DLHCT_DoKienQuoc

ONG LC HOC KET CAU TS. O KIEN QUOC

Chng 1. M AU 1

CHNG 1

M AU

1.1 Nhiem vu mon hoc ong lc hoc ket cau la mot lnh vc cua c hoc, nghien cu cac phng

phap phan tch phan ng (noi lc, ng suat hoac chuyen v, van toc, gia toc) trong ket cau khi chu tac dung cua tai trong ong.

1.2 Tai trong ong Khai niem:

Tai trong ong la tai trong thay oi theo thi gian ve tr so, phng, v tr, gay ra ng suat, chuyen v cung thay oi theo thi gian. Phan loai:

- Tai trong tien nh (Deterministic Loads): la tai trong biet trc c qui luat bien oi theo thi gian P = P(t). Th du: Tai trong ieu hoa, chu ky, khong chu ky, xungc mo ta theo qui luat cho trc.

- Tai trong ngau nhien (Random, Stochastic Loads): la tai trong biet trc c qui luat xac suat va cac ac trng xac suat nh gia tr trung bnh, o lech chuan Th du: lc gio, lc song, lc ong at.

Bai toan ong lc hoc ket cau chu tai trong ngau nhien c giai quyet bang ly thuyet dao ong ngau nhien (Random Vibration Theory). Cac thong tin can tm bao gom ng suat, chuyen v, phan lc cung mang tnh ngau nhien vi cac ac trng xac suat tr trung bnh, o lech chuan

Noi chung, cac tai trong trong thc te eu mang tnh chat ngau nhien mc o khac nhau, va c xac nh bang phng phap thong ke toan hoc.

Cac quan iem phan tch ong lc hoc:

Phan tch tien nh, phan tch ngau nhien va phan tch m (Fuzzy Analysis).

1.3 ac thu cua bai toan ong

Xet dam n gian, chu tai trong tnh va ong nh hnh ve:


Chng 1. M AU 2

Bai toan tnh: noi lc c xac nh t s can bang vi ngoai lc, khong can dung phng trnh ng an hoi nen mang tnh chat n gian. Do o, ng suat va chuyen v khong phu thuoc thi gian.

Bai toan ong: ngoai lc bao gom lc quan tnh phu thuoc vao phng trnh ng an hoi y = y(x,t). V vay, dan ti phng trnh vi phan ao ham rieng, phc tap ve toan hoc, khoi lng tnh toan ln, phai bat au t viec xac nh y(x,t).

Bai toan tnh (bao gom ca bai toan on nh) la trng hp ac biet cua bai toan ong khi lc quan tnh c bo qua.

1.4 Bac t do cua ket cau

Bac t do ong lc hoc (Number of dynamics degrees of freedom) cua ket cau la so thanh phan chuyen v phai xet e the hien c anh hng cua tat ca cac lc quan tnh.

Bac t do c nh ngha trong s lien quan en lc quan tnh va do o lien quan en khoi lng. So khoi lng cang nhieu th cang chnh xac nhng cung cang phc tap.

Chu y: Bac t do ong lc hoc khac vi bac t do trong bai toan tnh (so chuyen v nut cua ket cau).

Th du: cho ket cau nh hnh ben, neu P la tai trong tnh th so bac t do la 3, neu P la tai trong ong th so bac t do la vo cung.

Trong thc te, cac ket cau eu co khoi lng phan bo nen co vo han bac t do, viec giai bai toan rat phc tap nen tm cach ri rac hoa he.

P(t)

P

Bai toan tnh

Bai toan ong

q(t)= y(t)

P


Chng 1. M AU 3

1.5 Cac phng phap ri rac hoa

1.5.1 Phng phap khoi lng thu gon (Lumped Mass)

Thay the he co khoi lng phan bo (a) thanh cac khoi lng tap trung (b) theo nguyen tac tng ng tnh hoc. ay la phng phap thng c dung trong he ket cau phc tap. Khoi lng thng c thu gon ve iem nut (th du nh he dan).

So bac t do cua he tuy thuoc vao gia thiet ve tnh chat chuyen v cua he va tnh chat quan tnh cua cac khoi lng mi. Chang han, xet he (b) la he phang:

Neu bien dang doc truc va mi co quan tnh xoay: 9 BTD (3BTD/mass).

Neu coi mi la mot iem (khong co quan tnh xoay): 6 BTD (2 chuyen v thang/mass).

Bo qua bien dang doc truc nen ch co chuyen v ng: 3 BTD (1 chuyen v ng/mass).

Chu y: o phc tap cua bai toan ong lc hoc phu thuoc vao so bac t do.

1.5.2 Phng phap dung toa o suy rong (Generalised Coordinates)

Gia s ng an hoi la to hp tuyen tnh cua cac ham xac nh i(x) co bien o Zi nh sau:

=

=1)(),(

iii xZtxy (*)

trong o:

i(x) : Ham dang (Shape Functions)

ZI(t) : Toa o suy rong (Generalised Coordinates)

P(t)

m(z)

P(t)

m m m1 2 3

(a)

(b)


Chng 1. M AU 4

Ham dang i(x) c tm t viec giai phng trnh vi phan ao ham rieng, hoac do gia thiet phu hp vi ieu kien bien. Khi tnh toan thng gi lai mot so so hang au tien cua chuoi (*) va he tr thanh hu han bac t do (Zi ong vai tro bac t do).

1.5.3 Phng phap phan t hu han (Finite Element Method - FEM)

ay la trng hp ac biet cua phng phap toa o suy rong, trong o:

- Zi la cac chuyen v nut (Toa o suy rong).

- i(x) la cac ham noi suy (Interpolation Functions) cac phan t - Ham dang.

