_____ Massachusetts Institute of Technology MIT Video COurse Video Course Study Guide Finite Element Procedures for Solids and Structures- Linear Analysis Klaus-JOrgen Bathe Professor of Mechanical Engineering, MIT Published by MIT Center for Advanced Engineering study Reorder No 672-2100
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_____tt~---Massachusetts Institute of Technology
MIT Video COurse
Video Course Study Guide
Finite ElementProcedures for Solids
and StructuresLinearAnalysis
Klaus-JOrgen BatheProfessor of Mechanical Engineering, MIT
Published by MIT Center for Advanced Engineering studyReorder No 672-2100
PREFACEThe analysis of complex static and dynamic problems involves in essence three stages: selection of a mathematicalmodel, analysis of the model, and interpretation of the results.During recent years the finite element method implemented onthe digital computer has been used successfully in modelingvery complex problems in various areas of engineering andhas significantly increased the possibilities for safe and costeffective design. However, the efficient use of the method isonly possible if the basic assumptions of the proceduresemployed are known, and the method can be exercisedconfidently on the computer.
The objective in this course is to summarize modern andeffective finite element procedures for the linear analyses ofstatic and dynamic problems. The material discussed in thelectures includes the basic finite element formulations employed, the effective implementation of these formulations incomputer programs, and recommendations on the actual useof the methods in engineering practice. The course is intendedfor practicing engineers and scientists who want to solve problems using modem and efficient finite element methods.
Finite element procedures for the nonlinear analysis ofstructures are presented in the follow-up course, Finite ElementProcedures for Solids and Structures - Nonlinear Analysis.
In this study guide short descriptions of the lectures andthe viewgraphs used in the lecture presentations are given.Below the short description of each lecture, reference is madeto the accompanying textbookfor the course: Finite ElementProcedures in Engineering Analysis, by K.J. Bathe, PrenticeHall, Inc., 1982.
The textbook sections and examples, listed below theshort description of each lecture, provide important readingand study material to the course.
Contents
Lectures
l. Some basic concepts of engineering analysis 1-1
2. Analysis of continuous systems; differential andvariational formulations 2-1
3. Formulation of the displacement-based finite element method_ 3-1
4. Generalized coordinate finite element models 4-1
5. Implementation of methods in computer programs;examples SAp, ADINA 5-1
6. Formulation and calculation of isoparametric models 6-1
highest order of (spatial) derivatives in problem-governing differential equation is 2m.
highest order of (spatial) derivatives in essential b.c. is (m-1)
highest order of spatial derivatives in natural b.c. is (2m-1)
Definition:
We call this problem a Cm-1
variational problem.
2·5
Analysis 01 continuous systems; differential and variatioD,a1 fOl'llolatiODS
Example - Variational formulation
We have in general
II=U-W
For the rod
fLII = J }EA
o
and
iL
au 2 B(--) dx - u f dx - u Rax Lo
u = 0oand we have 0 II = 0
The stationary condition 6II = 0 gives
rL au au rL.BJO(EA ax)(6 ax) dx -)0 6u t- dx
- 6uL
R = 0
This is the principle of virtualdisplacements governing theproblem. In general, we writethis principle as
or
(see also Lecture 3)
2·6
lIiIysis of ..IiDIGUS systems; differential and variatiooallormulatioDS
However, we can now derive thedifferential equation of equilibriumand the b.c. at x = l .
Writing a8u for 8au , re-ax ax
calling that EA is constant andusing integration by parts yields
dx + [EA ~ Iax x=L
- EA ~\dXx=o
Since QUO is zero but QU isarbitrary at all other points, wemust have
and
au IEAax- x=L=R
B a2uAlso f = -A p - and, at2
hence we have
2·7
Analysis of cODtiDaoas syst_ diIIereatial and variatioul fOlllalatiODS
The important point is that invokingo IT = 0 and using the essential
b.c. only we generate
• the principle of virtualdisplacements
• the problem-governing differential aquatio!)
• the natural b.c. (these are inessence "contained in" IT ,i.e., inW).
In the derivation of the problemgoverning differential equation weused integration by parts
• the highest spatial derivativein IT is of order m .
• We use integration by partsm-times.
