BG_DONG_LUC_HOC

8/3/2019 BG_DONG_LUC_HOC

1/146

TRNG I HC BCH KHOA - HQG HCMKHOA K THUT XY DNGB MN SC BN KT CU

PGS.TS. KIN QUC

BI GING MN HC

NG LC HC KT CU

DYNAMICS OF STRUCTURES

Ti liu tham kho

1.Clough R. W., Penzien J.,Dynamics of Structures, McGraw-Hill, 1993 (1975).2.Chopra A. K., Dynamics of Structures, Prentice-Hall, 2001, (1995).3.Buchhold H., Structural Dynamics for Engineer, Thomas Telford, 1997.4.Geradin M., Mechnical vibrations and Structural dynamics, Belgian, 1993.5.Rao S. S., Mechnical Vibrations,Addison-Wesley Publishing Company, 1990.


2/146

CHNG 1. MU

1.1 NHIM V MN HCng lc hc kt cu l mt lnh vc ca c

hc, nghin cu cc phng php phn tch phnng (ni lc, ng sut hoc chuyn v, vn tc,

gia tc) trong kt cu khi chu tc dng ca ccnguyn nhn ng.

1.2 TI TRNG NG

Khi nim:

Ti trng ng l ti trng thay i theo thigian v tr s, phng, v tr, gy ra ng sut,chuyn v cng thay i theo thi gian.

Phn loi:- Ti trng tin nh (Deterministic Loads): l

ti trng bit trc c qui lut bin i theothi gian P = P(t). Th d: Ti trng iu ha,chu k, khng chu k, xungc m t theoqui lut cho trc.

- Ti trng ngu nhin ( Random, Stochastic

Loads): l ti trng bit trc c qui lut xcsut v cc c trng xc sut nh gi tr trung


3/146

bnh, lch chun Th d: ti trng gi, sngbin, lc ng t.

Bi ton LHKC chu ti trng ngu nhin

c gii quyt bng l thuyt dao ng ngunhin (Random Vibration Theory). Cc thng tincn tm bao gm ng sut, chuyn v, cng mangtnh ngu nhin vi cc c trng xc sut gi trtrung bnh, lch chun

Ni chung, cc ti trng trong thc t umang tnh cht ngu nhin mc khc nhau,v c xc nh bng phng php thng kton hc.

Cc quan im phn tch ng lc hc:Phn tch tin nh, phn tch ngu nhin v

phn tch m(Fuzzy Analysis).

1.3 C TH CA BI TON NG

Bi ton tnh: ni lcc xc nh t s cnbng vi ngoi lc, khngcn dng ng n hi nnmang tnh cht n gin.ng sut v chuyn vkhng ph thuc thi gian.

Tnh

ng

q(t)= r y(t)

P(t)

P


4/146

Bi ton ng: ngoi lc bao gm lc quntnh ph thuc vo ng n hi y = y(x,t). Vvy, dn ti phng trnh vi phn, phc tp vton hc, khi lng tnh ln, phi bt u t vicxc nhy(x,t).

Nhn xt:Bi ton tnh (bao gm c bi ton n nh) l

trng hp c bit ca bi ton ng khi lc

qun tnh c b qua.

1.4 BC TDO CA KT CUBc tdo ng lc hc (Number of dynamics

degrees of freedom) ca kt cu l sthnh phn

chuyn v phi xt th hin c nh hngca tt c cc lc qun tnh.

Bc t do c nh ngha trong s lin quann lc qun tnh v do lin quan n khi

lng. S khi lng cng nhiu th cng chnhxc nhng cng cng phc tp.

Ch : Bc t do ng lc hc khc vi bc tdo trong bi ton tnh (s chuyn v nt ca ktcu).


5/146

Th d: cho kt cu nh hnh bn, nu P l ti trng tnh th sbc t do l 3, nu P l ti trng

ng th s bc t do l v cng.Trong thc t, cc kt cu u c khi lng

phn b nn c v hn bc t do, vic gii biton rt phc tp nn tm cch ri rc ha h.

1.5 CC PHNG PHP RI RC HA

1.5.1 Phng php khi lng thu gn (LumpedMass)

Thay th h c khi lng phn b (a) thnhcc khi lng t p trung (b) theo nguyn tctng ng tnh hc. y l phng phpthng c dng trong h kt cu phc tp.Khi lng thng c thu gn vim nt (thd nh h dn).

P(t)

m(z)

P(t)

m m m1 2 3

(a)

(b)

P


6/146

S bc t do ca h ty thuc vo gi thit vtnh cht chuyn v ca h v tnh cht qun tnhca cc khi lng mi. Chng hn, xt h (b) l h

phng:Nu bin dng dc trc v mi c qun tnh

xoay: 9 BTD (3BTD/mass).

Nu coi mi l mt im (khng c qun tnhxoay): 6 BTD (2 chuyn v thng/mass).

B qua bin dng dc trc nn ch c chuynvng: 3 BTD (1 chuyn vng/mass).Ch : phc tp ca bi ton ng lc hc

ph thuc vo s bc t do.

1.5.2 Phng php dng ta suy rng

(Generalised Coordinates)Gi sng n hi l t hp tuyn tnh ca

cc hm xc nh i(x) c bin Zi nh sau:

=

=1

)()(),(i

ii xtZtxy

(*)trong : i(x) : Hm dng(Shape Functions)

Zi(t): Ta suy rng(Generalised Coordinates)

Hm dng i(x) ctm t vic gii phng trnh

L

Z2

Z3

y(x,t)

1(x)

Z1

3(x)

2(x)

( ) sin 1,2,...,ii x

i nL

= =


7/146

vi phn o hm ring, hoc do gi thit ph hpvi iu kin bin. Khi tnh ton thng gi limt s s hng u tin ca chui (*) v h trthnh hu hn bc t do (Zi ng vai tr bc tdo).

1.5.3 Phng php phn thu hn(Finite Element Method - FEM)

y l trng hp c bit ca phng phpta suy rng, trong :- Zi l cc chuyn v nt (Ta suy rng).- i(x) l cc hm ni suy (Interpolation

Functions) cc phn t - Hm dng.

Thng cchm ni suy i(x)c chn gingnhau cho cc phnt (ng vi cngmt bc t do) v lhm a thc nnvic tnh ton c

n gin. c bit,do tnh cht cc b ca cc hm ni suy nn cc

321 4 5

v3=1

3v(c)3v(b)

a b c d

3=1

3(c)3(b)


8/146

phng trnh t lin kt (uncoupled) vi nhau lmgim nhiu khi lng tnh ton.

1.6 CC PHNG PHP THIT LPPHNG TRNH VI PHN CHUYNNG1.6.1 Nguyn l DAlembert

Xt khi lng mi (i=1,n) chu tc ng calc Pi(t) c chuyn vvi(t) v gia tc )(tvi . Nut thm lc qun tnh th khi lng mi s cn

bng:

0)()( = tvmtP iii

(1.1)

Nu h c n bc t do th s c n phng trnh

vi phn chuyn ng.

1.6.2 Nguyn l cng kh dCho khi lng mi (i=1,n) mt chuyn v kh

dvi , cng kh d Wca cc lc tc dng lnmi (cn bng) trn chuyn vvi phi trit tiu:

= 0)]()([ iiii vtvmtP

(1.2)Nguyn l cng kh d thch hp cho h phc

tp gm cc khi lng im v khi lng cqun tnh xoay. Cc s hng trong phng trnh

l cc v hng (scalar) nn lp phng trnh ngin so vi phng trnh vector.


9/146

Nu cho h cc chuyn v kh d iv ln lttheo cc bc t do s thu c n phng trnh vi

phn ca chuyn ng.K hiu cng kh dca ngoi lcPi(t) l W,

t (1.2) ta c bin phn cng kh d: == iiii vtvmvtPW )]()( (1.3)

1.6.3 Nguyn l Hamilton (page 344, [1])Xt h gm cc khi lng mi (i=1, n) c cc

chuyn vvi(t)hai thi im t1v t2, chuyn vc cc tr s vi(t1) v vi(t2) tng ng vi haing bin dng (b) v (c). ng bin dng (d)ng vi t = t1 + t < t2. ng bin dng thttun theo nh lut II Newton. ng lch trngvi ng tht ti hai thi im t1 v t2:

v1(t1) =v1(t2) =0 (1.4)ng nng ca h ti thi im t:

)(2

1

1

2

i

n

iii vTvmT

== = Bin phn ca ng nng T tng ng vi

bin phn ca chuyn vvi:

T=1 1

n ni

i i i i i i i i ii i i ii

dvT dv m v v m v m v v

V dt dt

= =

= =

(1.5)


10/146

Mt khc, ta c ng nht thc:

( )i i i i i id d

v v v v v vdt dt

= +

Nhn c hai v vimi v ly tng cho

ton h:

+=i

iiii

iiii

ii v

dt

dvmvvmvvm

dt

d )(

WTvvmdt

di

iii += )( (1.6)

Nhn hai v vi dtv ly tch phn tt1n t2:

+=2

1

2

1

)(t

t

t

tiiidtWTvvm

Theo trn v vi(t1) = vi(t2) = 0 vi mi i nn vtri trit tiu:

0)(2

1

= +t

t

dtWT (1.7)

Nu ngoi lc tc dng trn h gm lc boton (lc th) v lc khng bo ton (th d lc

m 1 m 2 m 3 m 4

vv v1 2 3v 4

v (t )11

v (t )1 2

t=t1

t=t2

t=t +t < t1 2

v(t +Dt)11

d v1

2d v

3dv 4dv

tht

(a)

(b)

(c)

(d)

1

1t t2t+t1

1 1v(t +t) v (t)1 21v (t )1

v (t)1

t

dv (t +t)1 1

ng lchv(t)+dv1 1

ng Newton (tht)


11/146

ma st) th bin phn ca cng ngoi lc Wctch ra hai thnh phn:

W = Wc + Wnc (1.8)i vi lc bo ton th cng ca lc bng

gim th nng ca h nn:Wc = -V (1.9)

vi Vl bin phn ca th nng.Th (1.9) vo (1.8):

W = -V + Wnc (1.10)Th vo (1.7):

0)(2

1

2

1

= +t

t

t

tnc dtWdtVT (1.11)

y l nguyn l bin phn ca Hamilton, trong

:T: ng nng ca h.V: Th nng ca h, gm th nng bin dng

n hi v th nng ca lc bo ton.Wnc: Cng ca lc khng bo ton (lc cn,

ma st, ngoi lc...)

nghaCng thc (1.7) c vit li:

0)(2

1

= +t

t

dtWT (1.12)

Nh vy, trong tt c cc ng chuyn ngtrong khong thi gian t t1n t2 th ng lm


12/146

cho tch phn 0)(2

1

= +t

t

dtWT c gi tr dng (cc

tiu) l ng chuyn ng tun theo nh lutNewton.

Bi ton tnh T = 0 th (1.7) trthnh:

02

1

=t

t

Wdt suy ra 0=W hay 0)( = ncWV (1.13)

y l nguyn l th nng cc tiu trong biton tnh (Nu mt h cn bng n nh th thnng ca h cc tiu).

Ch : Nguyn l Hamilton cng l mtphng php nng lng, trong khng dngtrc tip n lc qun tnh v lc bo ton. Dngthch hp cho h phc tp, khi lng phn b.

Nhn xt: C 3 phng php DAlembert,Virtual Work v Hamilton u dn n phngtrnh chuyn ng ging nhau (u cng mang

bn cht nh lut II Newton).

Phng trnh Lagrange

Gi q1, q2,...., qn l cc ta suy rng. Trong

cng thc (1.11) ta c: ),....,,,,....,,( 2121 nn qqqqqqTT =


13/146

),....,,( 21 nqqqVV =

1 1 2 2....nc n nW Q q Q q Q q = + + +

vi Qi l lc suy rng khng bo ton.Th vo (1.11):

2

1

1 1

1 1

1 1 1

1

( ... ....