Thng cac ham noi suy i(x) c chon giong nhau cho cac phan t (ng vi cung mot bac t do) va la ham a thc nen viec tnh toan c n gian. ac biet, do tnh chat cuc bo cua cac ham noi suy nen cac phng trnh t lien ket (uncoupled) vi nhau lam giam nhieu khoi lng tnh toan.

L

Z2

Z3

y(x)

1(x) Z1

3(x)

2(x)

i x L

i=1,2,...,n i(x) =sin

3 2 1 4 5

v3=1

3v(c) 3v(b)

a b c d

3=1

3(c) 3(b)


Chng 1. M AU 5

1.6 Cac phng phap thiet lap phng trnh vi phan cua chuyen ong

1.6.1 Nguyen ly DAlembert

Xet khoi lng mi (i=1,n) chu tac ong cua lc Pi(t) co chuyen v vi(t) va gia toc )(tvi&& . Neu at them lc quan tnh th khoi lng mi se can bang:

0)()( = tvmtP iiir&&

r (1.1)

Neu he co n bac t do th se co n phng trnh vi phan chuyen ong.

1.6.2 Nguyen ly cong kha d

Cho khoi lng mi (i=1,n) mot chuyen v kha d vi , cong kha d W cua cac lc tac dung len mi (can bang) tren chuyen v vi phai triet tieu:

= 0)]()([ iiii vtvmtPrr

&&r

(1.2)

Nguyen l cong kha d thch hp cho he phc tap gom cac khoi lng iem va khoi lng co quan tnh xoay. Cac so hang trong phng trnh la cac vo hng (scalar) nen lap phng trnh n gian so vi phng trnh vector.

Neu cho he cac chuyen v kha d iv lan lt theo cac bac t do se thu c n phng trnh vi phan cua chuyen ong.

Ky hieu cong kha d cua ngoai lc Pi(t) la W, t (1.2) ta co bien phan cong kha d:

== iiii vtvmvtPW )]()( && (1.3)

1.6.3 Nguyen ly Hamilton (page 344, [1])

Xet he gom cac khoi lng mi (i=1, n) co cac chuyen v vi(t) hai thi iem t1 va t2, chuyen v co cac tr so vi(t1) va vi(t2) tng ng vi hai ng bien dang (b) va (c). ng bien dang (d) ng vi t = t1 + t < t2. ng bien dang that tuan theo nh luat II Newton. ng lech trung vi ng that tai hai thi iem t1 va t2:

v1(t1) =v1(t2) =0 (1.4)

ong nang cua he tai thi iem t:

)(2

1

1

2i

n

i

ii vTvmT && == =

Bien phan cua ong nang T tng ng vi bien phan cua chuyen v vi:


Chng 1. M AU 6

T =

=

n

ii

i

vV

T

1

& = === i i

iiii

ii

n

iiii v

dt

dvm

dt

dvvmvvm &&&&

1

(1.5)

Mat khac, ta co ong nhat thc:

iiiiii vdt

dvvvvv

dt

d &&&&& +=)(

Nhan ca hai ve vi mi va lay tong cho toan he:

+=i

iii

i

iii

i

ii vdt

dvmvvmvvm

dt

d &&&& )(

WTvvmdt

d

iiii += )( & (1.6)

Nhan hai ve vi dt va lay tch phan t t1 en t2:

m 1 m 2 m 3 m 4

v v v 1 2 3 v 4

v (t ) 1 1

v (t ) 1 2

t=t 1

t=t 2

t=t + t < t 1 2 v(t + t) 1 1

v 1

2 v

3 v 4 v

that

(a)

(b)

(c)

(d)

1

1 t t 2 t + t 1

1 1 v(t + t) v (t ) 1 2 1 v (t ) 1

v (t) 1

t

v (t + t) 1 1

ng lech v(t)+ v 1 1

ng Newton (that)


Chng 1. M AU 7

+=2

1

2

1

)(t

t

t

tiiidtWTvvm

Theo tren v vi(t1) = vi(t2) = 0 vi moi i nen ve trai triet tieu:

0)(2

1

=+t

t

dtWT (1.7)

Neu ngoai lc tac dung tren he gom lc bao toan (lc the) va lc khong bao toan (th du lc ma sat) th bien phan cua cong ngoai lc W c tach ra hai thanh phan:

W = Wc + Wnc (1.8)

oi vi lc bao toan th cong cua lc bang o giam the nang cua he nen:

Wc = -V (1.9)

vi V la bien phan cua the nang.

The (1.9) vao (1.8):

W = -V + Wnc (1.10)

The vao (1.7):

0)(2

1

2

1

=+ t

t

t

t

ncdtWdtVT (1.11)

ay la nguyen ly bien phan cua Hamilton, trong o:

T: ong nang cua he.

V: The nang cua he, gom the nang bien dang an hoi va the nang cua lc bao toan.

Wnc : Cong cua lc khong bao toan (lc can, ma sat, ngoai lc...)

Y ngha

Cong thc (1.7) c viet lai:

0)(2

1

=+t

t

dtWT (1.12)

Nh vay, trong tat ca cac ng chuyen ong trong khoang thi gian t t1

en t2 th ng lam cho tch phan 0)(2

1

=+t

t

dtWT co gia tr dng (cc tieu) la

ng chuyen ong tuan theo nh luat Newton.


Chng 1. M AU 8

Bai toan tnh T = 0 th (1.7) tr thanh:

02

1

=t

t

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

ay la nguyen ly the nang cc tieu trong bai toan tnh (Neu mot he can bang on nh th the nang cua he cc tieu).

Chu y: Nguyen ly Hamilton cung la mot phng phap nang lng, trong o khong dung trc tiep en lc quan tnh va lc bao toan. Dung thch hp cho he phc tap, khoi lng phan bo.

Nhan xet: Ca 3 phng phap DAlembert, Virtual Work va Hamilton eu dan en phng trnh chuyen ong giong nhau (eu cung mang ban chat nh luat II Newton).