Total Potential IT
IUse oIT = 0 and essential "b.c.
~
2·8
Principle of VirtualDisplacements
IIntegration by parts
~Differential Equation
of Equilibriumand natural b.c.
_ solveproblem
_solveproblem
balysis of aDa. syst-: diBerential and variatiouallnaiatiOlS
Weighted Residual Methods
Consider the steady-state problem
(3.6)
with the B.C.
B.[</>] = q., i =1,2, •••1 1
at boundary (3.7)
The basic step in the weightedresidual (and the Ritz analysis)is to assume a solution of theform
(3.10)
where the fi are linearly independent trial functions and the aiare multipliers that are determined in the analysis.
Using the weighted residual methods,we choose the functions fi in (3.10)so as to satisfy all boundary conditionsin (3.7) and we then calculate theresidual,
nR = r - L2mCL a· f.] (3.11 )
1 =1 1 1
The various weighted residual methodsdiffer in the criterion that they employto calculate the ai such that R is small.In all techniques we determine the aiso as to make a weighted average ofR vanish.
2·9
Analysis 01 C.tinnoDS systems; differential and variational 10000nlations
Galerkin method
In this technique the parameters ai aredetermined from the n equations
f f. R dD=O ;=1,2, ••• ,nD 1
Least squares method
(3.12)
In this technique the integral of thesquare of the residual is minimized withrespect to the parameters ai '
aaa.
1
;=1,2, ••• ,n
[The methods can be extended tooperate also on the natural boundaryconditions, if these are not satisfiedby the trial functions.]
RITZ ANALYSIS METHOD
Let n be the functional of the
em-1 variational problem that isequivalent to the differentialformulation given in (3.6) and (3.7).In the Ritz method we substitute thetrial functions <p given in (3.10)into n and generate n simultaneous equations for the parameters ai using the stationarycondition on n ,
2·10
an 0aa. =1
;=1,2, ••• ,n (3.14)
Analysis of continuous systems; differential and variational formulations
Properties
• The trial functions used in theRitz analysis need only satisfy theessential b.c.
• Since the application of oIl = 0generates the principle of virtualdisplacements, we in effect usethis principle in the Ritz analysis.
• By invoking 0 II = 0 we minimizethe violation of the internal equilibriumrequirements and the violation ofthe natural b.c.
• A symmetric coefficient matrixis generated, of form
Fig. 3.19. Bar subjected toconcentrated end force.
2·11
Analysis of COitiDlOIS systems; differeatial ad ,ariali" fOllDaialiODS
Here we have
1180
IT = 1 EA(~)2 dx2 ax
o- 100 uIx = 180
and the essential boundary conditionis u Ix=O = 0
Let us assume the displacements
Case 1
u = a1x + a2 iCase 2
~u = I1JO 0< x < 100
100 < x < 180
We note that invoking oIT = 0we obtain
1180
oIT = (EA ~~) o(~~) dx - 100 OU Ix=180
o = 0
or the principle of virtualdisplacements
£180
(~~u)( EA ~~) dx = 100 OU Ix=180o
JET T dV = IT. F.- - 1 1
V
2·12
Analysis of continuous systems; differential and variational formulations
Exact Solution
Using integration by parts weobtain
~ (EA ~) = 0ax ax
EA ~ = 100axx=180
The solution is
u = 1~O x ; 0 < x < 100
100 < x < 180
The stresses in the bar are
a = 100; 0 < x < 100
a = 100 ; 100 < x < 180(l+x-l00)2
40
2·13
Analysis of continuous systems; differential and variational formulations
Performing now the Ritz analysis:
Case 1
f180
dx+ I (1+ x-l00)22 40
100
Invoking that orr = 0 we obtain
E [0.4467
116
and
116
34076
128.6a1 = ---=E=--- a - 0.341
2 - - E
Hence, we have the approximatesolution
u =12C.6 0.341
E x - E2x
2·14
a = 128.6 - 0.682 x
Analysis of continuous systems; differential and variational formulations
Case 2
Here we have
100
E J 1 2n=2 (100 uB)a
f180
dx+ I (1+x-l00)22 40
100
Invoking again on =0 we obtain
E [15.4 -13][~:] = [~oo]240
-13 13
Hence, we now have
10000 11846.2U = E Uc EB
and
o = 100 0< x < 100
1846.2 = 23.08 x> 100o =80
2-15
Aulysis of COilinDmas systems; diUerenliai and varialiOlla1I01'1BDlaIiGlS
u
EXACT
~-- --- -::.:--~~~.-.-.