... ... ) 0

t

nnt

n n n nn n

T T Tq q q

q q q

T V Vq q q Q q Q q dt

q q q

+ + + +

+ + + =

(*)

Tch phn cc s hng cha vn tc iq tng phn:22 2

1 11

( )

tt t

i i ii i it tt

T T Tq dt q q dt

q q t q

=

(1.14)

Th vo biu thc (*):

=

+

+

=

2

1

0)(1

t

t

n

i

ii

iii

dtqQ

q

V

q

T

q

T

t

(1.15)V qi l ty nn:

iiii

Qq

V

q

T

q

T

t=

+

)(

(1.16)

y l phng trnh Lagrange, dng c cho h

tuyn tnh v phi tuyn.


14/146

CHNG 2. HE MOT BAC T DO

2.1 THIET LAP PHNG TRNH CHUYEN ONG2.1.1 Mo hnh he mot bac t doSingle Degree of Freedom system SDOF

Concentrated PropertiesKhoi lng: m

o cng: kHe so can: cLc kch ong:

p(t)Chu y: He mot bac t do co cac ac trng phan bo

m, k, c,p(t) eu co the a ve mo hnh co cac actrng vat lytap trung (he mot bac t do suy rong).

2.1.2 Cac phng phap thiet lap

phng trnh chuyen ong2.1.2.1 Nguyen ly DAlembertp(t) +fS+fI+fD=0

hay

)(tpkvvcvm =++ (2.1)2.1.2.2 Nguyen ly cong kha d

(t)

D

S

I

Lc tac dung

c

k

v(t)

(t)m

Mo hnh SDOFs


15/146

Cho khoi lng chuyen v kha d v. Cong kha d:

W=p(t)v +fSv +fIv +fDv = 0

hay 0)]([ =+ vtpkvvcvm v v 0 nen thu c giong nh (2.1).

2.1.2.3 Nguyen ly Hamilton

ong nang cua he:

2

21 vmT = , bien phan

ong nang vvmT =

The nang bien dang

an hoi cua lo xo:2

2

1kvV= ,

bien phan vkvV =

Bien phan cong cua lc khong bao toan p(t)va fD (tc la cong kha d cua hai lc nay tren

chuyen v kha dv): vvcvtpWnc = )(

Theo nguyen ly Hamilton: 0])([2

1

=+t

t

nc dtWVT

0])([2

1

=+t

t

dtvtpvvcvkvvvm (2.2)

tch phan tng phan so hang th nhat:

O v

= kvs

Lc

Chuyen v


16/146

2 2

2

1

1 10

t tt

t

t t

mv vdt mv v mv vdt = (2.3)

the (2.3), (2.2): 0)]([2

1

= +t

t

vdttpkvvcvm (2.4)

Nhan xet: Ca 3 phng phap cho cung ket qua vcung da tren nh luat quan tnh cua Newton.

Trong trng hp cu the nay nguyen lyDAlembertla n gian nhat.

2.1.3 Anh hng cua trong lcPhng trnh chuyen ong:

W)t(pkvvcvm +=++ trong o Wla trong lng cua khoi cng.

Chuyen v v gom tong cua chuyen v tnh(Static Displacement) st gay bi trong lng Wvachuyen v ong v

stv v= + , v v= , v v=

Thay bieu thc cua lc an hoivkkkvf

sts+==

vao phng trnh chuyen ong:


17/146

Wtpvkkvcvm st +=+++ )(

Mat khac stkW = nen phng trnh cuoi cung:

( )mv cv kv p t + + =

Ket luan: Neu lay v tr can bang tnh hoc do trong lng P =

mg gay ra lam moc e tnh chuyen v th phng trnh viphan chuyen ong van co dang (2.1). Nh vay, trong lckhong anh hng en phng trnh vi phan chuyen ong.

2.1.4 Anh hng cua s rung ong goi ta

ck

m

v(t)

(t)

(W)

S D

(t)

I

st

nh hng cua trong lc

S D

I

(t)

W

Wv(t)v(t)

v (t)

vg(t)

v

vt

I

D 0.5fS0.5fS


18/146

Phng trnh can bang lc: 0=++ SDI fff trong o lc quan tnh: tI vmf = vi gt vvv +=

la tong cua v la chuyen v uon va vg la chuyen vgoi ta (mat at).

0=+++ kvvcvmvm g

hay: )(tPvmkvvcvm effg =++ (2.5)

Ket luan:geff vmtP =)( la tai trong do rung ong

goi ta. Nh vay s rung ong cua mat at tngng nh lc kch ong effP tac dung tai vat nang.2.1.5 He mot bac t do suy rong (GeneralisedSDOF System)

He co ac trng vat lyphan bo (m, EI), thcchat co vo han bac t do.Neu coi he ch dao ong

vi mot ham dang nao oth he tr thanh 1 bac tdo. Tm cac ac trng taptrung cho he 1 DOF.

Gia s he chu rung ong ngang vg(t) cua goi

ta (do ong at chang han). Dung nguyen ly

l

x

xN

vg(t)

v (x,t) e(t)

z(t)

m(x)

EI(x)

v(x,t)

chuyen vO

t


19/146

Hamilton e thiet lap phng trnh chuyen ong.at:

v(x,t) = (x) Z(t) (2.6)(x) - Ham dang (Shape Function)

Z(t)- Toa o suy rong (Generalised Coordinate)

ong nang cua he:

[ ] dxtxvxmTt

l 2

0),()(2

1= dxvtxvxmT

ttl

),()(0=

The nang uon:

[ ] dxtxvxEIVl

f2

0

),(")(2

1= dxvtxvxEIV

l

f "),(")(0

= (2.8)

o co ngan cua thanh:[ ] dxtxvte

l2

0

),('2

1)( = (2.9)

The nang lc doc: [ ] dxtxvN

NeVl

N2

0

),('2 ==

hay dxvtxvNVl

N =0

'),(' (2.10)

V he khong co lc khong bao toan (lc can, lckch thch) nen:

=2

1

0)(

t

tdtVT (*), vi V= Vf+ VN


20/146

The (2.7), (2.8) va (2.10) vao (*):

0'),('),("),(")(),()(2

1 0 0 0

=

+ dtdxvtxvNdxtxvtxvxEIdxvtxvxm

t

t

l l ltt

(2.11)Dung cac lien he:

)(tv = v + gv va )(tv = v

"v = z" va Zv "" =

v = Z va Zv '' = Zv = va v =Z (2.12)

The (2.12) vao (2.11)0)'(")()()()(

2

1 0 0 0 0

222 =

++ dtdxZNZdxxEIZZdxxmtvZdxxmZZt

t

l l l l

g (2.13)

Chu y rang tch phan l

dxxf0

)( khong phu thuoc t,

nen ong vai tro la cac hang so khi thc hien tchphan theo bien t. e lam xuat hien cac tha soZtrong 2 so hang au, tch phan tng phan:

22 2 2 2 2

1 1 1 1 11

( )tt t t t t

t t t t t t

dZ dZ Zdt Z dt Z Z dt Z Z Z Zdt Z Zdt

dt dt = = = = (2.14)

=2

1

2

1

2

1

)()()(

t

t

g

t

t

t

t

gg ZdttvZtvdtZtv (2.15)

The (2.14) va (2.15), phng trnh (2.13) tr thanh:

z z

vt

O

vg

vv


21/146

[ ] =+2

1

0)(****t

ttG ZdttpZkZkZm (2.16)

=

l

dxxmm0

2* )(

: Khoi lng suy rong=l

dxxEIk0

2* )")(( : o cng suy rong

=l

G dxNk0

2* )'( : o cng hnh hoc suy rong

=l

gt dxxmtvtp0

* )()()( : Tai trong suy rong

V Zbat ky nen lng trong ngoac triet tieu,thu c phng trnh chuyen ong he suy rong:

)()()(***tptZktZm t=+ (2.18)

vi *** Gkkk = : o cng suy rong ket hp (2.19)

Khi lc doc N at tr so ti han N = Ncr th0* =k . T o, suy ra cong thc tnh lc Ncr la:

=

l

l

cr

dx

dxxEI

N

0

2

0

2

)'(

)")((

(2.20)

ay la cong thc cua phng phapRayleigh.

Chu y:


22/146

Neu thanh chu lc kch thch phan bop(x,t) valc doc N(x) th cong thc tnh lc kch thch suy

rong (lc tap trung) p

*

(t) va o cng hnh hoc k

*

Glan lt la:

=l

dxxtxptp0

*)(),()( (2.21)

=l

G dxxxNk0

2* )](')[( (2.22)

=l

dxxxcC0

2* )]()[( (2.23)

Th du: ExampleE8.3, page 144, [1]

Thiet lap phng trnh vi phan dao ong cua

he mot bac t do suy rong.Cho biet phng trnh ngan hoi (ham dang ) cchon nh sau:

L

xx

2

cos1)(

= (a)

Giai:

Ap dung (2.17), khoi lngva o cng suy rong:

( ) LmdxLx

mdxmm

LL

228.02cos10

2

0

2*

=

==

(b)

(x,t)

c(x)

L

x

xN

vg(t)

v (x,t) e(t)

z(t)

m

EI

v(x,t)

chuyen vO

t


23/146

( )3

4

0

2

2

2

0

2*

322cos

4"

L

EIdx

L

x

LEIdxEIk

LL =

== (c)

Tai trong tng ng suy rong (bo qua dau tr):)(364.0

2cos1)()()(

00

* tvLmdxLxtvmdxmtvtP g

L

g

L

g =

== (d)

Bo qua lc doc truc, phng trnh can bang:

)(364.0)(32

)(228.03

4

tvLmtZL

EItZLm g =+

(e)

Neu xet lc docNth o cng hnh hoc suy rong:( ) =

==L L

G L

Ndx

L

x

LNdxNk

0

2

0

2

2*

82sin

2'

(f)

o cng suy rong ket hp: LN

L

EIkkk G

832

2

3

4*** ==

V vay tai trong ti han mat on nh thu c khi

cho o cng ket hp bang 0 la:

3

2

23

4

4

8

32 L

EIL

L

EINcr

== (h)

ay la tai trong mat on nh that s cho cotconsole chu tai trong phan bo eu, bi v ham

dang c rut ra t (a) la dang mat on nh thatcua ket cau. Thay (h) vao (f) ta co the bieu dieno cng hnh hoc bi:

crG N

N

L

EIk

3

4*

32

= (i)


24/146

thay vao (e) ta co phng trnh can bang bao gomanh hng cua lc doc truc la:

)(364.0)(132)(228.0 3

4

tvLmtZN

N

L

EItZLm g

cr =

+

(j)Do o, bat ky hnh dang nao thoa man ieu kienbien hnh hoc eu c rut ra t ham dang )(x .

Neu ham nay c cho bi dangparabolic 22

)(L

xx =

Khi nay o cng an hoi suy rong tr thanh:

3

0

2

2

* 42

L

EIdx

LEIk

L

=

= LN

dxL

xNk

L

G3

42

0

2

2

* =

= Tai trong ti han c rut ra t ** Gkk = la:

23

3

4

34

L

EIL

L

EINcr == (l)

gia tr nay ln hn 21% so vi gia tr t (h).


25/146

2.2 DAO ONG T DO2.2.1 Nghiem cua phng trnh chuyen ong

Phng trnh chuyen ong cua he 1 bac t do(ke ca suy rong) co dang:

)()()( tpkvtvctvm =++

Neu khong co lc kch thchp(t) = 0 th:

0)()( =++ kvtvctvm (a)Nghiem co dang: v(t) = GestThe vao (a) ta c:

(ms2 + cs + k) Get= 0 (b)

at mk

=2

th (b) dan ti:

s2 +m

c+ 2 = 0 (c)

(c) la phng trnh ac trng,

nghiem s cua (c) tuy thuocvao he so can c.