Phng trnh Lagrange

Goi q1, q2,...., qn la cac toa o suy rong. Trong cong thc (1.11) ta co: ),....,,,,....,,( 2121 nn qqqqqqTT &&&=

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

nqnqqnc

QQQW +++= ....21 21

, vi Qi la lc suy rong khong bao toan.

The vao (1.11):

0).............(2

1

111

1

1

1

1

1

=++

++

+

++

t

t

nnn

n

n

n

n

n

dtqQqQqq

Vq

q

Vq

q

Tq

q

Tq

q

Tq

q

T &

&&

& (*)

Tch phan cac so hang cha van toc iq& tng phan:

= 2

1

2

1

2

1

11

1

1

1

)(((t

t i

t

t

t

t

dtqq

T

tq

q

Tdtq

q

T&

&&

& (1.14)

The vao bieu thc (*):

=

+

+

=

2

1

0)(1

t

t

n

i

ii

iii

dtqQq

V

q

T

q

T

T

& (1.15)

V qi la tuy y nen:

i

iii

Qq

V

q

T

q

T

T=

+

)(&

(1.16)

ay la phng trnh Larange, dung c cho he tuyen tnh va phi tuyen.


Chng 1. M AU 9


Chng 2. HE MOT BAC T DO 9

CHNG 2

HE MOT BAC T DO 2.1 THIET LAP PHNG TRNH CHUYEN ONG 2.1.1 Mo hnh he mot bac t do

Mo hnh n gian nhat cua he mot bac t do (Single Degree of Freedom system - SDOFs), gom cac ac trng vat ly tap trung (Concentrated Properties):

Khoi lng: m o cng: k He so can: c Lc kch ong: p(t)

Chu y: He mot bac t do co cac ac trng phan bo ve m, k, c, p(t) eu co the a ve mo hnh co cac ac trng vat ly tap trung (he mot bac t do suy rong). 2.1.2 Cac phng phap thiet lap phng trnh chuyen ong

2.1.2.1 Nguyen ly DAlembert

p(t) + fS + fI + fD =0

hay )(tpkvvcvm =++ &&& (2.1)

p(t)

f f

f

D

S I

Cac lc tac dung

c

k

v(t)

p(t) m

Mo hnh SDOFs



2.1.2.2 Nguyen ly cong kha d

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

W = p(t)v + fS v + fI v + fD v = 0

hay 0)]([ =+ vtpkvvcvm &&&

v v 0 nen thu c phng trnh chuyen ong giong nh (2.1).

2.1.2.3 Nguyen ly Hamilton

ong nang cua he: 22

1vmT &= , bien phan ong nang vvmT && =

The nang bien dang an hoi cua lo xo: 22

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 d v): 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:

=2

1

2

1

2

10

t

t

t

t

t

t

vdtvmvvmdtvvm &&321

&&&& (2.3)

the (2.3) vao (2.2) ta c: 0)]([2

1

= +t

t

vdttpkvvcvm &&&

(2.4)

v v tuy y nen bieu thc: 0)( =+ tpkvvcvm &&& co dang giong vi (2.1)

O v

f = kv s

Lc

Chuyen v



Nhan xet: Ca 3 phng phap cho cung ket qua v cung da tren nh luat quan tnh cua Newton. Trong trng hp cu the nay nguyen ly DAlembert la n gian nhat.

2.1.3 Anh hng cua trong lc

He tren hnh co phng trnh chuyen ong: W)t(pkvvcvm +=++ &&&

trong o W la trong lng cua khoi cng.

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

vv st +=

Thay bieu thc cua lc an hoi vkkkvf sts +==

vao phng trnh chuyen ong: Wtpvkkvcvm st +=+++ )(&&&

Mat khac stkW = nen phng trnh cuoi cung thu c:

)(tpvkvcvm =++ &&&

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 vi phan chuyen ong van co dang (2.1).

Nh vay, trong lc khong anh hng en phng trnh vi phan chuyen ong.

c k

m

v(t)

p(t)

(W)

S f fD

p(t)

fI

st

Anh hng cua trong lc

fS fD

fI

p(t)

W

W v(t) v(t)

v (t)



2.1.4 Anh hng cua s rung ong goi ta

Da vao hnh ve, ta co phng trnh can bang lc:

0=++ SDI fff trong o lc quan tnh co gia tr: tI vmf &&= vi gt vvv += la tong cua v la chuyen v

uon va vg la chuyen v goi ta (mat at). Thay cac gia tr lc vao phng trnh can bang ta co:

0=+++ kvvcvmvm g &&&&& hay:

)(tPvmkvvcvm effg =++ &&&&& (2.5)

Ket luan: Trong phng trnh tren geff vmtP &&=)( la tai trong do rung ong goi ta.

Nh vay s rung ong cua mat at tng ng nh lc kch ong effP tac dung tai

vat nang. 2.1.5 He mot bac t do suy rong (Generalised SDOF System)

Xet he co ac trng vat ly phan bo (m, EI), thc chat co vo han bac t do. Neu coi he ch dao ong vi mot ham dang nao o th he tr thanh 1 bac t do. Can tm cac ac trng vat ly tap trung cho he 1 bac t do o.

l

x

x N

vg(t)

v (x,t) e(t)

z(t)

m(x)

EI(x)

v(x,t)

chuyen v O

t

vg(t)

v

vt

fI

fC 0.5fS 0.5fS



Gia s he chu rung ong ngang vg(t) cua goi ta (do ong at chang han). Dung nguyen ly Hamilton e thiet lap phng trnh chuyen ong. at:

v(x,t) = (x) Z(t) (2.6)

trong o:

(x) - Ham dang (Shape Function)

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

ong nang cua he:

[ ] dxtxvxmT tl 20

),()(2

1&=

dxvtxvxmT ttl

&& ),()(0

= (2.7)

The nang uon:

