[SOLVED] C algorithm math matlab theory 28 3 Vol.28 No.3 2015 6 Journal of Vibration Enginering Jun.2015

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28 3 Vol.28 No.3 2015 6 Journal of Vibration Enginering Jun.2015
– *
1 1 1 2 (1. 116024;
2. 116024)
: ;
:;;;
: O 3 1 3 .7 : A : 1 0 0 4 -4 5 2 3 (2 0 1 5 )0 3 -0 3 3 7 -0 8
DOI:10.16385/j.cnki.isn.1004-4523.2015.03.001

[1-12]
Kane[1] : [2-6] ;
* :2013-12-26; :2014-06-12
: (11372057)
[7 ] [8] Shi P [9]
[10] [11]

338 28
[12] () [1 3 -1 6 ]
1 1 .1
()=d()/ds
1
1 pe = (EA+EI)ds (6)
{g 1 g 2 } {e1e2}e1 0 {t 1t 2 }
1
Fig.1A plane beam element undergoes large range of
motion 1s

12
= 1 + 1 .2
(4 )

F = EA;M = EIz (5) F ;M ;EA EIz
l
0

() s =N
()
T ;
=[1 2]N=[0 1-]8
( ) ()
= s/l (7 )
= N 9 (7 ) (9 )
7
(1 ) r1 s ds
(6 )
pe = ()TK (10)
K
EAl 0 0
EIl/3EIl/6 (11) //
r = r 1 +
K = 0 0

: = d s / d s 1 ; d s

(7)
d/ds= (1+)t1 (2)
x=e1=(1+)cos
(3) [-]
EIl 6EIl 3 1 .3

(13) w=e2 = 1+sin
N= 0 (2 2)l/2 2l/2 {()
s
s= Nds=N ()
0
( )
12

0l
[]
: = T ; g (A B ) = A T B + B T A ;
3 :- 339
(12)
= 1 + N (1 4 ) (3 ) s
s
( ) x= 1+ cosds
0
s(15) ()
^^
H=N1 N2 N2 0 0 (21)
l2 -1N2=2 3-2 () ()
cos 1 + 0 sin 1 + 0
A= () ()(23)
ss
ds- (w x )ds
10

18 s =sinc =cos=1+ 0:x0 =0w0 = 00 =0;(14)(15) x w

1
/2
s
0l ()
=(x w)T =Hl (ll)
(19) l=xlcwls (20)

T

() ()

Dl= (l
0 0 0 N2 ^N2 ^2^
N l1 N
1 =( ) 2 =
s
r(s) r A (22)
[ ]
sT +cNl ll
013 T()

0
w=1+ sinds
:
0 :sin cos 1 2 / 2 (3 )
////() w x = (w s ) (x s )= tan 16
[]
(15) x = s +

I= 0 -1

=1+
sin 1 + 0 cos 1 + 0 () ()
0
(22)

r(s)=r1+ 1AI+A (24)
02 w = w/x ds
0
2
( )
(17)
() rs=r1+1AI+A-1A+21AI 25
0
()
(19)

(12) s (1 5 )
(15) xw
x(0)=x0x(0)=c ;x(l)=xl =x(l)=c
=Hl=HDl;=HDl+HDl (26)
T ClT -Snl
cT sN
– l ;
Dl =
SlT +Cnl
ll
-(gSnlT +Cnl) TT
– s (N T )+c N N
g l l l l) (27)
T() (gCnlT -Snl)
T() T
(cg Nl T -sNlNl)
ll
T = 1 0 0
() () ;() () w0=w0w0=s wl=wlwl=s
lll
; ; TSl = sdsSnl = sNdsSnl = sN Nds
000
lll
Cl = cdsCnl = cNdsCnl = cN Nds
(28)
(26)
() () rs =r1 +AIHl 1 +AHDl 29

() () rs =r1+AIHl1+AHDl 30
000 ;;T
(26) (25)
rs =r1+ 1AIHl+AHDl+Aar 31

340 28

() ar =HDl- 21Hl+21IHDl 32

1 1
NH = Hd;NHH = HTHd;

(14)

0
0
l T
() () pa = [r r+Iz]ds 39
Iz
fqg= I mg (50)
(30)~ (34)
AINHl
(AN D )Tm (51)
1T q1 q q g f=Hlg
p a = (q ) (M q + M + F ) +
[13]
T q1