Imaginary

1

1 Real

eit

t

O

e = cos t isinti t

Cong thc Euler:


26/146

2.2.2 Dao ong t do khong can c = 0Khi o (c) co nghiem: s = i do o nghiem

cua (a) la:v(t) = G1e

it+ G2e-it

hay viet lai di dang thc:

v(t) =Asint+Bcost (d)

vi A, B c xac nh t ieu kien ban au: B =v(0), A =

)0(v

nen:

v(t) =

)0(vsint+ v(0)cost (2.24)

Co the viet (2.24) di dang khac:v(t) =cos(t- )

(2.24')

Vi bien o2

2 )0()]0([

+=

v

v

va pha ban au

= tan-1)0(

)0(

v

v

(2.25)

chu ky: T=f

12=

(2.26)


27/146

2.2.3 Dao ong t do co canc 0

Nghiem cua (c): s =2

2

22

mc

m

c

(2.27)Dang dao ong phu thuoc vao tr so cua he so

can c (vao bieu thc di dau can co dau dng,am hay bang khong)

- Can ti han (Critical damping) c = ccrccr= 2m th

02

2

2

=

m

ccr

s = =mccr

2

v(t)

v(0)

T=

v(0)

2

t

v(0)v(0)

v(t)

tO


28/146

Phng trnh chuyen ong:

v(t)=(G1+ G2t)e-it=[v(0)(1+t)+ )0(v t]e-t (2.28)

o th chuyen ong co dang nh hnh ve, khongco dao ong.

- Can t (Underdamping):

c < ccr =2m.at =

crc

c=

mc

2

trong o la t so can (damping ratio).

The vao (2.27):

s = - 22)( = -iD

vi D = 21 : tan so dao ong co can, trong

thc te cac ket cau co


29/146

trong o:

[ ]22

)0()0()0(

vvv

D

+

+=

= tan-1)0(

)0()0(

v

vv

D

+(2.30)

o th chuyen ong vi v(0) 0, )0(v = 0.

Xac nh t so can :

Phng trnh dao ong t do theo ieu kien au:

v ( t )

t

2

3 4

D

D

D

D

v 0

O

- te

v 1

v 2


30/146

v(t)= e-t(D

vv

)0()0( +

sinDt+v(0)cosDt) (2.31)

Chu ky dao ong co can: T =D

2

The vao (2.29):

)2exp()exp(1 Dn

n Tv

v

==+

o giam Loga:

2

1 122ln

===+ Dn

n

v

v

=21

2

2 , vi nho.

21......!2

)2(21

2

2

1

++++===+

eev

v

n

n

Do o: =1

1

2 +

+

n

nn

v

vv

(2.32)

Chnh xac hn: =mn

mnn

vm

vv

+

+2

(t mt

mn

n ev

v =+

) (2.33)

Cong thc (2.32) va (2.33) dung xac nh t so can bang thc nghiem.

He so can: c = 2m (2.34)


31/146

- Can nhieu (Overdamping)

Khi > 1 (c > ccr) th khong co dao ong, tng t

khi c = ccrcang ln th chuyen ong ve v tr can bang cang

cham.

2.3 PHAN NG VI TAI TRONG IEU HOA2.3.1 He khong canLc kch thch: tptp sin)(

0=

Phng trnh: tptkvtvmo

sin)()( =+ (a)

Nghiem thuan nhat: tBttvh

cossin)( += Nghiem rieng dang (on nh): tGtv

psin)( =

The vao (a) rut ra:21

1

=k

pG o vi:

=

Vay nghiem tong quat:

tk

ptBtA

tvtvtv

o

ph

sin1

1cossin

)()()(

2++

=+=

(2.35)

A, B xac nh t ieu kien ban au. Neu0)0()0( == vv , de dang tm c:


32/146

0,1

12

=

= Bk

pA o

(2.36)

the vao (2.35) ta c:)sin(sin

1

1)(

2tt

k

ptv o

= (2.37)

T so phan ng (Response Ratio):

)sin(sin1

1)()(

)( 2 ttk

p

tv

v

tv

tRost

===

Trong thc te, lc can lam cho so hang saubien mat sau mot khoang thi gian ngan. Khi ohe so ong (Manification Factor) se la:

2

)(1

1)(

==

st

tpv

tvMF (2.38)

2.3.2 He co can

Phng trnh chuyen ong:

tm

ptvtvtv o sin)()(2)( 2 =++ (2.39)

Nghiem tong quat: tBtAetvDD

t

h cossin()( += )

Nghiem rieng: tGtGtvp cossin)( 21 +=


33/146

The vao (2.39) va ong nhat 2 ve, thu c:

2222

222

2

1

)2()1(

2)2()1(

1

+

=+

=

k

pG

k

pG

o

o

(2.40)

V nghiem qua o tat rat nhanh, nen he ch daoong theo nghiem rieng. Dung vector quay tren

gian oArgrand, ta tm c:1

2 2 2 122

2[(1 ) (2 ) ] tan

1

op

k

= + =

(2.41)

va phng trnh dao ong on nh:

)sin()( = ttv (2.42)

- He so ong (Dynamic Magnification Factor):

Imaginary

Realtt

2

(1 ) +(2)2 2k

o1

(1 ) +(2)ko

2 22

2

2

Bieu dien dao ong bang vect quay


34/146

222 )2()1(

1

+==

kp

Do

(2.43)

Khi >> th khong co chuyen ong.

0 1 2 3

900

1800 = 0

Phase Angle

Frequency ratio

= 0.05

= 0.2

= 0.5

= 1

=0

=0.2

=0.5

=0.7

=1.0

1

2

3

4

D

0 1 2 3

k m


35/146

2.3.3 S cong hng (Resonance)

Khi 1==

th xay ra cong hng. Luc nay

he so ong theo (2.43) la:

21

1==D (2.44)

Neu he khong can, tc la = 0 thD=1

oi vi he co can khac 0, thDmax xay ra khi:

2max

2

12

1

210

=

==

D

d

dDdinh

(2.45)

Nh vay:Dmax khac D=1

Tuy nhien, vi he co t so can be th co thecoi:

21

1max = =DD (2.46)

2.3.4 S co lap dao ong (Vibration Isolation)S co lap dao ong can thiet trong 2 trng hp:


36/146

- Thiet b may moc truyen rung ong co haixuong ket cau .

- Ket cau (b rung) truyen dao ong co haicho thiet b tren.

1. Xet motor quay, tao ra lc kch ong:tptp

osin)( =

Chuyen ong on nh(Steady-State Displacement):

)sin()( = tDk

ptv o

Van toc: )cos()( = tDkptv o

(t) = p0 sin tv

Phan lc nen


37/146

Lc an hoi: )sin()( == tDptkvfos

Lc can: )cos(2

)cos()(

===

tDp

t

k

Dcptvcf

o

o

D

VfS(t) vafD(t) lech pha 90o, nen bien o phan lc

nen la:

( )[ ]2

12

max

2

max

2

max 21 +=+= DpfffoDS

Ty so truyen lc (TransmissibilityRatio-TR), cnh ngha:

( )

( ) ( )[ ] )21(

21

21

22

2max

+=

+==

D

Dp

fTR

o (2.47)

TR =D neu = 0 (khong can)

o th cho thay cac ng cong eu:at cc ai tai=1

Cung i qua iem co= 2

Vi > 2 th TR < 1

Ty so can lam giam hieu qua cua viec co lapdao ong khi> 2 ==> Khong nen dung damper


38/146

2. Xet khoi lng m, chu kch ong cua goi taChuyen ong tng oi cua m so vi goi ta

cho bi phng trnh:

)sin()( 2 = tDvtv go

vt

m

vg (t)=vg sin t

Ty so truyen dao ongVibra. Transmi. Ratio

0 1

= 0.33

= 0.2

= 0

2 32

1

2

3

= 0.25

TR


39/146

Chuyen ong toan bo vt bang tong vector cuavg va v:

( ) )sin(21)(2

+= tDvtv got

Ty so truyen:

( )2max 21 +== DTR

v

v

go

t

(2.48)

Ty so truyen dao ong giong nhau cho ca 2trng hp.

Chu y: Neu khong co damper th:

1

12 =TR (2.49)

Th du:

Xe c mo hnh mot bac t do, chuyen ong v= 72.4km/h. o cng lo xo: 100lb gay chuyen v

0.8 in, =0.4. Coi kch ong ng la ieu hoa vacau rat nhieu nhp.


40/146

Giai

o cng lo xo:cm

kGcm

kG

in

lbk 4.233

203.0

4.45

08.0

100===

Chu ky dao ong t nhien cua xe:

)(572.081.94.223

18162

.2 s

gkT =

==

Chu ky kch ong bang thi gian i het mot nhp

cau: )(606.01.20

2.12 sv

LTp

===

Ty so chu ky: 994.0606.0

572.0===

pT

T

Bien o dao ong ng cua oto la:

L = 40fl = 12,2m

vt

v=45miles/h=72,4km/h=20,1m/s

W=4000lb=1816kG

1,2in=3,05cm

mat cau


41/146

( )

( ) ( )

( )( ) ( )

)(009.5

944.04.02944.01

944.04.2105.3

21

21

21

22

2

21

222

2

max

cm

TR vvv gogot

=

++

=

+

+==

Neu xe khong co damper (= 0) th:

)(69.27944.01

05.3

1

122max

cmvv got =

=

=

ln gap 5.5 lan khi co damper. ieu o noi len scan thiet cua damper e han che s dao ong ng

cua oto khi chay tren mat ng ln song.

Bai tap 4-3, page-77, [1]Xet lai bai toan tren, nhng nhp L = 36 ft =

10.97 m. Xac nh:a.Toc o gay cong hng cho xe: Tp = T= 0.572 sv =L/Tp = 10.97/0.572 = 19.18 m/s = 69km/h.

b. Bien o toan phan tvmax

cua xe khi cong hng:


42/146

( )( ) ( )

( )( )

( )

( ))(88.4

4.02

4.02105.3

2

21

2

21

21

21

1

2

221

2

2

2

1

222

2

max

cm

TR

T

T

vv

vvv

gogo

gogo

t

p

=

+

=+

=

+=

=

++==

===

c. Bien o toan phan khi toc o v = 45mph =72.4km/h =20.1m/s

( )( ) ( )

( )( ) ( )

546.1048.14.02048.11

48.104.021

21

21

)(048.1546.0

572.0

)(546.01.20

97.10

2

1

22

2

2

1

222

2

2

=

++

=

+

+=

===

===

TR

sT

T

svLT

p

p

)(72.4546.105.3max

cmTRv vgot ===


43/146

2.4 PHAN NG VI TAI TRONG CHU KY2.4.1 Khai trien tai trong thanh chuoi Fourier

Tai trong p(t) co chu ky Tp c khai trien chuoi

Fourier:

pn

n

pn

noT

nbt

T

naatp

2sin

2cos)(

11

=

=++= (2.50)

vi cac he so c xac nh nh sau:

0

0

0

1( )

2 2( )cos( )

2 2( )sin( )

p

p

p

T

o

p

T

n

p p

T

n

p p

a p t dt T

na p t t dt

T T

nb p t t dt

T T

=

=

=

(2.51)

v(t)

tO

Tp Tp Tp


44/146

2.4.2 Phan ng vi tai trong chu ky (tuan hoan)Khi mot tai trong chu ky c phan tch ra

chuoi Fourier(2.50) th phan ng cua he c xacnh theo nguyen ly chong chat. Bo qua nghiemqua o, trong trng hp he khong can, phan ngnh sau:

- Vi so hang tai trong 2sin( )np

nb t

T

th phan ng

cua he theo (2.37) la:

12

1( ) sin( )

1

nn

n

bv t n t

k

=

vi

1

n

T

nT

p

n

n

=== ;PT

2

1

= : tan so vong c

ban cua tai trong.

- So hang tT

na

p

n

2cos , phan ng c xac nh

tng t:tn

k

atv

n

n

n 12cos

1

1)(

=

- So hang ao - tai trong hang so, gay chuyen v

tnh: ka

vo

o =


45/146

- Phan ng toan bo

( )

+

+=

=1112

sincos

1

11)(

n

nn

n

otbtnaa

k

tv

(2.52)

2.4.3 Dang phc cua nghiem theo chuoiFourierCong thcEuler:

nxinxenxinxeinxinx sincos,sincos =+=

Suy ra:2

cosinxinx

eenx

+=

i

eenx

inxinx

2sin

=

Tai trong:

( )

)22

(

sincos)(

1

111

i

eeb

eeaa

tbtnaatp

inxinx

nn

inxinx

no

nnno

=

=

+

++

=++=

=

++

+= 1 22)( n

nninxnninxo

ibae

ibaeatp (*)

Dung (2.51):


46/146

( )

==

=

p

p

T

tin

p

T

p

nn

n

dtetpT

dtttntpT

ibac

0

011

1)(1

sincos)(1

2

o

T

p

o

T

tin

p

nn

nn

adttpT

cn

dtetpT

ibacc

p

p

===

=+

==

0

0

)(1:0

)(1

21

Co the viet (*) lai:

tin

n

tin

nn

tin

nno

ectphay

ececctp

1

11

)(

)(11

=

=

=

++=

(2.53)

vi

=

pT

tin

p

ndtetp

Tc

0

1)(1

(2.54)

la pho ri rac cua he soFourier.

