Post on 08-Apr-2018
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HARMONIC AND TRANSIENTHARMONIC AND TRANSIENT
VIBRATION ANALYSISVIBRATION ANALYSIS
by:Shivaprasad.P 080922004
Kiran. R 080922018
Narendra. Pai 080922022
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CONTENTSCONTENTS
IntroductionTheory
Mini projectReferences
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IntroductionIntroduction
y Vibration and time response analyses canbe subdivided into the following threerelated categories.
y Modal or natural frequency analysisy Frequency response analysisy Transient response analysis
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y Harmonic response analysis gives you the ability topredict the sustained dynamic behaviour of yourstructures, thus enabling you to verify whether or notyour designs will successfully overcome resonance,
fatigue, and other harmful effects of forced vibrations.
y Harmonic response analysis is a technique used todetermine the steady-state response of a linear
structure to loads that vary sinusoidally (harmonically)with time.
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y Harmonic response analysis is a linear analysis. Somenonlinearity, such as plasticity will be ignored, even ifthey are defined.
y Figure shows Homogenous, particular, and generalsolutions for an under damped case.
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y Transient Analysisy Transient dynamic analysis is a technique
used to determine the dynamic responseof a structure under a time-varying load.
y Cases where such effects play a majorrole are under step or impulse loading
conditions, for example, where there is asharp load change in a fraction of time.
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y It should be noted that a transientanalysis is more involved than a static orharmonic analysis.
y It requires a good understanding of thedynamic behaviour of a structure.
y A modal analysis of the structure should
be initially performed to provideinformation about the structure's dynamicbehaviour.
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y The basic equation of motion solved by atransient dynamic analysis is
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TheoryTheory
y The free vibration equation of motionmay be written in matrix form as follow
!
!
00
xx
kkk
k)kk(
x
xm0
0mKxxM
2
1
32
2
2
21
2
..1
..
2
1..
Where the scalar mass matrix (M) modifies the vectorial
acceleration matrix ( ) and the scalar stiffness matrix (K)modifies the vectorial position matrix (x)
..
x
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Two-degree of freedom system
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The solution of harmonic motion in the samefrequency is
[[ titi
2
1
2
1
XeeX
X
x
x
x
Where X1 and X2 are the amplitude of oscillation ofm1 and m2, respectively. X is known as the systemseigen vectoror eigen mode.
? A
!
!
0
0
)(
)(
2
1
2
2
32
2
2
1
2
212
X
X
mkk
k
k
mkkXMK
[
[[
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y Nontrivial solutions may only be satisfiedwhen the determinant in Eq. (A) equalzero.
? A 0k)mkk)(mkk(MKdet 22223212212 ![[![
(A)
Eq. (A) yield a quadratic in 2
. The roots of this quadratic arethe natural frequencies of the system, also known as its eigenvalues. The deflected shape of any given time will be a linearcombination of all its eigen modes as follows
\!\! XXx
ii (B)
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y When a system is being excited by knownoscillatory frequencies, it may not be feasible todesign its natural frequencies out of this
operating range.
Damped single-degree of freedom system subject to harmonic excitation
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y The equation of motion for the abovesystem follows.
tsinFkxxcxm 0
...
[!
tsinm
Fxxm2x 0
2
n
...
[![^
The solution of the above equation is a sum of two terms
the transient solution, which is the decaying, free vibrationsolution, and the steady state solution, which is any solution tothe complete equation. A complete solution to the underdamped system is found as follows :
)tsin(A
k
F)tsin(Cex 0d
tn J[][! ^[
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y Where A is a non dimensional amplitude ratio ormagnification factorof the steady state solution, and isthe phase angle of the solution. These terms areprovided by the following equations.
2/12
n
22
n
21
[[
^
[[
!
[[
[[^
!J
2
n
n1
1
2
tan
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y For an n-degree of freedom system, the matrixequation equivalent to equation follows
y Where is now the driving frequency,C is a
system damping matrix, u(
) is a generalizedcoordinate displacement vector, and F() is theforcing function vector.
