05 continuous shaft

Dynamics of Continuous Structures
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Vibration of Continuous
Structures

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Course Contents
 SDOF
 M-DOF
 Cables/String
 Bars
• Shafts
• Vibration Attenuation
• Beams
• FEM for Vibration
• Plates
• Aeroelasticity

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Vibration of Shafts

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Objectives
• Derive the equation of motion for Shafts
• Estimate the Natural Frequencies
• Understand the concept of mode shapes
• Apply BC’s and IC’s to obtain structure
response

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Shafts
• This type of elements appear in rotary
machines (motors, Generators ...etc.)
• It resists deformations only by torsional
resistance

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Equation of Motion
calculusfrom,
mechanicssolidfrom,
),(
densitymasssshaft'the
sectioncrossareaofmomentpolar=
modulusshear=
dx
x
d
x
tx
GJ
J
G











(x,t)

x (x,t)+d
+d
x+dx

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Equation of Motion
dx
t
tx
Jdx
x
dx
2
2
),(
elementon themomentsSumming





 +
2
2
),(),(
yields;sexpressiontheseCombining
t
tx
J
x
tx
GJ
x 













(x,t)

x (x,t)+d
+d
x+dx
2
2
2
2
),(),(
constant
t
txG
x
tx
GJ






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


 G
c
t
tx
c
x
tx
 ,
),(),(
2
2
2
2
2
For no external torque the equation
of motion becomes:
, wave speed
We can follow the same procedure used to solve string and
bar equations

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Modes and Natural Frequencies
(  ( 
2
2
22
2
2
2
2
2
)(
)(
)(
)(
0
)(
)(
,
)(
)(
)(
)(
=and=where)()()()(
)()(),(
ctT
tT
x
x
x
x
dx
d
tT
tT
c
x
x
dt
d
dx
d
tTxctTx
tTxtx

























Solve by the method of separation of variables:
Substitute into the equation of motion to get:
Results in two second order equations
coupled only by a constant:

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Initial and boundary conditions
• Two spatial conditions (boundary conditions)
• Two time conditions (initial conditions)
• See table 6.3 for a list of conditions
• Clamped-free rod:
)(=,0)(and)(=,0)(
torque)(0boundaryFree0),(
)deflection(0boundaryClamped0),0(
00 xxxx
tG
t
t
x




 


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Example: Grinding Machine
• Top end of shaft is connected to pulley (x=0)
• J1 includes collective inertia of dive belt,
pulley and motor
x
J2
J1
Drive pulley, collective inertia
Grinding head inertia
Shaft of stiffness GJ, length l(x,t)

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bottomat
),(),(
at top
),(),(
conditionsBoundary
2
2
2
0
2
2
1
0
 



xx
xx
t
tx
J
x
tx
GJ
t
tx
J
x
tx
GJ








The minus sign follows from right hand rule.
Use torque balance at top and bottom to get the

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






G
c
tTtTxx
G
c
tT
tT
Gx
x
tTxtx

++

















0)()(,0)()(
)(
)(
)(
)(
)()(),(
22
2
2


Using separation of variables

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)()(
)0()0(
)(
)(
)0(
)0(
)()0()()0(
0atConditionBoundary
2
2
1
2
22
1
1










J
J
J
J
c
tT
tT
J
GJ
tTJtTGJ
x





Similarly the boundary condition at l yields:

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(x)  a1 sinx + a2 cosx  (0)  a2
(x)  a1cosx  a2sin x  (0)  a1
x  0 
(0)  
2
J1
J
(0)  a1  
J1
J
a2
x  l 
(l ) 
2
J1
J
(l )  a1cosl  a2sinl 
2
J1
J
a1 sinl + a2 cosl
 tan(l ) 
Jl (J1 + J2 )(l )
J1J2 (l )2
 (Jl )2 THE CHARACTERISTIC EQUATION

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shapemodefirstthe)(00
)(,0)(
ingshaft turntheofmodebodyrigidthe)(
0)(00,1forNote
and,,3,2,1,forsolveyNumericall
:solutionfirstitsas0has
)()(
))((
)tan(
111
1111
11
22
21
21
axbx
xbaxx
btatT
tTn
G
n
JJJ
JJJ
nnn

+
+



+













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,Hz039,114,Hz026,76
,Hz013,38,Hz0
Pa1025
m0.25=rad,/mkg5=
,kg/m2700=rad,/mkg10
0tan
43
21
9
2
32
21







ff
ff
G
J
JJ
x





05 continuous shaft

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05 continuous shaft