Imaginary

Real

=1

=1

n1t

inte

e-int

1

1

-n1t


47/146

Chu y:

cn = c-n v la hai so phc lien hp,tin

e 1 va tine 1

la hai so phc lien hp. Do o:tinnec 1 va tin

nec 1

cung la hai so phc lien hp, vaco tong la thc (Real).

Dang phc cua nghiem:Khi phan tch tai trong ra chuoi Fourier phc(2.53), phng trnh chuyen ong ng vi mot sohang - ham lc phc n v (Unit complex forcingfunction) di dang:

tinetkvtvctvm 1)()()( =++ (2.55)Nghiem on nh co dang:

tin

nenHtv 1)()(

1= (2.56)

The vao (2.56), (2.55) ta c:

( )1

1 12 2

1 1

1( ) ;2 1

H nk n in

= = + +

(2.57)

Dung nguyen ly chong chat tac dung:

= tin

necnHtv 1)()(

1

(2.58)


48/146

Chu y:

- )(1

nH va )(1

nH la so phc lien hp

- )( 1nH goi la ham truyen - Complex frequency response function hay la Transferfunction.

2.5 PHAN NG VI TAI TRONG XUNG2.5.1 Khai niem tai trong xung (Impulsive Loads)- La tai trong tacdung trong thi gian

tng oi ngan, otngot.- Phan ng (chuyen vchang han ) ln nhatcua he at c trong

thi gian rat ngan.- Lc can co vai tro nho, hap thu t nang lng cuaket cau. V vay ch xet he khong co can e ngian hoa.

tO

(t)


49/146

2.5.2 Xung hnh sinXet tai trong na

song hnh sin. Phan ngcua he c chia ra 2pha: Cng bc va t do.

+ Phase I:1

0 tt

Ket cau chu tac dungcua tai trong ieu hoa.

ieu kien ban au: v(0) = v(0) = 0 (trang thaingh). Phan ng gom 2 so hang (qua o va bnh

on) cho bi (2.37) :

)sin(sin1

1)(

2tt

k

ptv o

= (2.59)

+ Phase II: 01

= ttt

ieu kien ban au: )()0( 1tvtv == )()0(

1tvtv ==

Theo (2.24):

ttvt

tv

tv cos)(sin

)(

)( 11

+=

(2.60)

O

(t)

t

(t)=p sin to

P0

t1 t

Phase I Phase II1=


50/146

Tuy thuoc vao ty so t1/ T ma phan ng cc aithuoc vao Phase I hoac Phase II.

- Neu vmax thuoc Phase I : = 1tt (2.59)

0)coscos(1

1)(2

0 =

= ttk

p

dt

tdv

Hay tt coscos =

,...2,1,0,2 == ntnt (a)

The (a) vao (2.59) tm c vmax

ac biet: khi = , trong (a) lay dau (-) van=1 ta co :

+=

1

2t

- Neu vmax thuoc Phase II: khi > ( cang

ln th 2

1 =t cang nho)

Dung ieu kien ban au v(t1) va v (t1), ta co bieno dao ong t do (2.25)


51/146

[ ]2

1

2

2

1

2

1 cos221

)()(

+

=+

=

kp

tvtv

o

He so ong: )cos1(21 2

0

+

==p

kD

hay

2cos

1

22

=D (2.61)

2.5.3 Xung ch nhat+ Phase I:

10 tt

Tai trong at ot ngotva gi nguyen khong oitrong phase I. Nghiemrieng cho tai trong bac thang (Step loads) lachuyen v tnh:

kpv op =

Nghiem tong quat vi ieu kien au ngh (rest):

( )tk

ptv o cos1)( = (2.62)

+ Phase II: 01

= ttt

(t)

O

0

Phase I1

t

Phase II

t t


52/146

Dao ong t do theo (2.60)

ttvttv

tv

cos)(sin)(

)(1

1 +=

(2.63)

- vmax thuoc Phase I : 1tt

2,,0sin0

)( Tttt

dt

tdv=====

Neu 1tt tc la 21T

t

he so ong D = 2, vi2

11 T

t

- vmax thuoc Phase II: 01 = ttt

Bien o dao ong:

[ ]21

2

1

max)(

)(tv

tvv +

==

v v(t1) =

11

2sinsin t

Tk

pt

k

poo

=

nen:vmax=2

12

1

2

1

2 2cos12

sin

+ t

Tt

Tk

po

= Tt

k

ptTk

poo 1

21

1 sin22

cos12

=


53/146

He so ong:T

t

kp

VD

o

1max sin2

== vi2

11 T

t(2.64)

2.5.4 Xung tam giac

+ Phase I:1

0 tt ,

p(t)=po(1- t/t1)

Nghiem rieng:

=

1

1)(t

t

k

Ptv o

p

ng vi ieu kien ban au ngh, nghiem tong quatco dang:

+= 1cos

sin)(

11t

tt

t

t

k

ptv o

(2.65)

+ Phase II: 0t ieu kien ban au tai 0=t , hay t= t1 t (2.65)

Phase IO

(t)

Phase II

0

1t t t


54/146

+=

=

1

1

1

1

1

1

1

1

1sincos)(

cossin

)(

tt

tt

kptv

tt

t

k

ptv

o

o

(2.66)

Dao ong t do cua Phase II thu c bangcach the (2.66) vao (2.60) vmax tm t ieu kien

v

(t) = 0. Vi4.0

1

Tt

th vmax thuoc Phase I.He so ongD cho bang:

Tt

1

0.20 0.40 0.50 0.75 1.00 1.50 2.00

D 0.66 1.05 1.2 1.42 1.55 1.69 1.76

2.5.5 Pho phan ng (Response Spectra)

Khai niem: Pho phan ng la o th cua he so ongD theo ty so chieu dai xung Tt/

1,

=T

tDD 1

Y ngha: Dung tnh chuyen v cua ket cau chiu tacdung cua xung lc.


55/146

Chu y:Neu ket cau chu chuyen ong cua nen vg(t),

th tng ng vi chu lc xung pt(t) = -mv g(t),

vi tr so ln nhatpo = - mv g0. Khi nay he so ong

c nh ngha :kvm

vD

g

t

/0

max

= hoac

go

t

v

vD

max=

V v g0 o c nen se tnh c gia toc cc ai

cua ket caut

vmax

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.4

0.8

1.2

1.6

2.0

2.4

Rectangular

Half since waveTriangular

He so ong (Dynamicmanification factor), D

Ratiot1

T=

Impulse duration

Period


56/146

2.5.6 Tnh toan gan ung phan ng do lc xung

Gia x p(t) la lc xungtrong thi gian t1 rat be. He cochuyen v v(t), can bang lc:

kvtp

dt

vdmvm == )(

[ ]dtkvpvmt

=1

0

V v(t1) la lng be bac 2 so vi t1 : ( )2

1~ tv ,

nen bo qua. Do o:1

1 1

0

1( ) ( ) (0) ( )

t

v p t dt v t v v t m

= =

v(t1) = 0 va v (t1) ong vai tro ieu kien ban aucua Phase II.

Dao ong t do sau khi lc thoi tac dung:ttvt

tvtv

cos)(sin

)()(

1

1 +=

tdttpm

tvt

sin))((1

)(1

0

(2.67)

(t)

tt1O


57/146

Th du: Xem E6-3, page 97-98 va Bai Tap 6-5,page 99

Dung cong thc gan ung, phan tch phan ngcua he ket cau mot bac t do chu tai trong dangxung p(t) nh hnh ve. Biet cac ac trng vat lycua he ket cau nh sau: o cng k = 51.1 k/in,trong lng W = 2000 k.

Giai:

Tan so vong: sradWkg

/14.3== va

=1

0

.10)(t

skipdttp .

Chu ky dao ong cua he: sT 22

==

.

(t)

t

0.1s 0.1s 0.1s

0= 50k


58/146

V tai trong tac dung trong thi gian ngan

( 15.01 =T

t), nen dung cong thc gan ung (2.67):

10(386)( ) sin 0.614sin

2,000(3.14)v t t t = =

trong o gia toc trong trng cho bi 2/386 sing =

Phan ng at cc ai khi 1sin =t , ngha la:vmax = 0.614 in.

Lc an hoi cc ai: kipskvf

S4.31)14.06(1.51

maxmax,===

Gia tr chnh xac cua chuyen v cc ai c

xac nh t phng phap tch phan trc tiep la0.604 in.

Nhan xet: Nghiem thu c t phng phap xap xkha thch hp, sai so nho hn 2%.


59/146

2.6 PHAN NG VI TAI TRONG TONGQUAT

2.6.1 Tch phan Duhamel cho he khong can

Xet tai trong bat ky p(t). Xet thi iem =t ,

theo (2.67) ta co :)(sin

)()(

= tm

dptdv (2.68)

vi >t = tt

ay la dao ong t do cua he sau khi chuxung lc dp )( ; dv(t) khong phai la o thay oicua chuyen v v trong thi gian d .

Toan bo lch s tai trong co the xem nh baogom s noi tiep cua cac xung lc ngan, moi xung

lc tao ra mot phan ng vi phan theo (2.68). Phan

(t)

tO d

()

t-O

Phan ng dv(t) sau khi chu xung p( )d


60/146

ng toan bo la do chong chat cua cac xung lc taora, nen:

-Duhamel Integral =t

dtpm

tv0

)(sin)(1)( (2.69)

Ky hieu: )(sin1

)(

tm

th (2.70)

th (2.69) co dang:

- Tch phan cuon (Convolution Integral)

=t

dthptv0

)()()( (2.71)

Ham )( th c coi nh phan ng vi xung lc

n v.

Neu thi iem t = 0 (lc bat au tac dung),ket cau co ieu kien ban au khac khong:

0)0(,0)0( vv th (2.69) phai ke them dao ong

t do:

dtp

m

tovtov

tv

t

++=

0

)(sin)(1

cos)(sin)(

)(

(2.72)


61/146

2.6.2 Tch phan bang phng phap so choDuhamel Integral

Dung cong thc lng giac: sincoscossin)sin( ttt =

Phng trnh (2.69) viet lai:

=

t

t

dpm

t

dpm

ttv

0

0

sin)(1

cos

cos)(1

sin)(

v(t) = ttBttA cos)(sin)( (2.73)

Noi chung )(tA va )(tB c tch phan bang so.

Chang han, co the viet lai:

y()

=

O

y y y

yo

n

21

tn

t


62/146

==tt

y

dym

dpm

tA00

)(

)(1

cos)(1

)(

Phng phap SimpsonChia tra n phan (n chan)

n

t= khi o:

( ) ++++++

=

t

nyyyyyydy0

43210 1.....24243)(

2.6.3 Phan ng cua he co canTrong (2.31) dung ieu kien ban au do xung

lc dp )( tao ra :v(o) = 0, )0(v =m

dp )( , ta co:

= tt

m

dpedv

D

D

t ,)(sin)(

)( )( (2.74)

= t

Dt

D

dtepm

tv0

)( )(sin)(1)( (2.75)

Tng t he khong can, ta co:

)(sin1

)( )(

= tem

thD

t

D

(2.76)

ttBttAtvDD

cos)(sin)()( = (2.77)


63/146

=

=

t

Dt

D

t

Dt

D

deep

mtB

de

ep

mtA

0

0

sin)(1)(

cos)(1

)(

(2.78)

A(t) va B(t) c tnh bang so, chang hanphng phap Simpson.