? A )(F)(uKCiM2 [![[[ (C)
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Harmonic Response AnalysisHarmonic Response Analysisy Harmonic Response Analysis seeks the amplitude of
response and prescribed nodes vary sinusoidally with time.Several loads may be simultaneously applied and they may ormay not be in phase with one another.
y When 1st applied the harmonic loading does not produceharmonic response there is an initial transient that decays to0 over time because of damping.
y What then remains is steady state motion of the samefrequency as the loading.
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y Modal method the first step is to solve thevibration problem to obtain frequencies andmodes of the system usually only the lower
frequencies and the modes are needed.y For harmonic loading the solution has the
form.
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y Direct method Some harmonic responsecalculations are simplified by the use of complexnumbers. The harmonic response can be written inthe form.
y
Where i=-1 and D and R are complex amplitudes,respectively, of displacement and forcing function,both having circular frequency .
y The coefficient of D may be called a Dynamicstiffness matrix.
y
If the damping is 0 and M is Diagonal, the dynamicstiffness matrix can be regarded as K augmented bya spring of negative stiffness 2Mij attached to eachd.o.f i.
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y If damping is 0 , must not coincide withany natural frequency of the structure. Thenthe dynamic stiffness matrix becomes 0.
y
A disadvantage of above method is that if is close to a natural frequency, the dynamicstiffness matrix becomes ill conditioned.
y Another disadvantage is the expense of
solving of equation several times if forcingfunction of several different frequencies must be investigated are common.
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TRANSIENT RESPONSE ANALYSISTRANSIENT RESPONSE ANALYSIS
y Direct integration methods produce anequation of the form
In which A is non singular andindependent of time in linear problems,while F depends on quantities at instant n
and perhaps also at instant (n-1).
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Commonly used direct integration methodsare
1. Central difference method
2. Newmark method
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Central difference methodCentral difference method
y The basis of this method is a set of finitedifference formulas for first and secondderivatives, centered at instant n. For a vector D
rather than a single d.o.f these formulas areused.
The central difference method is continually stable, if t is toolarge, computed displacements become widely inaccurate and
grow without limit.
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Newmark methodNewmark methody The Newmark method uses finite difference expansions
in the time interval t, in which it is assumed that
tuu2
1uu 1n
..
n
..
n
.
1n
..
(
H
H
!
21n
..
n
..
n
.
n1n tuu21tuuu (
E
E(
!
Where , = Newmark integration parameters
t = t n-1 t n
u n = nodal displacement vector at time t n
n = nodal acceleration vector at time t n
n = nodal velocity vector at time t n
u n+1 = nodal displacement vector at time t n+1
n+1 = nodal acceleration vector at time t n+1
n+1 = nodal velocity vector at time t n+1
(E)
(D)
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Since the primary aim is the computation ofdisplacement{un+1}, the governing equation is evaluates at
time t n+1 as
? A ? A ? A _ aa1n1n.
1n
..
Fuuu !
The solution for the displacement at time is obtained by first rearranging equations(D) and (E),such that
!
n
..
3n
.
2n1n01n
..
uauauuau
!
1n
..
7n
..
6n
.
1n
.
uauauu
(G)
(H)
(F)
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Where
20t
1a
(E!
ta1
(E
H!
t
1a 2
(E!
12
1a 3
E!
1a 4 E
H!
E
H(! 2
2
ta
5
)1(ta6
H(!
ta 7 (H!
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Noting that {un+1} in equation (G) can be substitutedinto equation (H), equations for {un+1} and {un+1} can be
expressed only in terms of the unknown {un+1}. Theequation for {un+1} and {un+1} are then combined withequation (F) to form
? A ? A ? A_ a _ a! a1n10 FuaMa
? A _ a ? A _ a
n
..
3n
.
4n1n
..