[ ] dxtxvxEIVl

f2

0

),(")(2

1=

dxvtxvxEIVl

f "),(")(0

= (2.8)

o co ngan cua thanh:

[ ] dxtxvtel

2

0

),('2

1)( = (2.9)

The nang lc doc (chon v tr ban au cua N co the nang bang 0 ):

[ ] dxtxvNNeVl

N2

0

),('2 ==

hay dxvtxvNVl

N =0

'),('

(2.10)

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

=2

1

0)(

t

t

dtVT (*), vi V = Vf + VN

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 tch phan theo bien t. e lam xuat hien cac tha so Z trong 2 so hang au, ta tien hanh tch phan tng phan:

====2

1

2

1

2

1

2

1

2

1

2

1

)(

t

t

t

t

t

t

t

t

t

t

t

t

dtZZdtZZZZdtZdt

dZdt

dt

dZZdtZZ &&&&&&&&&&& (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:

[ ] =+2

1

0)(****t

ttG ZdttpZkZkZm && (2.16)

trong o: =l

dxxmm0

2* )( : Khoi lng suy rong

=l

dxxEIk0

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

=l

G dxNk0

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

z z

vt

O vg

v v



=l

gt dxxmtvtp0

* )()()( && : Tai trong tng ng suy rong

(2.17)

V Z bat ky nen lng trong dau ngoac triet tieu. Ta thu c phng trnh vi phan chuyen ong cua he mot bac t do 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 th 0* =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 phap Rayleigh.

Chu y:

Neu thanh chu lc kch thch phan bo p(x,t) va lc doc N(x) th cong thc tnh lc kch thch suy rong (lc tap trung) p*(t) va o cng hnh hoc k*G lan lt la:

=l

dxxtxptp0

* )(),()( (2.21)

=l

G dxxxNk0

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

=l

dxxxcC0

2* )]()[( (2.23)

p(x,t)

c(x)



Th du: Example E8.3, page 144, [1]

Thiet lap phng trnh vi phan dao ong cua he mot bac t do suy rong vi cac at trng vat ly (khoi lng, o cng ) phan bo eu theo chieu cao nh tren hnh ve. Cho biet phng trnh ng an hoi (ham dang ) c chon nh sau:

L

xx

2cos1)(

= (a)

Giai:

Ap dung (2.17) ta co khoi lng va o cng suy rong:

( ) LmdxL

xmdxmm

LL

228.02

cos10

2

0

2* =

==

(b)

( )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):

)(364.02

cos1)()()(00

* tvLmdxL

xtvmdxmtvtP g

L

g

L

g &&&&&& =

==

(d)

Bo qua lc doc truc, phng trnh can bang dao ong:

)(364.0)(32

)(228.03

4

tvLmtZL

EItZLm g&&&& =+

(e)

Neu xet lc doc truc N th ta co o cng hnh hoc suy rong:

L

x

x N

vg(t)

v (x,t) e(t)

z(t)

m

EI

v(x,t)

chuyen v O

t



( ) =

==L L

G L

Ndx

L

x

LNdxNk

0

2

0

2

2*

82sin

2'

(f)

o cng suy rong ket hp:

L

N

L

EIkkk G

832

2

3

4*** == (g)

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

EIN cr

== (h)

ay la tai trong mat on nh that s cho cot consol chu tai trong phan bo eu, bi v ham dang c rut ra t (a) la dang mat on nh that cua ket cau. Thay (h) vao (f) ta co the bieu dien o cng hnh hoc bi:

crG N

N

L

EIk

3

4*

32

= (i)

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

)(364.0)(132

)(228.03

4

tvLmtZN

N

L

EItZLm g

cr

&&&& =

+

(j)

Do o, bat ky hnh dang nao thoa man ieu kien bien hnh hoc cua ket cau eu c rut ra t ham dang )(x . Neu ham nay c cho bi dang parabolic

2

2

)(L

xx =

Khi nay o cng an hoi suy rong tr thanh:

3

0

2

2

* 42

L

EIdx

LEIk

L

=

=

va o cng hnh hoc suy rong:

L

Ndx

L

xNk

L

G3

42

0

2

2

* =

=

Trong trng hp nay, tai trong ti han c rut ra t ** Gkk = la:

23

3

4

34

L

EIL

L

EIN cr == (l)

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



2.2 DAO ONG T DO

2.2.1 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 thch p(t) = 0 th:

0)()( =++ kvtvctvm &&& (a)

Nghiem co dang: v(t) = Gest

The vao (a) ta c:

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

at m

k=2 th (b) dan ti:

s2 + m

c + 2 = 0 (c)

(c) la phng trnh ac trng, nghiem s cua (c) tuy thuoc vao he so can c.

2.2.2 Dao ong t do khong can - c = 0

Khi o (c) co nghiem: s = i do o nghiem cua (a) la:

v(t) = G1eit + 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(v& sint + v(0)cost (2.24)

Co the viet (2.24) di dang khac:

Imaginary

1

1 Real

e i t

t O

e = cos t isin t i t Cong thc Euler:



v(t) = cos(t - ) (2.24')

Vi bien o 2

2 )0()]0([

+=

v

v& va pha ban au = tan-1

)0(

)0(

v

v

&

(2.25)

chu ky: T = f

12=

(2.26)

2.2.3 Dao ong t do co can- c 0

Nghiem cua (c): s = 22

22

m

c

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 = ccr

ccr = 2m th 02

2

2

=

m

ccr

s = =m

ccr

2

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, khong co dao ong (Oscillation).

v(t)

v(0)

T =

v(0)

2

t

v(0) v(0)

v(t)

t O



- 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



v(t) = e-t (D

vv

)0()0( +& sinDt + v(0)cosDt) (2.31)

Chu ky dao ong co can: T = D2

The vao (2.29):

)2exp()exp(1 Dn

n Tv

v

==+

o giam Loga:

21 1

22ln

===+ Dn

n

v

v

= 21

2

2 , vi nho.