" Sol ution 2
---..I~ ..,-__--r-__--.,r--- ~X
15000E
10000E
5000-E-
100 180
CALCULATED DISPLACEMENTS
(J
50
100-I=:::==-==_==_:=os:=_=_=,==_=_=="" EXACT
"I~~ SOLUTION 1
I -< ,J SOLUTION 2
L._._. ~._._-+ ---,~--------r-------~X
100 180
CALCULATED STRESSES
2·18
balysis of coatiDloas systms; diBerenlial ud variational fonnllatioas
We note that in this last analysis
e we used trial functions that donot satisfy the natural b.c.
e the trial functions themselvesare continuous, but the derivatives are discontinuous at pointB. 1for a em- variational problemwe only need continuity in the(m-1)st derivatives of the functions; in this problem m =1 .
edomains A - Band B- e arefinite elements andWE PERFORMED AFINITE ELEMENTANALYSIS.
2·17
FORMULATION OF THEDISPLACEMENT-BASEDFINITE ELEMENTMETHOD
LECTURE 358 MINUTES
3·1
Formulation of the displacement-based finite element method
LECTURE 3 General effective formulation of the displacement-based finite element method
Principle of virtual displacements
Discussion of various interpolation and elementmatrices
Physical explanation of derivations and equations
Direct stiffness method
Static and dynamic conditions
Imposition of boundary conditions
Example analysis of a nonuniform bar. detaileddiscussion of element matrices
TEXTBOOK: Sections: 4.1. 4.2.1. 4.2.2
Examples: 4.1. 4.2. 4.3. 4.4
3·2
Formulation of the displacement-based finite element method
KINEMATICS, e.g. trussplane stressthree-dimensionalKirchhoff plateetc.
MATERIAL, e.g. isotropic linearelasticMooney-Rivlin rubberetc.
LOADING, e.g. concentratedcentrifugaletc.
BOUNDARY CONDITIONS, e.g. prescribed
1
displacementsetc.
FINITE ELEMENT SOLUTION
CHOICE OF ELEMENTS ANDSOLUTION PROCEDURES
YIELDS:GOVERNING DIFFERENTIALEQUATIONS OF MOTIONe.g.
..!.. (EA .!!!) = - p(x)ax ax
YIELDS:APPROXIMATE RESPONSESOLUTION OF MECHANICALIDEALIZATION
Fig. 4.23. Finite Element SolutionProcess
4·14
Generalized coordinate finite element models
SECTIONERROR ERROR OCCURRENCE IN discussing
error
DISCRETIZATION use of finite element 4.2.5interpolations
NUMERICAL evaluation of finite 5.8. 1INTEGRATION element matrices using 6.5.3IN SPACE numerical integration
EVALUATION OF use of nonlinear material 6.4.2CONSTITUTIVE modelsRELATIONS
SOLUTION OF direct time integration, 9.2DYNAMIC EQUILI-. mode superposition 9.4BRIUM EQUATIONS
SOLUTION OF Gauss-Seidel, Newton- 8.4FINITE ELEr1ENT Raphson, Quasi-Newton 8.6EQUATIONS BY methods, eigenso1utions 9.5ITERATION 10.4
ROUND-OFF setting-up equations and 8.5their solution
Table 4.4 Finite ElementSolution Errors
4·15
Generalized coordinate finite element models
CONVERGENCE
Assume a compatibleelement layout is used,then we have monotonicconvergence to thesolution of the problemgoverning differentialequation, provided theelements contain:
1) all required rigidbody modes
2) all required constantstrain states
~ compatibleLW layout
CD incompatiblelayout
~t:=
no. of elements
If an incompatible elementlayout is used, then in additionevery patch of elements mustbe able to represent the constantstrain states. Then we haveconvergence but non-monotonicconvergence.