2.6.4 Phan tch phan ng trong mien tan so(Frequency Domain)- Y ngha: Phan tch phan ng trong mien tan so cou iem hn trong mien thi gian khi au vao la

bat ky, khong tuan hoan (chu ky m rong ra ).ac biet la vi au vao (input) ngau nhien.

- Cong thc: e tien theo doi, viet lai (2.53) va

(2.54): (p

T

21

= : tan so c ban cua tai trong).

=

===

n

ti

nn

tin

n

necectp1)( (a)

==2

2

2

2

)(1

)(1

1

p

p

n

p

p

T

T

ti

T

Tp

tin

p

ndtetp

Tdtetp

Tc (b)


64/146

at:

2

1211

=== +

pp

nnTT

(c)

)(2

)(1 nnp

n ccT

c = (d)

The (d) vao (a):

=

=

n

ti

n ectpn

)(2

)( (a)

khi pT th d (theo (c)), do o (a)

viet lai:

=

dectpti

n

n)(2

1)(

The (d) vao (b):

==

2

2

)()(

pT

pT

n

dtetpcTc

ti

npn

pho

lien tuc cua he soFourier.

cho pT :

= dtetpc tin n )()( (2.80)

Phng trnh (2.79) va (2.80) la cap bien oiFourier(Fourier Transform Pair)

ieu kien ton tai bien oi Fourier:


65/146

Tng t nh (2.58) phan ng cua he bieu dien biphng trnh:

=

decHtv ti)()(21)( (2.81)

Ham truyen:

)12(

11)(

22 ++=

++=

ikkcimH

vi= (2.82)

2.7 PHAN TCH PHAN NG CUA KET CAU

PHI TUYEN2.7.1 Khai niemKet cau phi tuyen: Phng trnh chuyen ong laphng trnh vi phan phi tuyen. Yeu to phi tuyenco the nam trong quan he gia lc - chuyen v,trong he so can, do chuyen v ln hoac thay oi so tnh trong qua trnh chuyen v. Chang han, phanng cua mot toa nha (building) do ong at manhgay ra h hong nghiem trong. S o tnh thay oi

trong qua trnh b h hong. Khong dung cnghiem cua phng trnh tuyen tnh.


66/146

Cac cach giai quyet - co hai cach:

- Phng phap giai tch: Tm nghiem giai tch

cua phng trnh vi phan phi tuyen. Do khokhan ve toan hoc, nen ch giai c nhngphng trnh gan tuyen tnh (a tuyen ), ng vicac bai toan vat ly tng oi n gian. TheoChopra [2] th ket qua cua phng phap giai

tch khong thoa man cho phi tuyen bat ky.- Phng phap so: Tm nghiem di dang so(nghiem ri rac) co the xet en moi yeu to phituyen, vi quy luat phc tap von co cua baitoan. Phng phap tch phan tng bc (Step-by-Step Integration) cua Newmark c coi laphng phap manh (powerful), tng oi ngian, oi hoi t khoi lng tnh toan, nhng choket qua tot.

2.7.2 Phng trnh so gia cua can bang(Incremental Equation of Equilibrium)Xet he SDOF nh hnh ve cac ac trng m, k, c vap(t) co the la cac ai lng suy rong. Gia thiet k(v)va )(vc la cac ham phi tuyen.


67/146

Theo nguyen ly lembertD , phng trnh can

bang cua cac lc tac dung len he thi iem t:)()()()( tptftftf

SDI=++ (a)

thi iem tt + ( t be):

)()()()( ttpttfttfttfSDI

+=+++++ (b)

Lay (b)-(a):)()()()( tptftftf

SDI=++ (2.83)

So gia cua cac lc xac nh nh sau:)()()()( tvmtfttftf

III=+=

)()()()()( tvtctfttftf DDD += )()()()()( tvtktfttftf

SSS+=

)()()( tpttptp += (2.84)

v(t)

(t)

c

k

m(t)

(t)

(t)

(t)D

S

I

(a)(b)


68/146

Trong o c(t) va k(t) la o doc tiep tuyen aumoi thi oan t (o doc cat tuyen khong xac nhc v cha biet c tr cua )( ttf

D+ va

)( ttfS

+ thi iem tt + ):

(t)

O

(t+ t)

(t)

Tai trong p(t)

t t+ t t

(t)

t

(e)

d) Chuyen v

v

fs

v

s(v)

v

Dif

v

v

)(vfD

c) Van toc

1+iv

fsiv

vi+1vi

He so can c(t) o cng k(t)


69/146

t

D

v

ftc

)( ,t

S

v

ftk

)( (c, k= const neu

dao ong tuyen tnh) (2.85)

The (2.84) vao (2.83):

)()()()()()( tptvtktvtctvm =++ (2.86)

ay la phng trnh so gia can bang cua bai toanong lc hoc. Co 3 an trong (2.86).

Can bieu dien )(tv , )(tv theo )(tv . Trong o)(tv la bien c ban c tm trc.

2.7.3 Tch phan tng bcGia thiet:

Cac tnh chat cua he nh k, c la hang so trongmoi t .

Gia toc bien oi bac 1 v bac 2 , v bac 3.

tt+t

v

)(tv

)(tv

Gia toc

t t+t

v

)(tv

)(tv

Van toc

t t+t

v

)(tv

)(tv

Chuyen v

bac 2 bac 3


70/146

So gia cua van toc va chuyen v c tnh bicac phng trnh sau:

2)()()( ttvttvtv += (a)

6)(

2)()()(

22t

tvt

tvttvtv

+

+= (b)

Chon )(tv la an so c ban. Giai (b) cho )(tv va

the vao (a) tnh c )(tv , the hien nh sau:

)(3)(6

)(6

)(2

tvtvt

tvt

tv

= (c)

)(

2

)(3)(3

)( tvt

tvtv

t

tv

= (d)

Chu y rang )(tv la an so c ban.

The (c) va (d) vao (2.86) ta c:

)()()()(2

)(3)(3

)(

)(3)(6

)(6

2

tptvtktvt

tvtvt

tc

tvtvt

tvt

m

=+

+

hay: )(~/)(~)()(~)()(~ tktptvtptvtk == (2.87)

vi )(36

)()(~

2

tct

mt

tktk

+

+= (2.88a)


71/146

++

+

+= )(2

)(3)()(3)(6

)(~ tvt

tvtctvtvt

mtpp

(2.88b)

Chu y:Phng trnh (2.87) tng t nh phng trnh

can bang tnh. Tnh chat ong c ke do hieu ngquan tnh va can trong cac so hang )(~ tk va )(~ tp .

The )(tv cua (2.87) vao (d) se tm c )(tv .ieu kien ban au cua bc tnh tiep theo:

)()()(

)()()(

tvtvttv

tvtvttv

+=+

+=+(e)

2.7.4 Tom tat trnh t tnhTrong moi bc thi gian (Time Step) t , trnh

t tnh gom:1. ieu kien ban au )(tv va v(t) c biet, t

bc tnh trc o hoac ieu kien ban au cua baitoan, tc la )0(v va v(0).

2. Xac nh c(t) va k(t) cho bc tnh t o thva tnhfD(t) vafS(t).


72/146

3. ieu kien ban au ve gia toc xac nh biphng trnh can bang lc:

[ ])()()(1)( tftftpmtv SD = (2.89)

4. So gia cua )(~ tp va o cng )(~ tk c tnht (2.88).

5. So gia chuyen v )(tv va so gia van toc)(tv xac nh t (2.87) va (d).

6. Chuyen v )( ttv + va van toc )( ttv + tnhtheo (e).

Bc tnh tiep theo c lap lai nh tren. oi

vi ket cau tuyen tnh, phng phap tnh nay cungdung c va viec tnh n gian hn.

Van e chnh xac cua phng phap

o chnh xac cua phng phap phu thuoc vaoo ln t . Co 3 yeu to phai xet khi chon t :

1. Mc o (Rate) bien oi cua tai trongp(t).

2. o phc tap cua tnh chat phi tuyen cua o cngva he so can.

3. Chu ky dao ong Tcua he.


73/146

vi quy luat p(t) tng oi n gian th t phuthuoc vao T.

Thng 10Tt cho ket qua ang tin cay.

Th du:Xem E8-1, page 124, [1].V du nay dung e minh hoa cho k thuat giai

tay ap dung phng phap tng bc c mo ta

tren oi vi dao ong phi tuyen, tnh toan phanng cua khung an hoi deo mot bac t do chu taitrong thay oi nh hnh ve.

Khi phan tch s dung bc thi gian 0.1s th

cha at o chnh xac cao, tuy nhien gia tr cuaphan ng ong cung thoa man muc ch phan tch.

0 0.2 0.4 0.6 0.8 t, s

1

2

(t), kips

5

87

5

3

Load history

c = 0.2 kip.sec/in

(t), kips

m = 0.1 kip.sec2/in

k = 5 kips/in

v(t)s

v

vmax1.2in

Inelastic displacement

Elastoplastic stiffness

6kips

6kips


74/146

Trong ket cau nay, he so can c gia thietvan la hang so, v vay ay la bai toan phi tuyen o

cng (o cng thay oi trong qua trnh chuyenong).

( ))(66

1.0

3

1.0

6)()(

~2

tkcmtktk +=++=

trong o k(t) co gia tr 5 kips/in oi vi oan an

hoi hoac bang 0 oi vi oan deo.

V vay so gia cua tai trong la:

vvtp

vcmvcm

tptp

31.06.6)(

2

1.033

1.0

6)()(~

++

=

++

++=

So gia cua van toc c cho bi:

vvvv 05.0330 = Nghiem c the hien tren o th sau:


75/146


76/146

CHNG 3. HE NHIEU BAC T DO

3.1 THIET LAP PHNG TRNH CHUYEN ONG3.1.1 La chon bac t doY ngha: thc te ket cau thng la he phan bo,

co vo han bac t do. a ve s o mot bac t doch thch hp trong mot so trng hp ac biet, khihe hau nh ch dao ong vi mot dang nhat nh.e thu c ket qua chnh xac hn, ta phai a heket cau ve he ri rac nhieu bac t do. So bac t doc chon da vao bai toan cu the.

Cac cach chon bac t do: co hai cach- Chon bien o dao ong tai mot so iem ri

rac: bao gom phng phap don khoi lng vaphant hu han (FEM) e ri rac hoa.

- Chon toa o suy rong, la bien o cua mot sokieu (pattern) bien dang cua he.

3.1.2 Phng trnh can bang onge n gian ta xet he lien tuc nh hnh ve, vi

cac bac t do la chuyen v tai cac iem 1, 2, 3, ...,N.


77/146

Tai moi iem (nut) co cac lc tac dung: taitrongpi(t), lc quan tnhfIi, lc canfDi, va lc anhoifSi. Phng trnh can bang nut i:

fIi+fDi +fSi =pi(t) , i = 1, 2, 3, ...,NDang ma tran:

[fI] + [fD] + [fS] = [p(t)] (3.1)

trong o:

[fI] =

IN

I

I

f

f

f

2

1

, [fD]=

DN

D

D

f

f

f

2

1

, [fS]=

SN

S

S

f

ff

2

1

, [p(t)] =

)(

)()(

2

1

tp

tptp

N

- Lc an hoi

Dung nguyen l cong tac dung, ta co:fSi = ki1v1 + ki2v2 + .... + kiNvN vi i = N,1

v1(t)v2(t) vi(t) vN(t)

1 2 i

(x,t)

m(x)EI(x)

chieu dngchuyen v

chieu dngcua lc

chuyen v

Di

Ii m i

vi(t)

i(t)

Si


78/146

vi kij la lc tai nut i do chuyen v vj = 1 gay ra.

Chu y:Lc an hoi can bang vi lc nut nham duytr ng an hoi (ngc chieu vi lc nut).

Dang ma tran:

SN

S

S

f

f

f

2

1

=

NNNN

N

N

kkk

kkk

kkk

21

22221

11211

Nv

v

v

2

1

(3.2)

hay: [fS] = [K][v] (3.3)

[K] goi la ma tran cng (Stiffness Matrix).