21......!2

)2(21

22

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 c xac nh theo cong thc:

c = 2m (2.34)

- Can nhieu (Overdamping)

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

cang ln th chuyen ong ve v tr can bang cang cham.



2.3 PHAN NG VI TAI TRONG IEU HOA

2.3.1 He khong can

Lc kch thch: tptp sin)( 0=

Phng trnh dao ong: tptkvtvmo

sin)()( =+&& (a)

Nghiem thuan nhat (qua o): tBtAtvh

cossin)( +=

Nghiem rieng dang (on nh): tGtvp

sin)( =

The vao (a) rut ra:21

1

=

k

pG o vi:

=

Vay nghiem tong quat:

tk

ptBtAtvtvtv oph

sin1

1cossin)()()(

2++=+= (2.35)

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

0,1

12

=

= Bk

pA o

(2.36)

the vao (2.35) ta c: )sin(sin1

1)(

2tt

k

ptv o

=

(2.37)

T so phan ng (Response Ratio):

)sin(sin1

1)()()(

2tt

kp

tv

v

tvtR

ost

===

Trong thc te, lc can lam cho so hang sau bien mat sau mot khoang thi gian ngan. Khi o he 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: tBtAetv DDt

h cossin()( += )

Nghiem rieng: tGtGtvp

cossin)(21

+=

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 dao ong theo nghiem rieng. Dung vector quay tren gian o Argrand, ta tm c:

2

121

222

1

2])2()1[(

=+=

tgk

po (2.41)

va phng trnh dao ong on nh:

)sin()( = ttv (2.42)

- He so ong (Dynamic Magnification Factor):

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

Imaginary

Real t t

2

(1 ) +(2) 2 2 p k

o 1

(1 ) +(2) k p o

2 2 2

2

2

Bieu dien dao ong bang vect quay

=0

=0.2

=0.5

=0.7

=1.0 1

2

3

4

D

0 1 2 3

k m



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 th D=1

oi vi he co can khac 0, th Dmax 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 the coi:

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:

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

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

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

p(t) = p0 sin t

f

v

Phan lc nen



tptp o sin)( =

Chuyen ong on nh (Steady-State Displacement):

)sin()( = tDk

ptv o

Van toc: )cos()( = tDk

ptv o&

Lc an hoi: )sin()( == tDptkvf os

Lc can: )cos(2)cos()( === tDptk

Dcptvcf o

oD &

V fS(t) va fD(t) lech pha 90o, nen bien o phan lc nen la:

( )[ ] 212max2max2max 21 +=+= Dpfff oDS Ty so truyen lc (Transmissibility Ratio-TR ), c nh ngha:

( )

( ) ( )[ ] )21(

21

21

22

2max

+=

+==

D

Dp

fTR

o TR = D neu = 0 (khong can) (2.47)

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 lap dao ong khi > 2 ==> Khong nen dung damper

2. Xet khoi lng m, chu kch ong cua goi ta

Chuyen ong tng oi cua m so vi goi ta cho bi phng trnh:

)sin()( 2 = tDvtv go

Chuyen ong toan bo vt bang tong vector cua vg va v:

( ) )sin(21)( 2 += tDvtv got

Ty so truyen dao ong Vibra. Transmi. Ratio

0 1

= 0.33

= 0.2 = 0

2 3 2

1

2

3

= 0.25

TR

vt

m

vg (t)=vg sin t



Ty so truyen:

( )2max 21 +== DTRvv

go

t

(2.48)

Ty so truyen dao ong giong nhau cho ca 2 trng 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 va cau rat nhieu nhp.

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

2.12s

v

LT

p===

Ty so chu ky:

Bien o dao ong ng cua oto la:

994.0606.0

572.0===

pT

T

L = 40fl = 12,2m

vt

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

W=4000lb=1816kG

1,2in=3,05cm

mat cau



( )( ) ( )

( )( ) ( )

)(009.5944.04.02944.01

944.04.2105.3

21

212

1

22

221

222

2

maxcmTR vvv gogo

t=

+

+=

+

+==

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 s can 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 s

v = L/Tp = 10.97/0.572 = 19.18 m/s = 69km/h.

b. Bien o toan phan tvmax cua xe khi cong hng:

( )( ) ( )

( )( )

( ) ( ))(88.4

4.02

4.02105.3

2

21

2

21

21

21

1

2221

2

2

21

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

22

1

222

2

2 =

+

+=

+

+=

===

===

TR

sT

T

sv

LT

p

p

)(72.4546.105.3max cmTRv vgot ===



2.4 PHAN NG VI TAI TRONG CHU KY

2.4.1 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:

dttT

ntp

Tb

dttT

ntp

Ta

dttpT

a

pT

pp

n

pT

pp

n

pT

p

o

)2

sin()(2

)2

cos()(2

)(1

0

0

0

=

==

=

(2.51)

2.4.2 Phan ng vi tai trong chu ky

Khi mot tai trong chu ky c phan tch ra chuoi Fourier (2.50) th phan ng cua he c xac nh theo nguyen ly chong chat. Bo qua nghiem qua o, trong trng hp he khong can, phan ng nh sau:

- Vi so hang tai trong dttT

nb

p

n )2

sin( th phan ng cua he theo (2.37) la:

)sin(1

1)( 12 tnk

btv

n

n

n

==

vi

1n

T

nT

p

nn === ;

PT

21 = : tan so vong c ban cua tai trong.

v(t)

t O

Tp Tp Tp



- So hang tT

na

p

n

2cos , phan ng c xac nh tng t:

tnk

atv

n

n

n 12cos

1

1)(

=

- So hang ao - tai trong hang so, gay chuyen v tnh: k

av oo =

- Phan ng toan bo

( )

+

+=

=1112

sincos1

11)(

nnn

n

o tbtnaak

tv

(2.52)

2.4.3 Dang phc cua nghiem theo chuoi Fourier

Cong thc Euler:

nxinxenxinxe inxinx sincos,sincos =+=

Suy ra: 2

cosinxinx ee

nx+

= i

eenx

inxinx

2sin

=

Tai trong:

( ) )22

(sincos)(11

11i

eeb

eeaatbtnaatp

inxinx

nn

inxinx

non

nno

=

=

+

++= ++=

++

+=

=

1 22)(

n

nninxnninx

o

ibae

ibaeatp (*)

Dung (2.51):

( ) ==

=pp T

tin

p

T

p

nnn dtetp

Tdtttntp

T

ibac

00

111)(

1sincos)(

1

2

o

T

p

o

Ttin

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

n dtetpT

c0

1)(1 (2.54)

la pho ri rac cua he so Fourier.