4·16
Geuralized coordinate finite e1eJDeDt models
7 "
/ " 'r->
,;
/
(
""1-- - --
,IIII
III
iI
(a) Rigid body modes of a planestress element
......~_Q
II
II
(b) Analysis to illustrate the rigidbody mode condition
Table 5.2. Sampling points andweights in Gauss-Legendre numeri-cal integration.
Now let,
ri be a sampling point and
eli be the corresponding weight
for the interval -1 to +1.
Then the actual samplingpoint and weight for theinterval a to bare
a + b + b - a r. and b - a el.-2- 2 1 2 I
and the ri and eli can betabulated as in Table 5.2.
8·9
Numerical integrations, modeling considerations
In two- and three-dimensional analysiswe use
+1 +1f f F(r,s) dr ds =I: "1-1 -1 1
or
+1f F(ri's) ds-1
(5.131)
+1 +1f f F(r,s)drds= I: ,,;,,/(ri'sj)-1 -1 i ,j
(5.132 )
and corresponding to (5.113),a·IJ• = a. a. , where a. and a.
I J I Jare the integration weights forone-dimensional integration.Similarly,
+1 +1 +1f f 1F(r,s,t}drdsdt-1 -1 -1
= ~a.·a.·a.kF(r.,s.,tk)LJ 1 J 1 Ji,j,k
(5.133 )and a··k = a. Q. Qk .IJ I J
8·10
Numerical integrations, modeling considerations
Practical use of numerical integration
.The integration order required toevaluate a specific element matrixexactly can be evaluated by studying the function f to be integrated.
• In practice, the integration isfrequently not performed exactly,but the· integration order must behigh enough.
Considering the evaluation of theelement matrices, we note thefollowing requirements:
a) stiffness matrix evaluation:
(1) the element matrix doesnot contain any spurious zeroenergy modes (i.e., the rank ofthe element stiffness matrix isnot smaller than evaluatedexactly) ; and
(2) the element contains therequired constant strain states.
b) mass matrix evaluation:
the total element mass must beincluded.
c) force vector evaluations:
the total loads must be included.
8·11
Numerical integrations, modeling considerations
Demonstrative example
2x2 Gauss integration"absurd" results
3x3 Gauss integrationcorrect results
Fig. 5.46. 8 - node plane stresselement supported at B by aspring.
Stress calculations
(5.136)
• stresses can be calculated atany point of the element.
• stresses are, in general, discontinuous across elementboundaries.
8-12
Numerical integrations. modeling considerations
thickness = 1 cmA -p 3xl07 2
1~[E = N/cm
<3> e. CD \) = 0.3
I 1> p 300 N=c·
:... ..,- -of3c.m. 3 Coft'1.
A
8 ...
'100 N!Crrt'l./
(a) Cantilever subjected to bendingmoment and finite element solutions.
Fig.5.47. Predicted longitudinalstress distributions in analysis ofcantilever.
= a .
8·13
Numerical integratiODS. modeling coDSideratioDS
'?, A,
~ @ B <D 4l,, C.
I, s_ a~ "?
v = 0.3
P = lOON
" "8 & <DCo 174+ /lA/e-t'-
Co
A A
® B 8 <DCo c
~I".00 "'Ie-."
(b) Cantilever subjected to tip-shearforce and finite element solutions
Fig. 5.47. Predicted longitudinalstress distributions in analysis ofcantilever.
Some modeling considerations
We need
• a qualitative knowledge of theresponse to be predicted
• a thorough knowledge of theprinciples of mechanics andthe finite element proceduresavailable
• parabolic/undistorted elementsusually most effective
and we have obtained the 3x3unreduced matrix in (8.3)
SoIltiOl of finite elemelt eqlilihrilDl equations in static aualysis
5 -4 0 VI
:1-4 6 -4 U2
1 -4 6 -4 U3 :10 1 -4 5 U4
14 -!§ U2"5 5
-!§ 29 -4 U3 05 -5
-4 5 V4 0
Fig. 8.1 Physical systemsconsidered in the Gauss eliminationsolution of the simply supported beam.
9-7
Solutiou of finite element eqDilihriom equations in static analysis
SUBSTRUCTUR ING
• We use static condensation on theinternal degrees of freedom of asubstructure
• the result is a new stiffness matrixof the substructure involvingboundary degrees of freedom only
-?-?-~-o--o
e--c>---nl -6
50x50
Example
......--.- L
32x32
Fig. 8.3. Truss element withlinearly varying area.