- Lc can- ket qua tng t nh lc an hoi

DN

D

D

f

f

f

2

1

=

NNNN

N

N

ccc

ccc

ccc

21

22221

11211

Nv

v

v

2

1

(3.4)

vi cij la lc tai nut i do jv = 1 gay ra, goi la he soanh hng can.

hay: [fD]= [C][ v ] (3.5)

trong o: [C] la ma tran can (Damping Matrix)

- Lc quan tnh


79/146

IN

I

I

f

f

f

2

1

=

NNNN

N

N

mmm

mmm

mmm

21

22221

11211

Nv

v

v

2

1

(3.6)

vi mij : lc tai nut i do jv = 1 gay ra, la he so anhhng khoi lng,

hay: [fI]= [M][v ] (3.7)

trong o: [M] la ma tran khoi lng (Mass Matrix)

HeNphng trnh vi phan chuyen ong:

[M][v ] + [C][v] + [K][v] = [p(t)](3.8)

Phng trnh tren la phng trnh mang tnhchat tong quat cua bai toan ong lc hoc.

Tuy thuoc vao [p(t)] ma ta co cac trng hpphan tch ong lc hoc cua he:

Phan tch dao ong t do,

Phan tch phan ng cua he vi tai trong ongnh tai gio, ong at, song bien...

3.1.3. Anh hng cua lc doc (nen)Lc doc lam tang them chuyen v nut, nen se

co vai tro nh lc nut tac dung theo chieu cua

chuyen v nut, ky hieu bi ma tran [fG]. Khi nayphng trnh can bang nut (3.1) tr thanh:


80/146

[fI] + [fD] + [fS] - [fG] = [p(t)] (3.9)

Lc nut [fG] tng ng vi vai tro cua lc doc,c bieu dien bi cac he so cng hnh hoc(Geometric - Stiffness Coefficients) nh sau:

GN

G

G

f

f

f

2

1

=

GNNGNGN

NGGG

NGGG

kkk

kkk

kkk

21

22221

11211

Nv

v

v

2

1

(3.10)

vi kGij la lc tai nut i do vj = 1 gay ra, co anhhng cua lc doc

hay: [fG] = [KG][v]

(3.11)trong o: [KG] la ma tran cng hnh hoc(Geometric - Stiffness Matrix)

Phng trnh (3.9) tr thanh:

[M][v ] + [C][v] + [K][ v] [KG][v] = [p(t)] (3.12)hay: [M][v ] + [C][v] + [K][ v] = [p(t)] (3.13)

vi: [K] = [K] [KG] la ma tran o cng tong hp(Combined Stiffness Matrix)

Nh vay, lc doc lam giam o cng cua ket cau(lam cho ket cau mem i).


81/146

3.2 XAC NH CAC MA TRAN TNH CHAT

3.2.1 Tnh chat an hoi

3.2.1.1 o mem cua ket cau

Goi: fij la chuyen v tai i dopj = 1 gay ra. Taphp cacfij (i = 1,N) tao nen ng an hoi dopj =

1 gay ra (hnh ve). Chieu dng cua chuyen v valc theo chieu dng cua truc toa o.

Chuyen v tai iem i do cac lc pj (j = 1,N)theo nguyen ly cong tac dung:

vi=f

i1p

1+f

i2p

2+ .... +f

iNp

Ni = 1, N

Dang ma tran:

1j2j ij

jj

Nj

1 2 3

j


82/146

Nv

v

v

2

1

=

NNNN

N

N

fff

fff

fff

21

22221

11211

Np

p

p

2

1

(3.15)

hay: [v] = [f][p](3.16)

trong o:

[f] : Ma tran o mem cua ket cau (FlexibilityMatrix)

[p]: Ma tran tai trong nut, co cung chieu dngvi chuyen v nut. Lc an hoi can bang lc nut

[p] = [fS], (3.16) tr thanh:[v] = [f][fS] (3.17)

1 2 3

S1 S2 S3

1v v2 v31 i

1 i

jv=1

1=k

1j=ki ij =kN Nj

=kj jj

k1j

kij

kjj kNj

jv=1


83/146

3.2.1.2 o cng cua ket cau

He so cng kij (c minh hoa tren hnh ve)lacac lc nut do chuyen v vj = 1 gay ra (cac chuyenv khac vi = 0, vi i j). kijchnh la phan lc tai nutneu at them cac lien ket.

Thng ma tran o cng [K] c suy ra t matran o mem [f] hoac dung phng phap phan t

hu han (FEM).3.2.1.3 Cac khai niem c s

- The nang bien dang: (bang cong ngoai lc)

]][[

2

1]][[

2

1

2

1

1

pvvpvpUTT

i

N

i

i ====

(3.18)

Theo (3.16) vao (3.18) ta c:

]][][[2

1pfpU

T= (3.19)

Hoac the (3.3) vao (3.18), vi chu y rang [p] = [fS]:

]][][[2

1

vKvUT

= (3.20)V U > 0 nen suy ra:

[vT][K][v] > 0 va [pT][f][p] > 0 (3.21)[K] va [f] thoa (3.21) vi moi [v], [p] 0 nen lacac ma tran xac nh dng (Positive Definite),khong suy bien va nghch ao c.


84/146

Thiet lap quan he [K], [f], (3.3): [fs] = [K].[v]hay [K-1][fs] = [v]Mat khac (3.17): [v] = [f].[fs]

suy ra: [f] = [K-1] hoac [K] = [f-1] (3.22)Thng xac nh ma tran cng thong qua ma tranmem theo (3.22).

- nh ly Betti:Cong kha d cua lc trang thai (a) tren chuyen

v trang thai (b) bang cong kha d cua lc trangthai (b) tren chuyen v trang thai (a)

[paT][vb] = [pb

T][va] (3.23)

hay [paT

][f][pb] = {[pbT

][f][pa]}

T

= [paT

] [f

T

] [pb]suy ra: [f] = [fT] Ma tran oi xng (3.24)Mot cach tng t, ma tran cng oi xng:

K = KT (3.25)

a1

va1

a2

a2v

a3

va3

b1

b1

v b2

b2

v b3

b3

v

1 2 3Trang thai (a)

Trang thai (b)


85/146

3.2.1.4 Thiet lap ma tran o cng bangphngphap phan t hu han (FEM)

He c quan niem gom nhieu phan t noi vinhau tai mot so hu han nut. Tnh chat cua hec tm bang cach chong chat cac phan t motcach thch hp.

Xet phan t dam thang co 2 nut nh hnh ve:

Co hai bac t do moi nut: bao gom chuyen vthang va goc xoay.

Ham dang i(x) ch chuyen v vi= 1 gay ra, concac chuyen v nut khac eu bang 0. Ham i(x)phai thoa man ieu kien bien, nhng thng chonham chuyen v trong dam co o cng EI = constdo chuyen v nut vi = 1 gay ra. o la cac ham athcHermitbac ba nh sau:

1(x) = 1 - 32

L

x+ 2

3

L

x(a)

3(x) =x(1-L

x)2 (b)

2(x) = 32

L

x- 2

3

L

x(c)

4(x) = 1

2

L

x

L

x(d) (3.26)

EI(x)L

xa b

v(x)1v v3

2v4v

1

av =v =11

=v =1a 3

(x)1

3(x)


86/146

Dung bon ham noi suy nay, chuyen v cua damxac nh theo cac chuyen v nut:

v(x) = 1(x) v1+ 2(x) v2 + 4(x) v3 + 4(x) v4 (3.27)

trong o:

4

3

2

1

v

v

v

v

=

b

a

b

a

v

v

(3.27)

He so cng cua phan t la cac phan lc nut dochuyen v nut gay ra. e n gian ta xet phan tdam nh hnh ve. He sok13, tc la phan lcpa tren

hnh ve c xac nh nh sau:

Dung nguyen l cong kha d: WE= pava = k13v1

Momen do a = 1 gay ra la:M(x) = EI(x) ''3 (x)

Cong kha d cua noi lc: WI= v1 dxxxEIL

)()()(''

3

0

''

1

=v3 =11 3

(x)v =v1a

k13 = pa =pa

v(x)=(x)v11

(chuyen v kha d)


87/146

Cho WI=WEsuy ra: k13 = dxxxxEIL

)()()(''

30

''

1 (3.28)

Tong quat hoa:

kij = dxxxxEI jL

i)()()(

''

0

'' : o cng suy rong (3.29)

v kij = kji nen ma tran o cng oi xng.

Vi dam co o cng euEI = const, ta co:

4

3

2

1

S

S

S

S

f

f

ff

=3

2

L

EI

22

22

233

233

33663366

LLLL

LLLL

LLLL

4

3

2

1

v

v

vv

(3.30)

Neu dam co o cng EI(x) thay oi th (3.30)la gan ung. o chnh xac se cao hn, neu chiadam ra cac phan t nho hn.

He so o cng kij cua ket cau bang tong cac heso cng tng ng cua cac phan t noi vao nut.

Chang han, neu cac phan t m, n, p cung noi vaonut i th he so cng cua ket cau tai nut i la:

kii =)(m

iik + )(n

iik + )( p

iik (3.31)

trong o )(mii

k , )(nii

k , )( pii

k la he so cng cua phan t a

bien oi sang he toa o chung(t toa a phng).


88/146

Th du:

Xet he nh hnh ve, gom 3 phan t noi tai 2nut. Bo qua bien dang doc truc, he co 3 bac t do:v1, v2va v3

Cac he so o cng cua he c xac nh bangcach lan lt cho cac chuyen v cng bc n vvi = 1 va cong lc nut ng vi cac phan t. Ma

tran o cng ket cau:

)26(2

311x

L

EIk = )3(

2321

LL

EIk = )3(

2331

LL

EIk =

)6(2

)2(2)2(

42)2(

2 23

2

3

2

32233L

L

EILx

L

EIxL

L

EIkk =+==

)2(2

)2()2(

42 23

2

332L

L

EIL

L

EIxk ==

2L

L EI EI

4EI

v1v2 v3

EI EI

4EI

k11k21 k31

v1=1

EI EI

4EI

k12k22 k32

v2=1


89/146

=

3

2

1

22

22

3

3

2

1

623

263

33122

v

v

v

LLL

LLL

LL

L

EI

f

f

f

S

S

S

Chu y: Bai toan ong lc hoc cua he phan bothng oi hoi nhieu bac t do hn so vi bai toantnh, do anh hng cua lc quan tnh. Tuy nhien,khi a chon cac bac t do cho bai toan ong roi th

viec xay dng ma tran cng giong nh trng hpbai toan tnh.

3.2.2 Tnh chat khoi lng

3.2.2.1 Ma tran khoi lng thu gon (Lumped MassMatrix)

Ta xem khoi lng phan bo cua cac phan tc thu gon ve cac nut theo nguyen tac tnh hoc,ta co he gom cac khoi lng tap trung. Ma trankhoi lng thu gon la ma tran ng cheo:

1m m2 m3

1 2 3


90/146

[M] =

Nm

m

m

00

0

0

00

2

1

(3.32)

trong o: mij = 0 vi i j, v gia toc tai khoi lngnao ch gay ra lc quan tnh tai khoi lng o.

3.2.2.2 Ma tran khoi lng tng thch (Consistent

- Mass Matrix)

Xet phan t dam co hai bac t do moi nut.Dung cac ham noi suy i(x) nh ma tran cng.

Gia s dam chu tac dung cua gia toc goc bangn v tai nut a,

3v = a

= 1, gia toc chuyen ong ngang cua dam la:

L

m(x)

v(x)v

1a 3

v v4b

2v

v = v

=v =1a 3

a

(chuyen v kha d)

v(x)= (x) v

m =p

1

13 a

1(x)

1 1

.. ..


91/146

)()(33xvxv = (3.33)

Lc quan tnh: )()()()()(33xvxmxvxmxf

I == (3.34)

Cho dam chu chuyen v kha d v(x) = 1(x)v1. Can bang cong kha d cua lc nut va lc quan

tnh, ta co: pava = dxxvxfL

I)()(

0

hay m13 = dxxxxmL

)()()(3

0

1

KL suy rong mij = dxxxxm jL

i)()()(

0

(3.35)

v mij = mji, nen ma tran tng thch oi xng.