Chu y:

cn = c-n v la hai so phc lien hp, tine 1 va tine 1 la hai so phc lien hp. Do o:

tin

nec1 va tinnec 1

cung la hai so phc lien hp, va co 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 so hang - ham lc phc n v (Unit complex forcing function) di dang:

tinetkvtvctvm 1)()()( =++ &&& (2.55)

Nghiem on nh co dang:

tinn enHtv 1)()( 1 = (2.56)

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

( )

=++

= 11

2

121

;12

1)(

innknH (2.57)

Dung nguyen ly chong chat tac dung:

=

tin

necnHtv1)()( 1

(2.58)

Chu y:

- )( 1nH va )( 1nH la so phc lien hp

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

Imaginary

Real

=1

=1

n1t

in t e

e -in t

1

1

-n1t



2.5 PHAN NG VI TAI TRONG DANG XUNG

2.5.1 Khai niem tai trong xung (Impulsive Loads)

- La tai trong tac dung trong thi gian tng oi ngan, ot ngot. - Phan ng (chuyen v chang han ) ln nhat cua he at c trong thi gian rat ngan. - Lc can co vai tro nho, hap thu t nang lng cua ket cau. V vay ch xet he khong co can e n gian hoa. 2.5.2 Xung hnh sin

Xet tai trong na song hnh sin. Phan

ng cua he c chia ra 2 pha: Cng bc va t do.

+ Phase I: 10 tt

Ket cau chu tac dung cua tai trong ieu hoa.

ieu kien ban au: v(o) = v& (o) = 0 (trang thai ngh). Phan ng gom 2 so hang (qua o va on nh) 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):

ttvttv

tv

cos)(sin)(

)( 11 +=

& (2.60)

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

- Neu vmax thuoc Phase I :

= 1tt

t O

p(t)

O

p(t)

t

p(t)=p sin t o

P0

t1 t

Phase I Phase II 1=



(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 (-) va n=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 bien o dao ong t do (2.25)

[ ]2

1

2

2

1

2

1 cos221

)()(

+

=+

=

kp

tvtv

o&

He so ong: )cos1(21 20

+

==

p

kD

hay

2cos

1

22

=D

(2.61)

2.5.3 Xung ch nhat

+ Phase I: 10 tt

Tai trong at ot ngot va gi nguyen khong oi trong phase I. Nghiem rieng cho tai trong bac thang (Step loads) la chuyen v tnh:

k

pv op =

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

( )tk

ptv o cos1)( = (2.62)

+ Phase II: 01 = ttt

p(t)

O

p0

Phase I 1 t

Phase II

t t



Dao ong t do theo (2.60)

ttvttv

tv

cos)(sin)(

)( 11 +=

& (2.63)

- vmax thuoc Phase I : 1tt

2

,,0sin0)( T

tttdt

tdv=====

Neu 1tt tc la 2

1

Tt he so ong D = 2, vi

2

11 T

t

- vmax thuoc Phase II: 01 = ttt

Bien o dao ong: [ ]212

1max )(

)(tv

tvv +

==

&

v v& (t1) = 112

sinsin tTk

pt

k

p oo

= nen:

vmax =2

12

1

2

1

2 2cos12

sin

+ tT

tTk

po =T

t

k

pt

Tk

p oo 12

1

1 sin22

cos12

=

He so ong: T

t

kp

VD

o

1max sin2

== vi 2

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 op

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

+= 1cos

sin)(

11 t

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 I

O

p(t)

Phase II

p0

1 t t t



+=

=

1

1

1

11

1

1

1

1sin

cos)(

cossin

)(

tt

t

t

k

ptv

tt

t

k

ptv

o

o

&

(2.66)

Dao ong t do cua Phase II thu c bang cach the (2.66) vao (2.60) vmax

tm t ieu kien v& (t) = 0. Vi 4.01 T

t th vmax thuoc Phase I.

He so ong D cho bang:

Tt1 0.20 0.40 0.50 0.75 1.00 1.50 2.00

D 0.66 1.05 1.20 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 ong D theo ty so chieu dai xung

Tt /1 ,

=

T

tDD 1

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

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 wave Triangular

He so ong (Dynamic manification factor), D

Ratio t1

T = Impulse duration

Period



Chu y:

Neu ket cau chu chuyen ong cua goi vg(t), th se tng ng vi chu lc xung pt(t) = -m v&& g(t), vi tr so ln nhat po = - m v&& 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 cau tvmax&&

2.5.6 Tnh toan gan ung phan ng do lc xung

Gia x p(t) la lc xung trong thi gian t1 rat be. He co chuyen v v(t), can bang lc:

kvtpdt

vdmvm == )(

&&&

[ ]dtkvpvmt

=1

0

&

V v(t1) la lng be bac 2 so vi t1 : ( )21~ tv , nen bo qua. Do o:

==1

0

11 )()0()()(1

t

tvvtvdttpm

v &&&

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

Dao ong t do sau khi lc thoi tac dung co phng trnh:

ttvttvtv

cos)(sin)(

)( 11 +=

&

tdttpm

tvt

sin))((1

)(1

0

(2.67)

p(t)

t t1 O



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

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

Giai:

Tan so vong: sradW

kg/14.3== va =

1

0

.10)(

t

skipdttp .