We have for the element.
9·8
[
17
~~ -206L
3
-20
48
-28
SoIali. oIliDile e1emeal eqailihrilDl eqaaliODS ia stalic aaalysis
First rearrange the equations
EA, [ '76"L 3
-20Static condensation of U2 gives
EA, Ir76L 3
3] [-20]- [lJ[-2025 -28 48
or
ll. EA, [ 19 L -1
and
9·9
Solution of fiDile elemeul equilibrilll equati. in slatic aDalysis
Multi-level Substructuring
I' L 'I' L~ , L ,I. L .1A 2A 4A, I SA, I I 16A,
, , \ '~
-\&-o=2:E~f' · 'n-~ -U Ur;, U6 U7 Us UgI U2 U3 u. Rs
Bar with linearly varying area
-I I 1-
U, - u3u2
---I • .-U, u3
(a) First-level substructure
---I I I I 1-
U, - Usu3
_I I • I 1-
U, Us
(b) Second-level substructure
_I I I I I I I I 1-U, - ug
Us.Rr;,
-. I I I I I I I 1-U, ug
(c) Third-level substructure andactual structure.
Fig. 8.5. Analysis of bar usingsubstructuring.
'-10
Solution of fiDile e1eDleul equilihrio equti. ill static analysis
Frontal Solution
Elementq Element q + 1 Elementq + 2 Elementq + 3
--------m m+3
~I N:"
Element 1 Element 4
4
Wave front Wave frontfor node 1 for node 2
Fig.8.6. Frontal solution of planestress finite element idealization.
• The frontal solution consists ofsuccessive static condensation ofnodal degrees of freedom.
• Solution is performed in theorder of the element numbering .
• Same number of operations areperformed in the frontal solutionas in the skyline solution, if theelement numbering in the wavefront solution corresponds tothe nodal point numbering in theskyline solution.
9·11
Solution of finite element equilibrium equations in static analysis
L D LT FACTORIZATION
- is the basis of the skyline solu-tion (column reduction scheme)
- Basic Step
L- 1 K = K--1 - -1
Example:
5 -4 a 5 -4 a
4 -4 6 -4 a ~4 165 5 5
=1 a -4 6 -4 a _16 29 -4-5 5 5
a a a a -4 5 a -4 5
We note
4 4
-1- 5 -5L = 1 ~1 1-1 a a
5 S-
o a a a a a
9·12
Solution of finite element equilibrium equations in static analysis
Proceeding in the same way
-1 -11.2 1.1 K:= S
x x x x x
x x x x
S x ....... x upper:= triangular
x x matrix
x
x
Hence
or
Also, because ~ is symmetric
where
0:= di agona1 rna t r i x d .. := s ..11 11
9·13
Solution of finite eleJDent equilihriDII equations in static analysis
In the Cholesky factorization, we use
where
t = L D~
SOLUTION OF EQUATIONS
Using
9·14
K = L 0 LT
we have
L V = R
o LT U = V
where
-IV := L- -n-l
and
(8.16)
(8.17)
(8.18)
(8.19)
(8.20)
Solution of finite element equilibrimn equations in static analysis
COLUMN REDUCTION SCHEME
5 -4 1
6 -4 1
6 -4
5
~4 5 4
5 -55
14 -4 14 -4- 556 -4 6 -4
55
~5 4 1 5 4 1
-5 5 -5 5
14 8 1 14 85 7 5 -7
15 -415 -4T T
5 5
9·15
Solation of finite element eqailihriam eqaati. in static analysis
X=NONZERO ELEMENT0= ZERO ELEMENT
_~ COLUMN HEIGHTS
SYMMETRIC
o 0 000o 0 000
'-----,
X 000 Xo 0 000o 0 x 0 0o X 000X X X X 0
X X X XX XX
X
X
ELEMENTS IN ORIGINAL STIFFNESS MATRIX
Typical element pattern ina stiffness matrix
SKYLINE
o 0 000o 0 000
L...-_
X 0 0 0 XX 0 0 0 XX 0 X 0 XX X X 0 XX X X X X
X X X XX X X
X XX
ELEMENTS IN DECOMPOSED STIFFNESS MATRIX
Typical element pattern ina stiffness matrix
9-16
SYMMETRIC
Solution of finite element equilibrium equations in static analysis
Solution of finite element equilibrium equations in dynamic analysis
Modeling of a structural vibrationproblem
1) Identify the frequencies contained in the loading, using aFourier analysis if necessary.