- Neu dam co khoi lng phan bo eu th ta co:

4

3

2

1

I

I

I

I

f

f

f

f

=420

mL

22

22

432213

341322

221315654

132254156

LLLL

LLLL

LL

LL

4

3

2

1

v

v

v

v

(3.36)

Ma tran khoi lng cua ket cau cung c chongchat t ma tran cua phan t, tng t nh ma trancng.


92/146

Th duThanh lap ma tran khoi lng cho ket cau nh

hnh ve theo hai phng phap. Qua trnh tnh cache so khoi lng c ch ro tren cac hnh ve.

Ma tran khoi lng thu gon:

[M] =

0

0

840

210

Lm

m11m21 m31 11 =v

m12m22 m32

12

=v

2L

L m m

v1

v2 v3

1.5m v1

v2 v31.5mL

0.5mL

0.5mL 0.5mL

0.5mL

1.5mL

m11= 4mLm22 = m33 = 0


93/146

m22 = m33 = 0 v gia thiet rang khoi lng thu gonkhong co quan tnh xoay, tc la cac gia toc goc tainut khong gay ra momen quan tnh.

Ma tran khoi lng tng thch:

768210

25.1)2156(420

11

LmLxmx

Lmm =+=

LLm

LLm

mm 11210

)22(420

3121===

222

332226

210)2(4

420

25.14

420L

LmL

LxmL

Lmmm =+==

22

32)18(

210)2()3(

420

25.1L

LmLx

Lxmm ==

[M] =

22

22

261811182611

1111786

210LLLLLL

LLLm

Nhan xet

Bai toan ong lc hoc ng vi ma tran khoi

lng thu gon n gian hn v:- [M] thu gon dang ng cheo, trong khi [M]

tng thch co nhieu he so khac 0 ngoai ngcheo. Cac he so cua [M] thu gon ng vi cacchuyen v xoay cung bang 0, cang lam cho bai

toan n gian hn.


94/146


95/146

3.2.4.1 Tai trong nut tng ng tnh hoc

Xem nh tai trong at tren dam phu co mattruyen lc at tai nut. Lc truyen vao nut se thaythe cho tai trong at tren phan t. Nh vay khongtruyen mo men tap trung vao nut.

3.2.4.2 Tai trong nut tng thch

(x,t) q(x,t)F(t)

i(t) j(t)

Lc nut tng ng

a

3

1

4b

2L

v(x)= (x) v1 1

v = va 1

(x,t)

Tai trong suy rong


96/146

Tai trong nut c tnh theo nguyen l chuyenv kha d, dung cac ham noi suy i(x). Th du:

p1(t) = dxxtxp

L

0 1)(),(

Tai trong suy rong pi(t) = dxxtxp

L

i0

)(),( (3.38)

Neu tai trong co dang phan ly (trng hp naythng gap trong thc te)

p(x,t) =(x)(t)

th lc nut suy rong tr thanh:

pi(t) = (t) dxxxL

i0

)()( (3.39)

Chu y rang, vi cac ham noi suy i(x) (i = 1,4)ta co 2 lc nut va 2 mo men nut tai 2 au dam.

3.2.5 o cng hnh hoco cng hnh hoc

the hien khuynh hnglam tang chuyen v uoncua lc nen N. He so

cng hnh hoc chnh lalc nut doNtao ra. Gia

iv jv

i

x

v

NO

N

N

i

i

Li

iv

jviGi=

v -i vjL i

iN

iNGj= L iv -j vi


97/146

thiet rang lc nenNdo tai trong tnh gay ra la chuyeu; phan do lc ong gay ra co the bo qua c.V vay, coiN khong oi trong qua trnh dao ong.

(Neu N(t) thay oi theo thi gian th [KG] cungthay oi theo thi gian. Bai toan tr nen phituyen).

Xap x tuyen tnh: 1 BTD/nut

Gia s lc doc trong phan t i laNi. Coi phant i thang th lc nutfGi vafGj c xac nh theolc nenNi tren hnh ve. Viet lai dang ma tran:

=

j

i

i

i

Gj

Gi

v

v

l

N

f

f

11

11(3.40)

Ma tran cng hnh hoc cua ket cau dam:

+

+

+

+

=

n

i

n

n

n

n

n

N

i

i

i

i

i

i

i

i

Gn

Gi

G

G

v

v

v

v

L

N

L

N

L

N

l

N

l

N

l

N

l

Nl

N

l

N

l

N

l

N

l

N

l

N

l

N

f

f

f

f

2

1

1

1

1

1

1

1

1

1

2

2

2

2

1

1

1

1

1

1

1

1

0

0

2

1

00

0

0

00

(3.41)

co dang 3 vet cheo. Viet dang k hieu:]][[][ vKf

GG= (3.42)


98/146

+ o cng hnh hoc tng thch:

Dung khai niem phan t hu han, ta thu ccong thc:

( ) ( ) ( )dxxxxNkji

L

oGij

'' = (3.43)

Neu phan t co lc docN(x)= N = const, dungcac ham noi suy trc ay, ta thu c ma trancng hnh hoc phan t:

=

4

3

2

1

22

22

4

3

2

1

433

433

333636

333636

30

v

v

v

v

LLLL

LLLL

LL

LL

L

N

f

f

f

f

G

G

G

G

(3.44)

=

22

22

433

433

333636

333636

30][

LLLL

LLLL

LL

LL

L

NKe

G

[ eGK ] la ma tran o cng cua phan t (oi xng).

Bieu o N(x)

PG2

b PG4

PG1

PG3

a


99/146

Ma tran [KG] cua ket cau suy ra t [e

GK ] tng

t nh [K], [M].

3.2.6 La chon cach thiet lap ma tran tnh chatCo 2 cach tnh gan ung cac ma tran khoi

lng, o cng hnh hoc, tai trong:- Phng phap s cap ch xet chuyen v thang.

- Phng phap tng thch xet ca chuyen vthang va chuyen v xoay.

Ve nguyen tac, phng phap tng thch choo chnh xac cao hn, v xet ay u va he thonghn cac phan nang lng lien quan en s lam

viec ong cua ket cau. Tuy nhien, trong thc te tho chnh xac cua phng phap tng thch khongtroi bao nhieu so vi phng phap s cap, nhngkhoi lng tnh toan th ln hn nhieu. ieu ochng to chuyen v xoay cua nut ong vai tro kemquan trong so vi chuyen v thang.

Phng phap s cap de dang hn, v cac matran xuat phat de tnh hn va so bac t do phai xetcung t hn.

Neu phng phap thu gon khoi lng cdung vi ma tran cng thiet lap bang FEM(tc la


100/146

ke en bac t do chuyen v xoay) th co the loaitr cac chuyen v xoay nay trong phng trnhchuyen ong. Khi o ma tran cng cung c rut

gon lai, goi la Static Condensation (kch thc matran cng thu nho lai). e minh hoa, ta viet laiphng trnh (3.2) trong o a sap xep lai cacchuyen v thanh 2 nhom: vt la thanh phan chuyenv thang va vo la thanh phan chuyen v xoay.

Phng trnh chuyen ong c viet lai dangma tran chia khoi (ma tran con):

{ }

{ }

{ }

{ }

{ }

{ }

=

=

0][][

][][St

S

Stt

t

ttt f

f

f

v

v

KK

KK

(3.45)

Trong o { } { }0=Sf , tc la cac moment nut anhoi bang 0, neu tac ong tren he ch la lc chkhong co moment tap trung at ngay tai nut.

Trong (3.45) co the bieu dien cac chuyen v xoay{ }

v theo chuyen v thang { }

tv :

{ } { }ttvKKv ][][

1

= (3.46)

Phng trnh th nhat cua ma tran con t (3.45):

{ } { } { }Sttttt fvKvK =+ ][][ { } { }

SttttttfvKKKK = ][]][[][ 1


101/146

hay { } { }Sttt fvK =][ (3.47)

trong o ][]][[][][ 1ttttt

KKKKK

= (3.48)

la ma tran o cng tng ng vi chuyen v thang(ma tran cng rut gon).

Nh vay, cac chuyen v xoay trong FEMco theloai tr va so bac t do thc s phai giai quyetgiam xuong. o la u iem ln cua phng phap

khoi lng thu gon.

Th du:

Trong th du tren, ta co:

3

2

1

s

s

s

f

f

f

=3

2

L

EI

22

22

623

263

3312

LLL

LLL

LL

3

2

1

v

v

v

][

K = 32

L

EI

22

22

62

26

LL

LL=

L

EI4

31

13

2L

L

EI EI

4EI

v1

v2 v3


102/146

1][

K =EI

L

32

31

13

Bieu dien chuyen v xoay theo chuyen v thang

(3.46):

v =

3

2

v

v= -

EI

L

32

31

133

2

L

EI

L

L

3

31v =-

L8

3

1

1

1v

Ma tran cng rut gon theo (3.48):

tK =

3

2

L

EI [ ]

L

LLL

8

38

3

3312 =3

2

L

EI

4

39

3.3 DAO ONG T DO KHONG CAN3.3.1 Phan tch tan so dao ong

T phng trnh (3.8), phan tch dao ong t donen vect tai trong ngoaip(t) = 0, ta co:

{ } { } { } { }0)(][)(][)(][ =++ tvKtvCtv Bo qua thanh phan lc can [C]= [0]

{ } { } { }0)(][)(][ =+ tvKtv (3.49)Do tnh chat tuan hoan nen chon nghiem co dang:

{ } { } )sin()( += tvtv (3.50)

trong o: { })(tv -the hien dang dao ong; { }v - labien o dao ong.


103/146

{ } { } )sin()( 2 += tvtv

Thay vao (3.49) tren ta co:

{ } { } { }0)sin(][)sin(][2

=+++ tvKtvM hay: { } { }0]][][[ 2 = vMK (3.51)

V { } 0 v , nen nh thc cua ma tran vuong N x Nphai triet tieu:

0]][][[det2

= MK (3.52)ay la phng trnh ai so bac N, do o co N

nghiem 21, 2

2 , ..., 2

N. Ly thuyet ma tran chng

minh: ma tran vuong thc, oi xng va xac nh

dng co cac tr rieng thc va dng.Vect tan so rieng nh sau:

{ }

=

N

2

1

(3.53)

T i ta se tm c chu k hay tan so dao ong tnhien cua cong trnh:

T = 2/ va f =T

1


104/146

Th du (E12-1)Tnh tan so rieng cua khung san cng: khoi

lng va o cng nh hnh ve (a). Cac he so cngtnh tren hnh ve (b).

Cac ma tran cua khung:

=0,20

5,1

00,1

][M (kip.s

2

/in)

=

520

231

011

600][K (kip.s/in)

Phng trnh ac trng (3.52):

1,0 kip.s2/in

1,5

2,0

v1

kin600

1200

1800

v1 =1 K = 60011

K = - 60021

K =031

K = -60012

K = 180022

K = -120032

K = 013

K = -120023

K = 300033

v =12

V =13

(a) (b)

v2

v3


105/146

0

2520

25,131

011

600

0,20

5,1

00,1

520

231

011

600][][22

=

=

=

B

B

B

MK

vi600

2=B

B3 5,5B2 + 7,5B 2 = 0

Nghiem la:B1 = 0,351 B2 = 1,61 B3 = 3,54

Do o: [] =

=

1,46

1,31

5,14

3

2

1

(rad/s).

3.3.2 Phan tch hnh dang mode cua dao ong

T phng trnh (3.51), ng vi moi tan sonta co mot vect rieng { }nv . Nhng v nh thc

(3.52) triet tieu, nen hang cua ma tran ch conN-1,do o ch co N-1 thanh phan cua { }v oc lap.Thng chon thanh phan au tien { } 1

1=

nv , khi o

vect chuyen v tr thanh:


106/146

{ }

=

=

Nn

n

Nn

n

n

n

v

v

v

v

v

v

1

22

1

at: ][][][ 2)( MKEn

n = (3.53)

Phng trnh (3.51) c viet lai:

=

0

0

0

1

2

)()(

2

)(

1

)(

2

)(

22

)(

21

)(

1

)(

22

)(

11

Nn

n

n

NN

n

N

n

N

n

N

nn

n

N

nn

v

v

eee

eee

eee

(3.54)

Viet lai (3.54) dang k hieu dung ma tran con:

][][

][)(

00

)(

01

)(

10

)(

11

nn

nn

EE

Ee

{ }

nv

0

1=

{ }

0

0

Tng ng vi 2 phng trnh:

{ }

0]][[

0]][[][

0)(

10)(

11

0

)(

00

)(

01

=+

=+

nnn

n

nn

vEe

vEE

(a)

Giai he phng trnh (a) tren ta c:

{ } ][][ )(01

1)(

00

nn

onEEv

= (3.55)

Dang dao ong (mode shape) th n c nhngha bi vect (khong th nguyen)


107/146

=

=

Nn

n

kn

Nn

n

n

n

v

v

v

1

1][

22

1

(3.56)

viknv la thanh phan (chuyen v) moc e so sanh.