Chu ky dao ong cua he: sT 22 == .

V tai trong tac dung trong thi gian ngan ( 15.01 =T

t ), nen dung (2.67) phan

ng xap x c tnh:

ttv sin)14.3(000,2

)386(10)( =

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: kipskvfS 4.31)14.06(1.51maxmax, ===

p(t)

t

0.1s 0.1s 0.1s

p0= 50k



Gia tr chnh xac cua chuyen v cc ai c xac nh t phng phap tch phan trc tiep la 0.604 in.

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

2.6 PHAN NG VI TAI TRONG TONG QUAT

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)(

)(

= t

m

dptdv (2.68)

vi >t = tt

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

Toan bo lch s tai trong co the xem nh bao gom s noi tiep cua cac xung lc ngan, moi xung lc tao ra mot phan ng vi phan theo (2.68). Phan ng toan bo la do chong chat cua cac xung lc tao ra, nen:

=t

dtpm

tv

0

)(sin)(1

)(

- Duhamel Integral (2.69)

Ky hieu:

)(sin1

)(

tm

th (2.70)

th (2.69) co dang:

=t

dthptv

0

)()()( - Tch phan cuon (Convolution Integral) (2.71)

p(t)

t O d

p( ) t- O

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



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:

dtpm

tovtov

tvt

++=0

)(sin)(1

cos)(sin)(

)(&

(2.72)

2.6.2 Tch phan bang phng phap so cho Duhamel Integral

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

Phng trnh (2.69) viet lai:

=tt

dpm

tdpm

ttv00

sin)(1

coscos)(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:

==tt

y

dym

dpm

tA00 )(

)(1

cos)(1

)(

44 34421

Phng phap Simpson

Chia t ra n phan (n chan)

n

t= khi o:

( ) ++++++

=t

nyyyyyydy0

43210 1.....24243

)(

2.6.3 Phan ng cua he co can

Trong (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

D

t

D

dtepm

tv0

)( )(sin)(1

)(

(2.75)

Tng t he khong can, ta co:

)(sin1

)( )(

= tem

th Dt

D

(2.76)

y( )

= O

y y y y

o

n

2 1

t n

t



ttBttAtv DD cos)(sin)()( = (2.77)

=

=

t

Dt

D

t

Dt

D

de

ep

mtB

de

ep

mtA

0

0

sin)(1

)(

cos)(1

)(

(2.78)

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

2.6.4 Tch phan phan ng trong mien tan so (Frequency Domain)

- Y ngha: Phan tch phan ng trong mien tan so co u 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): (pT

21 = : tan so c ban cua

tai trong).

=

=

==n

ti

n

n

tin

nnecectp

1)( (a)

==2

2

2

2

)(1

)(1

1

p

p

n

p

p

T

T

ti

T

T p

tin

p

n dtetpT

dtetpT

c (b)

at:

2

1211

=== +

pp

nnTT

(c)

)(2

)(1

nn

p

n ccT

c

= (d)

The (d) vao (a):

=

=

n

tin ectp

n

)(2

)( (a)

khi p

T th d (theo (c)), do o (a) viet lai:

=

dectpti

nn)(

2

1)(

The (d) vao (b):

==2

2

)()(

pT

pT

n dtetpcTcti

npn pho lien tuc cua he so Fourier.

cho p

T :

= dtetpc tin n )()(

(2.80)

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



ieu kien ton tai bien oi Fourier:



Xet he SDOF nh hnh ve cac ac trng m, k, c va p(t) co the la cac ai lng suy rong. Gia thiet k(v) va )(vc & la cac ham phi tuyen.

Theo nguyen ly AlembertD , phng trnh can bang cua cac lc tac dung len he moi thi iem t:

)()()()( tptftftf SDI =++ (a)

thi iem tt + ( t be):

)()()()( ttpttfttfttf SDI +=+++++ (b)

Lay (b)-(a):

)()()()( tptftftf SDI =++ (2.83)

So gia cua cac lc xac nh nh sau: )()()()( tvmtfttftf III &&=+=

)()()()()( tvtctfttftf DDD &+= (2.84)

)()()()()( tvtktfttftf SSS +=

)()()( tpttptp +=

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

D+ va

)( ttfS

+ thi iem tt + ):

p(t)

O

p(t+ t)

p(t)

Tai trong p(t)

t t+ t t

p(t)

t

(e)

d) Chuyen v

v

fs

v

fs(v)

v&

Dif

v&

v

)(vfD &

c) Van toc

1+iv&

iv&

vi+1 vi

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



t

D

vd

dftc

&)( ,

t

S

dv

dftk

)( (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 toan ong lc hoc.

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

2.7.3 Tch phan tng bc

Gia thiet:

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

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

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

2)()()(

ttvttvtv

+= &&&&& (a)

6)(

2)()()(

22 ttv

ttvttvtv

+

+= &&&& (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

tvtvt

tv &&&&

= (d)

t t+ 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



Chu y rang )(tv la an so.

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

)()()()(2

)(3)(3

)()(3)(6

)(62

tptvtktvt

tvtvt

tctvtvt

tvt

m =+

+

&&&&&&

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

vi )(36)()(~2

tct

mt

tktk

+

+=

(2.88a)

++

+

+= )(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 ng quan 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 tnh

Trong 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 bai toan, tc la )0(v& va v(0).

2. Xac nh c(t) va k(t) cho bc tnh t o th va tnh fD(t) va fS(t).

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

[ ])()()(1)( tftftpm

tv SD =&& (2.89)

4. So gia cua )(~ tp va o cng )(~ tk c tnh t (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 +& tnh theo (e).



Bc tnh tiep theo c lap lai nh tren. oi vi ket cau tuyen tnh, phng phap tnh nay cung dung c va viec tnh n gian hn.