3n
.
2n0uauauaCuauauaM
Once a solution is obtained for {un+1
}, velocities andaccelerations are updated as described in equation (G)and (H)
2
2
1
4
1
u HE 0
2
1"EH
2
1uH
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In the ANSYS implementation, the selected values for theNewmark s parameters are as follows
214
1KE ! K!H
2
1
Where = amplitude decay factor .
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Mode Superposition MethodMode Superposition Method
Solve a vibration problem to determine the lowerfrequencies and modes of structure.
Evaluate the modal time dependent loads pi from thephysical time dependent loads R,
Solve the uncoupled modal equations to determine zi=zi(t) for each i, and finally recover the physical d.o.f D byusing following equation.
Most of the computational effort made in mode
superposition method occurs in solving for frequenciesand modes of the system
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MINI PROJECTMINI PROJECT
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y PROBLEM DESCRIPTION :y For a Gear of 20 teeth of carbon steel with involute profile
is to be analyzed formaximum working load.
Given Data :Module : 10Addendum : 1MDeddendum :1.157MPressure angle : 20o
Fillet Radius : 0.3 MYoungs Modulus : 2.1*105 MpaPoissons Ratio : 0.3Maximum Allowable Stress : 250 MPa
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y Assuming Niemann's formula for the strength ofgears as the basis and assuming for carbon steel gears,the allowable stress is taken as 250 MPa themaximum working load P on the gear is computedand found to be 520 N per unit width of the gear.
y A load of 500 N per unit width of the gear is appliedat three positions, at a distance of about 0.2 module,1 module and 1.5 module from the top of the toothwith the boundary fixed.
y For transient analysis the tooth is assumed to run at
200rpm.y A load of 500 N per unit width is assumed to travel
along the flank.
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y Element Type:y Plane 42 :y is used for 2-D modeling of solid structures.y The element can be used either as a plane
element (plane stress or plane strain) or as anaxisymmetric element.y The element is defined by four nodes having two
degrees of freedom at each node: translations inthe nodal x and y directions.
y The element has plasticity, creep, swelling, stressstiffening, large deflection, and large straincapabilities.
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y PLANE42 GEOMETRY
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y Boundary Condition:y The gear tooth can be analyzed as a
cantilever beam.y The node below the root radius ,at
bottom lines ,are fixed in x and ydirection.
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y The model was created and meshedy Analysis type was selected as transient
using ANTYPE command.y Full method was selected for solution in
analysis optionsy Load steps were generated as Specifiedy Solver is the run.
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Meshed ModelMeshed Model1
X
Y
Z
MAY 8 2009
07:18:14
ELEMENTS
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Application of Boundary conditionsApplication of Boundary conditionsAnd Forces.And Forces.
1
X
Y
Z
1
X
Y
Z
1
X
Y
Z
1
X
Y
Z
1
X
Y
Z
MAY 8 2009
07:19:46
A-E-L-K-N
U
F
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Results: Transient AnalysisResults: Transient AnalysisDeformed Shape :Deformed Shape :1
X
Y
Z
MAY 8 2009
07:22:03
DISPLACEMENT
STEP=4
SUB =6
TIME=.400E-03
DMX =.016069
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Stress at end of last load stepStress at end of last load step1
MN
MX
X
Y
Z
2.803
118.369233.935
349.502465.068
580.634696.201
811.767927.333
1043
MAY 8 2009
07:23:43
NODAL SOLUTION
STEP=4
SUB =6
TIME=.400E-03
SEQV (AVG)
DMX =.016069
SMN =2.803
SMX =1043
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Stress at Node 83Stress at Node 831
0
125
250
375
500
625
750
875
1000
1125
1250
Stress ( MPa)
.8
1.2
1.6
2
2.4
2.8
3.2
3.6
4
4.4
4.8
TIME
(x10**-4)
MAY 8 2009
07:37:48
POST26
SEQV_3
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Displacement At Node 83Displacement At Node 83
1
0
.4
.8
1.2
1.6
2
2.4
2.8
3.2
3.6
4
(x10**-3)
Stress ( MPa)
.8
1.2
1.6
2
2.4
2.8
3.2
3.6
4
4.4
4.8
TIME
(x10**-4)
MAY 8 2009
07:39:32
POST26
UX_4
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ConclusionConclusion
y At node 83 maximum Stress occurs i.e.1042.9 Mpa
y Since the Gear tooth Experiences thestress more than its maximum allowablestress, It not advisable to run the gear
tooth at allowable Load.