2) Choose a finite element meshthat accurately represents allfrequencies up to about fourtimes the highest frequencyw contained in the loading.
u
3) Perform the direct integrationanalysis. The time step /':, t forthis solution should equal about120 Tu,where Tu = 2n/wu 'or be smaller for stability reasons.
Modeling of a wave propagationproblem
If we assume that the wave lengthis Lw ' the total time for thewave to travel past a point is
(9.100)
where c is the wave speed. Assumingthat n time steps are necessary torepresent the wave, we use
(9.101 )
and the "effective length" of afinite element should be
10·12
c /':,t (9. 102)
SoIaliOi .. filile 81••1eqailihriDl eqaali_ in dJUlDic ualysis
SUMMARY OF STEP-BY-STEP INTEGRATIONS
-INITIAL CALCULATIONS ---
1. Form linear stiffness matrix K,mass matrix M and dampingmatrix ~, whichever appl icable;
Calculate the following constants:
Newmark method: 0 > 0.50, ex. 2:. 0.25(0.5+0)2
2aO
= , / (aAt )
a4 = 0/ ex.- ,
as = -a3
a,=O/(aAt)
as = I1t(O/ex.- 2)/2
ag = I1t(' - 0)
a3 = , / (2ex. )- ,
a7 =-a2
Central difference method:
a, = '/2I1t
... 0 O· 0··2. Inltlahze !!., !!., !!. ;
For central difference methodonly, calculate I1tu frominitial conditions: -
3. Form effective linear coefficientmatrix;
in implicit time integration:
in explicit time integration:
M= a~ + a,f.
10·13
Solution of finite element equilibrium equations in dynamic analysis
4. In dynamic analysis usingimplicit time integrationtriangularize R:.
--- FOR EACH STEP ---
(j) Form effective load vector;
in implicit time integration:
in explicit time integration:
(ii) Solve for displacementincrements;
in implicit time integration:
in explicit time integration:
10·14
SoI.ti. of filile elOl.1 equilihriDl equations in dynamic analysis
Newmark Method:
Central Difference Method:
10·15
MODE SUPERPOSITIONANALYSIS; TIMEBISTORY
LECTURE 1148 MINUTES
11·1
Mode slperpClilion analysis; lillie bistory
LECTURE 11 Solution of dynamic response by modesuperposition
The basic idea of mode superposition
Derivation of decoupled equations
Solution with and without damping
Caughey and Rayleigh damping
Calculation of damping matrix for givendamping ratios
Selection of number of modal coordinates
Errors and use of static correction
Practical considerations
TEXTBOOK: Sections: 9.3.1. 9.3.2. 9.3.3
Examples: 9.6. 9.7. 9.8. 9.9. 9.10. 9.11
11·2
Mode superposition analysis; time history
Mode Superposition Analysis
Basic idea is:
transform dynamic equilibrium
equations into a more effectiveform for solution,
using
!L = 1:. !(t)nxl nxn nxl
P = transformation matrix
! (t ) =general ized displacements
Using
!L(t) = 1:. !(t)
on
MU+ c 0 + K U = R
we obtain
(9.30)
(9.1)
~ R(t) + f i(t) + R!(t) ~(t)
(9.31)where
C fT ~ f ;
R = PT R (9.32)
11·3
(9.34)
Mode sDperJMlilion ualysis; tiDle history
An effective transformation matrix fis established using the displacementsolutions of the free vibration equilibrium equations with dampingneglected,
M 0 + K U = 0
Using
we obtain the generalized eigenproblem,
(9.36)
with the n eigensolutions (w~, p..,) ,2 2
( ul2 ' ~) , ... , (wn ' .P.n) , and
11·4
T 1== 0'<P 1" M'" "- _.:t:..J
i = j
i ., j
2< W- n
(9.37)
(9.38)
Mode superposition analysis; time history
Defining
(9.39)we can write
and have
(9.40)
Now using
!L(t) = ! ~Jt)
¢T M¢ = I (9.41)
(9.42)
we obtain equilibrium equationsthat correspond to the modalgeneralized displacements
!(t) + !T ~! !(t) + r;i ~(t) = !T !S.(t)
(9.43)
The initial conditions on ~(t) areobtained using (9.42) and theM - orthonormality of ¢; i.e.,at time 0 we have
(9.44)
11·5
Mode SUperpClitiOD aualysis; tilDe bislory
Analysis with Damping Neglected
(9.45)
i.e., n individual equations ofthe form
2.x .(t) + w. x. (t) = r. (t )1 1 1 1
where
with
T aX'I = lj). M U1 -1 - -t=O
• .T O'X'I =-'-- cp.M U1 -1 - -t=O
i = ',2, ... ,n
(9.46)
(9.47)
Using the Duhamel integral we have
=-' jtr1·(T) sinw.(t-T)dTw. 1
1 0 (9.48)
+ a.. sin w.t + 8. cos w·t111 1
where a.i and 8i are determined
from the initial conditions in (9.47).