Ma tran dang dao ong (Mode shape matric) la taphp cuaNvect dang dao ong:

[]= [[1] [2]... [N]] =

NNNN

N

N

21

22221

11211

(3.57)

Nh vay khi xac nh c [i] ta se biet c hnhdang dao ong cua mode th i.

Th du (E12-2)Xet lai th du trc, tm cac dang chnh cua

dao ong. Lay chuyen v tren cung bang 1. Haichuyen v tang di cua mode n tm theo (3.55):

n

n

3

2

= -

n

n

B

B

252

25,13

0

1

vi Bn =600

2

n


108/146

Ket qua nh hnh ve.

3.3.3 Phan tch tan so theo ma tran memNhieu bai toan dung ma tran mem [f] tien hn

ma tran cng [K]. Khi o can xac nh tan so riengtheo [f].

Phng trnh (3.51) c viet lai va bien oinh sau:

{ } { }0]][][[ 2 = vMK

(3.51)

Nhan 2 ve [f]: { } { }0]]][[]][[1

[2

= vMfKf

v 1][][ = Kf nen ][]][[ IKf = , ta co:

{ } { }0]]][[][1[ 2 = vMfI (3.58)

1.000 1.000 1.000

-2.570

2.470

-0.601

-0.676

0.644

0.300

Mode 1 Mode 2 Mode 3

=14.51 =31.12 =46.13


109/146

do { } 0 v , nen phng trnh tan so:

0]]][[][1

[det2

= MfI

(3.59)

3.3.4 Anh hng cua lc hoc3.3.4.1 Dao ong t do

Phng trnh dao ong (3.49) ke en o cng hnh

hoc co dang:{ } { } { } { }0)(][)(][)(][ =+ tvKtvKtv

G (3.60)

hay { } { } { }0)(][)(][ =+ tvKtvM

Phng trnh tan so:

0]][][[det2 = MK (3.61)

Lc doc lam cho ket cau b mem hn, nencac tan so rieng cung thap hn. Ket cau thnglam viec bat li hn di tac dung cua tai trong

ong trong thc te. Tng ng, cac dang dao ongchnh (mode shapes) cung b thay oi do lc doc.

3.3.4.2 Tai trong ti han (gay mat on nh)

Khi lc doc at gia tr ti han N0 th ket cau

khong dao ong ( = 0). Lc quan tnh cung triettieu. Phng trnh (3.60) tr thanh:


110/146

{ } { } { }0)(][)(][0

= tvKtvKG (3.60)

][0G

K - Ma tran cng hnh hoc, ng vi lc doc

N0(x), vi cac he so xac nh bi:ij

Gk 0 = L

jiodxxxxN

0

'')()()( (3.62)

Goi tham so tai trong (load factor)G

=)(

)(0

xN

xN (3.63)

viN(x) la lc doc do tai trong ang xet gay ra th

ta co:ij

Gk 0 =ij

GG k do o: ][][

0 GGGKK = (3.64)

The (3.64) vao (3.60):

{ } { }0)(]][][[0

= tvKKGG

(3.65)

v { } { }0)( tv nen phng trnh xac nh tham so tai trong

G

0][][det0

=GG

KK (3.66)

Tai trong ti han thap nhat ng vi 1G = min la

co y ngha thc te. Dang mat on nh tng ngvi vector chuyen v

1v , c tm bang cach the 1

G

vao (3.65).


111/146

Mat on nh vi tai trong ieu hoaXet tai trong tac dung co dang:

tptpo

sin)( =

trong o: la tan so cua tai trong tac dung.Phng trnh can bang dao ong khong can:

tpvkkvvmoG

sin=+

Phng trnh nay co nghiem: tvtv sin)( = tvtv sin)( 2=

Thay cac nghiem nay vao tren ta co:

oGpvkvkvm =+ 2

o cng ong cua he c nh ngha bi:

mkk 2

Thay vao bieu thc tren va bieu dien o cng hnhhoc la mot ham cua he so tai trong G , ta co:

oGoGpvkk =

Neu bien o tac dung cua tai trong tien danen 0 th phan ng (chuyen v) van co the khac 0

neu nh thc cua ma tran vuong bang 0. V vayieu kien mat on nh oi vi ket cau chu taitrong ieu hoa la:

0=GOGkk

Khi tai trong thoi tac dung, phng trnh tac dungco the viet:


112/146

02 = vkmk

GOG

Ta thay s to hp cua tai trong mat on nhG

va tan so dao ong 2 se thoa man phng trnh trrieng. Nh vay khi chu tai trong ieu hoa ng vimot tan so nao o th he co the mat on nh ngayca khi bien o lc bang 0.

3.3.5 ieu kien trc giao (Orthogonality)3.3.5.1 Cac ieu kien c ban

Phng trnh dao ong (3.51) viet lai cho tan so

n va

m (gia thiet

n

m )

{ } { }nnn vMvK ][][2

= (3.67){ } { }

mmmvMvK ][][

2= (3.68)

Nhan trc { }Tmv cho (3.67):

{ } { } { } { }n

T

mnn

T

mvMvvKv ][][

2= (3.69)

Chuyen tr (3.69) ca hai ve, chu y][][],[][ MMKK

TT == v chung oi xng:

{ } { } { } { }m

T

nnm

T

nvMvvKv ][][

2= (3.70)

Nhan trc { }Tnv cho (3.68):

{ } { } { } { }mTnmmTn vMvvKv ][][ 2= (3.71)


113/146

T (3.70), (3.71) suy ra: { } { } 0][)( 22 = mT

nnmvMv

Vn

m

nen ta co ieu kien trc giao au tien:

{ } { } 0][ =mT

n vMv (3.72)The (3.72) vao (3.71) suy ra ieu kien th 2:

{ } { } 0][ =m

T

nvKv (3.73)

Bieu dien ieu kien trc giao theo mode, ta co:

{ } { }{ } { } 0][

0][

==

m

T

n

m

T

n

K

M

nmnm

(3.74)

Chu y: ieu kien trc giao ch dung cho 2 mode cotan so khac nhau: n m

3.3.5.2 Chuan hoa theo ma tran khoi lngVector bien o { }nv c chuan hoa theo ma

tran khoi lng thanhn

thoa man ieu kien:

{ } { } 1][ =nT

n M (3.75)

Goi { } { } nnT

n MvMv =][ = scalar. Th vectorchuan hoa se la:

{ }nnn

Mv = (3.76)

Khi o ma tran vuong gom N vectorn

se

thoa man:IM

T

= ][ (3.71)


114/146

Cac vectorn

c goi la cac vector trc chuan

(Orthonormal).

3.4 PHAN TCH PHAN NG ONGPhng phap dung e phan tch phan ng ong

c dung la phng phap chong chat mode. Noidung chnh cua phng phap nay la bien he dao

ong co he n phng trnh vi phan thanh dang heong co n phng trnh vi phan tach ri. e dungphng phap tren ta phai tm hieu toa o chuan,sau o thiet lap phng trnh chuyen ong tach ricua he khong can va co can.

3.4.1 Toa o chuan (Normal Coordinates)

v1

v2

v3

11v 12v 13v

21v

31v

22v

32v

23v

33v

v =Yv = Y v = Y v = Y1 1 1 2 32 2 3 3

= + + + ....


115/146

Vect chuyen v [v] cua heNbac t do co thetao ra bang cach to hp tuyen tnh cua N vect cs a biet nao o. Tuy nhien, neu chon cac vect

c s la cac dang chnh ( Mode Shapes) cua daoong t do th se co nhieu u iem do tnh trcgiao cua chung. Cac dang chnh ong vai tro tngt nh cac ham lng giac cua chuoi Fourier, vachuyen v cua he co the xap x kha tot vi mot soso hang cua chuoi.

Xet dam console nh hnh ve e minh hoa.Vect chuyen v ng vi ham dang [n] la ][ nv xac

nh bi cong thc:

][ nv = [n] Yn (t) (3.78)

trong o: Yn(t) la bien o (toa o suy rong) ng viham dang [n]

Chuyen v toan phan [v] c phan tch thanh

tong cac dang chnh nh sau:

[v]=[1]Y1 + [2]Y2+ ... +[n]Yn = =

N

nnn

Y1

]][[ (3.79)

Dang ma tran: [v] = [] [Y(t)]

[]: ma tran vuong cua cac dang chnh.


116/146

[Y] : vec t cac toa o suy rong, cung c goila cac toa o chuan. Cac thanh phan Yn cua vect[Y] co the tm de dang nh tnh trc giao cua cac

ham dang nh sau:Nhan 2 ve cua (3.79) vi [n]

T [M]:

[n]T [M][v] = [n]

T [M] [][Y] (3.80)

ap dung tnh trc giao [i]T [M][j] = 0 vi i j, ve

phai (3.80) c trien khai:[n]

T[M][][Y]=[n]T[M][1][Y1]+[n]

T[M][2][Y2] +...+ [n]

T [M][n][Yn] = ]][][[][ nnT

nYM (3.81)

The (3.81) vao (3.80):

[n]T [M][v] = [n]T [M][n][Yn]

hay Yn =]][[][

]][[][

n

T

n

T

n

M

vM

(3.82)

Nh vay, moi toa o chuan Yn , n =1..N, eu

c xac nh theo (3.82)

3.4.2 Phng trnh chuyen ong tach ri(uncoupled) cua he khong can

Phng trnh chuyen ong khong can cua henhieu bac t do:


117/146

)]([]][[]][[ tpvKv =+ (3.83)

The ]][[][ Yv = t (3.79) vao (3.83):

)]([]][][[]][][[ tpYKYM =+

(3.84)Nhan trc 2 ve cho [n]

T:

)]([][]][][[][]][][[][ tpYKYMT

n

T

n

T

n =+ (3.85)

Do tnh trc giao nen ta co:

)]([][]][[][]][[][ tpYKYM TnnnT

nnnT

n =+ (3.86)

at cac k hieu mi:

)]([][)(

]][[][

]][[][

tptP

KK

MM

Tnn

n

T

nn

n

T

nn

=

=

=

(3.87)

lan lt goi la: khoi lng, o cng va tai trongsuy rong cho dang dao ong chnh th n. Phngtrnh (3.86) c viet lai:

)()()( tPtYKtYM nnnnn =+ (3.88)ay la phng trnh dao ong cho he mot bac tdo cho dang chnh n.

T phng trnh ieu kien trc giao (3.67): ]][[]][[ 2 nnn vMvK =


118/146

The nnn Yv ][][ = vao va n gian i Yn cho 2 ve:

]][[]][[2

nnnMK =

(3.89)

Nhan trc [n]T cho 2 ve cua (3.89):

]][[][][][][2

n

T

nnn

T

nMK =

hay: Kn = n2 Mn (3.90)

Nh vay, viec dung toa o chuan a bien he Nphng trnh vi phan dao ong cua he co Nbac tdo ve dang gom N phng trnh vi phan tach rinhau. ng vi moi dang dao ong chnh th phanng ong cua he c xac nh bang cach chong

chat cac phan ng cua cac dang chnh (mode).Phng phap c goi la phng phap chong chatmode (Mode Superposition Method).

3.4.3 Phng trnh chuyen ong tach ri cua heco can+ Thiet lap phng trnh

Phng trnh chuyen ong cua he co can:

)]([]][[]][[]][[ tpvKvCv =++ (3.91)

Documents

BG_DONG_LUC_HOC