk eqeq
sin
2,1,3
32,1,
32,1,
111
eq
eq
eq
eqeq kk
kkk
kkk
Example 2.5: Q2.7
Solution :Now find the natural frequency
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2
33
2
22
2
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2
232
2
121
lklklkm
lkklkk
m
keq
n
Example 2.5: Q2.45
Draw the free-body diagram and derive the equation of motion
using Newton s second law of motion for each of the systems
shown in Fig
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Example 2.5: Q2.45
Solution F.B.D
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Example 2.5: Q2.45
Equation of motion:
The distance:
For mass m: --- (1)
For pulley Jo: --- (2)
According to static equilibrium: --- (3)
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orx 4
..xmTmg
rrkTrJ oo 44..
rk
mgrrkmgr oo
1644
Example 2.5: Q2.45
Equation of motion [cont]:
Substitute equations 1 and 3 into equation 1:
Use the relation to relate the translational
motion with the rotational one:
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016....
016....
1616
....
22
2
krrxmJmgrkrrxmmgrJ
rk
mgkrrxmmgJ
oo
o
.... rxrx
016..
22 krmrJo
P
End of chapter2 – part I
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