( ) (M q + M + F )
Mr Mr M q q = [
l
Mq = [
Fq = ()

r M M
0

Mr
;Mq = (Mq)T (41) ]
(10)
M Fr
(42)
pd = ()Tfd (53)
F
fd =K (54) 2
2 .1
r ; T
M = mI M = m l N H H l + mIz
(40 )
; p d = E A + E I z d s 5 2
()() ]
11
;T;() NN = NdNNN = NNd 45

1 .5

; () = 1 + N = 1 + N 33
0
1
0

() = 1 + N 34
r q
r1 r2
12 2

TT NHIH = HIHd
0

ll
IAIHll 1 .4
() m1 +m2 +l AIHl f2 +lp
B =

q = q = (35)
l
( )T ( )T ()
1 2
(14)(29)(31)(33)(34)
pf =[ri fi+imi]+ r psds i=1 0

; ;
q2 =q1+q2 =Bq1+Qq2 =Bq1+Q+
(46) p(s);fimi
(30)(33)
AHll Aalr
==(37)
l
f1+f2+ pds
N l
0
fq =
T T()T T ()
(36)
pf =(q1)Tfq+()Tf (47)
[]

AHlDl [0 1 ] [Nl ]

0
( ) ; ( ) p = AIH Tpdsp = AH Tpds
;Q =
(38)
TTT
(48) f=m2Nl+DlAHl f2+Dlp 49
0

[ ] ()
0
T
0
; r;r ;

T
M = m D l N H H D l + m I z N N N (4 3 )
M = mAINHl M = mANHDl
T
M = mlNHIHDl +mIzNN .
r2
( )

( )
F = m A N H D l N H l 1 + 2 I N H D l 1

T( 2 ) F = m l N H I H D l N H I H l 1 + 2 N H H D l 1

F = m D lT N H H D l N H H l 21 + 2 N TH I H D l 1

( ) (44)
36

3 :- 341
n

(10) (40)
(47) (55)

1
1 q0
1T 001 0 0 ()
:
(1) ;
(2) L
:L 1 n L (i ) i 1; 0
2
2 Fig.2 Regularly labeling of beam elements
:L = [0 1 2 ]
L
(3 6 ) i

; ;
q i1 = q 10 + i q i1 = iq 10 + i q i1 = iq 10 + i + i
q0 Mq0+M+F+ ()
T 01 () ()
(M q0 + M + F )= 0 59
M
i Mi i
n
00 Tq

=
M = i Mii+MiTi M = M
i=1 n
q ; 0 ()
( 0 )T [Tq q T(q
0
T q q
i=1 n
M = i Mii +MiTi +Ti ()
Mii +
Mi Ti
i=1
)]
(60)
F=i Mii+Fi-fi + [()
n
0 Tqqq
F= i Mii+Fi-fi

() n()
i=1
Tqqq 61
i=1
Tq
T i ( M i i + F i + K i i f i ) ]
(59)
M00 M0q10 Fi0
()

(55)
3
3 [10-1114-15]
3 .1
M L = 10 m u v 3

5

=0.1
4
i =L(i)+L(i);i =BL(i)L(i)+QL(i)TL(i) {;
i =BL(i)L(i)i =BL(i)L(i)+L(i)
: T T T T ; =1 2 n
() Ti i:i =Ti ;

0 =0;0 =0;0 =0;0 =0; {;;;;
(57) 0 =IB0 =I0 =0Q0 =0T0 =0
2 .2
:
n nnn
pia + pie = pif i=1 i=1 i=1
(58)
(56)
[ + =0 62
]
0
M M Fi
q10

342
28 : L = 10 m A = 1 m 2
3 Fig.3 Cantilever beam undergoes bending load
I =510-4 m4E=2.8107 Pa z
= 1 .2 k g / m 3
6 Fig.6 A flexible beam on a rotating base
(t ) 6 t2 15 2 2t
cos 10 t 15 s [(2)(15 )]
+
(t)=152
6t 45t > 15 s 5
(63) 30s7
4 Fig.4 Large deformation of the cantilever under bending
moments
M =2EIz/L
5
5
Fig.5The tip displacements of the cantilever with an
end bending load
5
3 .2
[14-15] 6[14]
7 Fig.7 Displacement curves at the fre end
7: 5 .1 4 1 0 4 m 0 [14-15]