And then
11-&
(9.49)
Mode sDperp.ition analysis; time history
4f----..-----:--..--r----,..----~---_r_---...,
equation•• 2 . 2 .x + E;,wx + W X = S 1n P t
staticresponse
~= \-0
31-__-+__+--+-+-__+-__-+ -+-__--.,
0 2....0.....uCtI.....-0CtI0
u
ECtIr:::::>-
0
2 3
Fig. 9.4. The dynamic load factor
Hence we use
uP =~¢. x· (t)-- ~--l 1
i =1
where
uP - U
The error can be measured using
(9.50)
11·7
Mode superposition analysis; time history
Static correction
Assume that we used pmodes to obtain ~p , then let
n
~_=LriUl~)i =1
Hence
Tr. = ¢. R1 -1-
Then
and
K flU fiR
Analysis with Damping Included
Recall, we have
!(t) + !T f!i(t) + fi !(t) = !T ~(t)
(9.43)
If the damping is proportional
T¢. C (po = 2w. E;,. cS. .-1 ---J 1 1 1J
and we have
(9.51)
x.(t) + 2w. E;,. x.(t) + w~ x.(t) = r1·(t)
1 1 1 1 1 1
i=l, ... ,n
(9.52)
11·8
Mode superposition analysis; time history
A damping matrix that satisfies therelation in (9.51) is obtained usingthe Caughey series,
(9.56)
where the coefficients ak ' k = , , ••• , p ,are calculated from the p simultane-ous equations
A special case is Rayleigh damping,
C = a ~1 + B K- -- --
example:
Assume ~, = 0.02
w, = 2
calculate a and B
We use
(9.55)
or
'/a + Bw:- 2w. ~.- - 1 1 1
2w. ~.1 1
11·9
Mode superposition analysis; time history
Using this relation for wl ' [,1 andw2 ' [,2 ' we obtain two equations
for a and 13:
a + 4ii = 0.08
a + 913 = 0.60
The solution is a = -0.336and 13 = O. 104 . Thus thedamping matrix to be used is
C = -0.336 M + 0.104 K
Note that since
2a + 13 w. = 2w. [, .1 1 1
for any i, we have, once a and13 have been established,
E,. =1
2a + SW.
1
2w.1
a 13= - + - w2w. 2 i
1
11·10
Mode sDperp.ition analysis; time history
Response solution
As in the case of no damping.we solve P equations
x. + 2w. E,. x· + w~ x. = r.1 111111
with
r·1
ITO
xi t = 0 "--. !i!i .!:L
• ITO'xi t = 0 = !i f1 .!:L
and then
PuP ~¢. x. (t)LJ-1 1
i =1
Practical considerations
mode superposition analysisis effective
- when the response lies in afew modes only, P« n
- when the response is to beobtained over many time intervals (or the modal responsecan be obtained in closed form).
e.g. earthquake engineeringvibration excitation
- it may be important tocalculate E p(t) or thestatic correction.