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Results : Harmonic AnalysisResults : Harmonic Analysis
The model was created and meshedThe Loads were Applied as specifed.Analysis type was selected as harmonic using
ANTYPE command.Full method was selected for solution in analysisoptionsFrontal solver was selected as solver and pre-
stress was assumed as zeroThe frequency range was given as 0-10 and no ofsubsteps were given as 20
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y The boundary conditions were given asthe ends fixed in all DOFs
y The amplitude of the harmonic load =
500 N was given at specifeid node.y The model was solvedy The results were analysed
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ResultResult1
-6
-4
-2
0
2
4
6
8
10
12
14
X Displacement
125
250
375
500
625
750
875
1000
1125
1250
1375
TIME
MAY 8 2009
08:15:10
POST26
UX_4
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1
-1.25
0
1.25
2.5
3.75
5
6.25
7.5
8.75
10
11.25
YDisplaceme
nt
125 250 375 500 625 750 875 1000 1125 1250 1375
TIME
MAY 8 2009
08:17:26
POST26
UY_6
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ConclusionConclusion
y Stress distribution at node 358
1
34.752
34.76
34.768
34.776
34.784
34.792
34.8
34.808
34.816
34.824
34.832
VALU
0
5
10
15
20
25
30
35
40
45
50
FREQ
MAY 8 2009
08:23:23
POST26
AMPLITUDE
SEQV_4
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Conclusion:Conclusion:
y Stresses where found within theMaximum allowable stress limits.
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y Bibliography:
y Singiresu.S.Rao ,Mechanical Vibration, FourthEdition, Pearson Education,2004
y
V.Ramamurthi ,Computer Aided Design inmechanical Engineering ,3rd Edition,McgrawHill Edition,1998
y O.C.Zienkiewicz,R.L.Taylor and J.Z.Zhu ,The
Finite Element Method,Its basis andfundamentals ,6th edition, ElsevierPublication.
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y Case study reference:
y Ulrich Sdmersen, Oliver Pietsch,Wilfried Reimche, Fiedrich-WilhelmBach, TRANSIENT VIBRATION SIGNATURES AT STEAM AND GASTURBINES, PANNDT Conference, June 2 2003 Rio de Janeiro - RJ Brasil.
y Peter Manis andWes Jones,Wind-Induced Harmonic Resonance:Considerations for Light Pole Design, Structure magazine, Published byNational Council of Structural Engineers Associations (NCSEA), March2008.
y Dr.-Ing. Jrg-Henry Schwabe, Doz. Dr.-Ing. habil. Helmut Kuch, IFFWeimare.V. and Dipl.-Ing. Steffen Mothes, F. C. Ndling Betonelemente GmbH +Co. KG, Harmonic vibration on concrete block making machines,26 29October 2004 CPI Congress,Germany
y VikVedantham, HARMONIC VIBRATION ANALYSIS Establishing,Identifying and eliminating harmful frequencies, 3DVision Technologies.
y Case Studies on Paper Machine Vibration Problems, Andrew K. Costain,Bretech Engineering Ltd. 70 Crown Street, Saint John, NB Canada E2L3V6
y Transient Analysis of Equipment Operation,Kelly Eberle, P.Eng. and GordonSun, Ph.D., P.Eng. Beta Machinery Analysis, GMC Today, October 2008.