3 :- 343

3 .3
8
8 Fig.8 The flexible pendulum
[10 -11 ]
CPU [10]
4
:
(1) ;
(2) ;
(3)
:
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139151. [2] .
[J]. 200436(1):118124. YANG HuiHONG JiazhenYU Zhengyue.Experi- mental investigation on dynamic stifening phenomenon [J].Acta Mechanica Sinica200436(1):119124.
[3] . [J]. 200335(2):253256. YANG HuiHONG Jia-zhenYU Zheng-yue.Experi- ment validation of modeling theory for rigid-flexible
coupling systems[J].Acta Mechanica Sinica200335 (2):253256.
[4] Cai G PHong J ZYang S X.Dynamic analysis of a flexible hub-beam system with tip mas[J].Mechanics
Research Communications200532(2):173190. [5] Cai G PLim C W.Dynamics studies of a flexible hub- beam system with significant damping efect[J].Jour-
nal of Sound and Vibration2008318(1):117. [6] .-
[1 0 ] [1 1 ] L = 1 .8 m E = 6 .895 109
P a A = 2 .5 c m 2 I z = 0 .1 3 c m 4 = 27 6 6 .6 7 kg/m3
5 [10] Matlab ode23tb : AbsT=0.1 RelT=
0 .0 1 2 .5 s C P U 1
1 CPU Tab.1 The CPU time comparison betwen absolute cordinate
method and the proposed method

C P U 9
[10 -11 ]
9 Fig.9 The tip deformation of the flexible pendulum
[10] 2 2 5 .0 5 s

6 .2 9 s

344 28
[J]. 200215(2):194198. Liu Jinyang Hong Jiazhen.Study on rigid-flexible coupling dynamic behaviour of flexible beam[J].Jour-
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[J]. 201362(4):305311. Fang JianshiZhang Dingguo.Analyses of rigid-flexi- ble coupling dynamic properties of a rotating internal cantilever beam[J].Acta Physica Sinica201362(4): 305311.
[8] .- [J]. 200628(6):
800804. Deng FengyanHe XingsuoYang Yongfenget al. Dynamics modeling for a rigid-flexible coupling system with onlinear deformation field[J].Journal of Mechan-
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[J]. 200539(5):827831. Li BinLiu Jinyang.Application of absolute nodal co- ordination formulation in flexible beams with large de- formation[J].Journal of Shanghai Jiaotong Universi-
ty200539(5):827831. [11].
[J]. 201345 (2):251256.
Chen SijiaZhang DingguoHong Jiazhen.A high-or- der rigid-flexible coupling model of a rotating flexible beam under lagre deformation [J]. Acta Mechanica Sinica201345(2):251256.
[12]. [J]. 2013 34(6):620629.
Zhang ZhigangQi ZhaohuiWu Zhigang.A large de- formation beam element based on curvature interpola- tion[J].Applied Mathematics and Mechanics2013 34(6):620629.
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ear Dynamics201267(3):1 8251 835.
Rigid-flexible dynamics analysis of a large deformation beam element based on interpolation of strains
ZHANG Zhi-gang1 QI Zhao-hui1 WU Zhi-gang12 (1.State Key Laboratory of Structural Analysis for Industrial EquipmentDalian University of Technology
Dalian 116024China; 2.School of Aeronautics and AstronauticsDalian University of TechnologyDalian 116024China)
Abstract:The dynamic stifeningphenomenon in flexible multi-body system dynamics is due to the deformation coupling. The first-order approximation model has ben sucesfuly applied in the rigid-flexible coupling modeling of smal deformation. Howeverit is found necesary to consider more deformation coupling efects in lager deformation cases.In this paperthe beam bending curvature and axial strain are selected as the element parameters.Then the recursion formulations are obtained
for the kinematic parameters of two end nodes of an element based on theories of large deformation and finite rotation.A planar beam element used for large deformation rigid-flexible dynamics analysis is proposedwhich can automaticaly take into acount the dynamic stif fening terms.Final lythe validity and ef fectivenes s of the proposed algorithm are verified through some nu- merical examples which involve the large deformations and rigid-flexible dynamics of beams.
Key words:rigid-flexible coupling;beam element;strain interpolation;dynamic stifening
: (1984 ) :15326175369;E-mail:[email protected] : (1964 ) E-mail:[email protected]

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[SOLVED] C algorithm math matlab theory 28 3 Vol.28 No.3 2015 6 Journal of Vibration Enginering Jun.2015
$25