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7/29/2019 C9.DIstributed Mass and Elasticity2 Compatibility Mode
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DYNAMIC OF STRUCTURES
CHAPTER 9
SYSTEMS WITH DISTRIBUTED MASS
AND ELASTICITY
Department of civil engineering, University of North Sumatera
Ir. DANIEL RUMBI TERUNA, MT;IP-U
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
We return to the partial diffential equation of simply beamsubjected to force
The displacement is given by a linier combination of the modes:
( )( )
( )( )
( ) )1(,,,
2
2
2
2
2
2
txpt
txuxm
x
txuxEI
x=
+
Subtituting Eq.(2) in Eq.(1) gives
( )txp ,
( ) ( ) ( ) )2(,1
tqxtxu rr
r
=
=
( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) )3(,11
txptqxxEItqxxm rr
rr
r
r =+
=
=
&&
Multiply each term by , integrate it over the length of the beam gives( )xn
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) )4(,01 01 0
dxxtxpdxxxEIxtqdxxxxmtqL
nr
r
L
nrr
r
L
nr =+
=
=
&&
This equation can be written as
nr =By use orthogonality properties of modes, all term left side vanish exceptthe one term for which . Thus, Eq.(4) becomes
( ) ( ) ( )[ ] ( ) ( ) ( ) ( )[ ] ( ) ( ) )5(,00
2
0
=+
L
nn
L
nn
L
nn dxxtxpdxxxEIxtqdxxxmtq &&
( ) ( ) ( ) )6(tptqKtqM nnnnn =+&&where
( ) ( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] dxxxEIxdxxxEIxKdxxxmM nL
nn
L
nn
L
nn
2
00
2
0
, ===
( ) ( ) ( ) )7(,0
=L
nn dxxtxptP
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
Equation (6) governs the modal coordinate , and the generalized
properties , , and depend only on the mode .Thus we have infinite number of equation like Eq.(6), one for each mode. Thepartial diffrential equation equation (1) in the unknown fungtion hasbeen transformed to an infinite set of ordinary diffrential equation (6) inunknowns .
Once the have been determined, the contribution of the mode to the
displacement is given by:
The generalized mass and generalized stiffness for modeare related
( )txu ,
nM nKthn
)8(2 nnn MK =
( )tqnthnnM nK tPn
th
n ( )xn
tqntqn
( ) ( ) ( ) )9(, tqxtxu nnn =
thn
txu ,
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
( ) ( ) ( ) ( ) )10(,,11
tqxtxutxu nn
n
n
n
=
=
==
( )xu
( ) ( ) ( ) ( ) ( ) ( )[ ] )11(== xuxEIxVxuxEIxM
4. Specialize
The bending moment and shear force at any section along the length of thebeam are related to displacement as follows:
The static relationships apply at each instant of time with replaced by
which is given by Eq.(10). Thus
xu
( )txu ,
( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) )12(,,11
=
=
==n
nnn
nn tqxxEItxVtqxxEItxM
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EXAMPLE
( ) )(sin2
22
aL
xnx
m
EI
L
nnn
==
0p
L
EIm,
x
0p
t
SOLUTION
1. Determined the natural frequencies and modes
2. Determined the generalized mass, stiffness and force
( ) ( )[ ] )(2
sin0
2
0
bmL
dxL
xnmdxxxmM
LL
nn =
==
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EXAMPLE
( ) ( ) ( )[ ] )(2
sin3
442
0
4
44
0
cL
EIndx
L
xn
L
nEIdxxxEIxK
L
n
L
nn
=
=
=
( ) ( ) ( ) )(, 00
epdxxtxpP n
L
nn ===
3. Set up modal equation
The n th modal equation is
( ) ( )( )
n
nnnn
nnnnn
M
ptqtqor
ftptqKtqM
0
2
)(
=+
=+
&&
&&
)(2
3
442 d
L
EInMKor nnn
==
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EXAMPLE
( ) ( ) ( ) )(sincos 0 gK
ptBtAtq
n
nnnnnn
++=
( ) 00 =nq
4. Solved modal equation
If at , and , we get0=t00
=nq&
)(00 hBK
pA n
n
nn =
=
Subtituting Eq.(h) in Eq.(g) lead to
( ) ( ) ( ) ( ) )(2cos1 443
00 inEI
LptK
ptq nnn
nn
==
Subtituting Eq.(i) in Eq.(10) , we obtained the displacement response txu ,
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
( )( )
( ) )(sincos12/2
,1
44
3
0 hL
xnt
n
L
EI
Lptxu
nn
n
=
=
4. Specialiaze for
where
)(
,...11,7,31
,...9,5,11
,...6,4,20
2j
n
n
nL
n
=
=
=
=
2/L=
Subtituting Eq. (j) in Eq.(h) gives
( )
)(...........7
sin2401
cos15sin
625
cos1
3sin
81
cos1sin
1
cos12,
75
31
4
3
0
kL
xt
L
xt
L
xt
L
xt
EI
Lptxu
+
+
=
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
Displacement at mid span is
The displacement for the first mode contribution only can be expressed as
)(...........2401
cos1
625
cos1
3sin
81
cos1
1
cos12,
2
75
31
4
3
0
ltt
L
xtt
EI
Lpt
Lu
+
+
+
+
=
( ) )(cos16.48
,2
1
3
0 mtEI
Lpt
Lu =
dynamic displacementamplication factor
Staticdisplacement
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
( )
)(...........7sin49cos15sin
25cos1
3sin
9
cos1sin
1
cos12,
75
31
2
0
kL
xtL
xt
L
xt
L
xtLptxM
+
+
=
The bending moment are obtained by subtituting Eq.(h) in Eq.(12)
The bending moment at mid span is
)(...........49
cos1
25
cos1
9
cos1
1
cos12
,2
75
31
2
3
0
ott
s
ttLp
t
L
M
+
+
+
+
=
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MODAL ANALYSIS OF FORCED DYNAMIC RESPONSE
The bending moment for the first mode contribution only can be expressed as
( ) )(cos193.4
,2
10 pt
EI
Lpt
LM =
Moment dynamicamplication factor
Static bendingmoment
Solution for by use Duhamel integral
( ) ( ) ( ) ( ) ( )
dtpm
dtpm
tqn
t
nn
n
t
nn
==
sin
1sin
1
00
0
tqn
( ) ( )[ ] ( )tn
pt
m
ptq n
nt
n
n
nn
cos1
2cos
44
0
02
0 ==