Van e chnh xac cua phng phap

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

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

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

3. Chu ky dao ong T cua he.

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

Thng 10

Tt 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 phan ng cua khung an hoi deo mot bac t do chu tai trong 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 cua phan ng ong cung thoa man muc ch phan tch.

0 0.2 0.4 0.6 0.8 t, s

1

2

p(t), kips

5

8 7

5

3

Load history

c = 0.2 kip.sec/in

p(t), kips

m = 0.1 kip.sec2/in

k = 5 kips/in

v(t) fs

v

vmax 1.2in

Inelastic displacement

Elastoplastic stiffness

6kips

6kips



Trong ket cau nay, he so can c gia thiet van la hang so, v vay ay la bai toan phi tuyen ve o cng (o cng thay oi trong qua trnh chuyen ong).

( ))(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:

vvtpvcmvcm

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:


Chng 3. HE NHIEU BAC T DO 46

CHNG 3

HE NHIEU BAC T DO

3.1 THIET LAP PHNG TRNH CHUYEN ONG

3.1.1 La chon bac t do

Y ngha: thc te ket cau thng la he phan bo, co vo han bac t do. a ve s o mot bac t do ch thch hp trong mot so trng hp ac biet, khi he hau nh ch dao ong vi mot dang nhat nh. e thu c ket qua chnh xac hn, ta phai a he ket cau ve he ri rac nhieu bac t do. So bac t do c 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 va phng phap phan t hu han (FEM) e ri rac hoa.

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

3.1.2 Phng trnh can bang ong

e 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.

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

1 2 i

p(x,t)

m(x) EI(x)

chieu dng chuyen v

chieu dng cua lc

chuyen v

fDi fIi m i

vi(t)

pi(t)

fSi



Tai moi iem (nut) co cac lc tac dung: tai trong pi(t), lc quan tnh fIi, lc can fDi, va lc an hoi fSi. Phng trnh can bang nut i:

fIi + fDi + fSi = pi(t) , i = 1, 2, 3, ..., N

Dang ma tran:

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

trong o:

[fI] =

IN

I

I

f

f

f

M

2

1

, [fD]=

DN

D

D

f

f

f

M

2

1

, [fS]=

SN

S

S

f

f

f

M

2

1

, [p(t)] =

)(

)(

)(

2

1

tp

tp

tp

N

M

- Lc an hoi

Dung nguyen l cong tac dung, ta co:

fSi = ki1v1 + ki2v2 + .... + kiNvN vi i = N,1

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 duy tr ng an hoi (ngc chieu vi lc nut).

Dang ma tran:

SN

S

S

f

f

f

M

2

1

=

NNNN

N

N

kkk

kkk

kkk

L

LLLL

L

L

21

22221

11211

Nv

v

v

M

2

1

(3.2)

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

trong o: [K] goi la ma tran cng cua ket cau (Stiffness Matrix).

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

DN

D

D

f

f

f

M

2

1

=

NNNN

N

N

ccc

ccc

ccc

L

LLLL

L

L

21

22221

11211

Nv

v

v

&

M

&

&

2

1

(3.4)



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

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

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

- Lc quan tnh

IN

I

I

f

f

f

M

2

1

=

NNNN

N

N

mmm

mmm

mmm

L

LLLL

L

L

21

22221

11211

Nv

v

v

&&

M

&&

&&

2

1

(3.6)

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

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

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

Thay (3.3), (3.5), (3.7) vao (3.1) ta thu c he N phng trnh vi phan chuyen ong viet di dang ma tran:

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

Phng trnh tren la phng trnh mang tnh chat tong quat cua bai toan ong lc hoc. Trong o: [p(t)] la vect tai trong ngoai, tuy thuoc vao [p(t)] ma ta co cac trng hp phan tch ong lc hoc cua he: phan tch dao ong t do, phan tch phan ng cua he vi tai trong ong nh 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 nay phng trnh can bang nut (3.1) tr thanh:

[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

M

2

1

=

GNNGNGN

NGGG

NGGG

kkk

kkk

kkk

L

LLLL

L

L

21

22221

11211

Nv

v

v

M

2

1

(3.10)

vi kGij la lc tai nut i do vj = 1 gay ra, co anh hng 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) (3.14)

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

3.2 XAC NH CAC MA TRAN TNH CHAT CUA HE KET CAU

3.2.1 Tnh chat an hoi

3.2.1.1 o mem cua ket cau

Goi: fij la chuyen v tai i do pj = 1 gay ra. Tap hp cac fij (i = 1,N) tao nen ng an hoi do pj = 1 gay ra (hnh ve). Chieu dng cua chuyen v va lc 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 = fi1p1 + fi2p2 + .... + fiN pN i = 1, N

Dang ma tran:

f 1j 2j f ij f jj f

Nj f

1 2 3 j j p



Nv

v

v

M

2

1

=

NNNN

N

N

fff

fff

fff

L

LLLL

L

L

21

22221

11211

Np

p

p

M

2

1

(3.15)

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

trong o:

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

[p]: Ma tran tai trong nut, co cung chieu dng vi chuyen v nut.

Lc an hoi can bang vi lc nut [p] = [fS], khi o (3.16) tr thanh:

[v] = [f][fS] (3.17)

3.2.1.2 o cng cua ket cau

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

Thng ma tran o cng [K] c suy ra t ma tran 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

pvvpvpU TTi

N

i

i === =

(3.18)

Theo (3.16) vao (3.18) ta c:

]][][[2

1pfpU T= (3.19)

p 1 p 2 p 3

S1 f S2 f S3 f

1 v v 2 v 3 1 i j

1 i j

j v=1 1 p=k 1j

p=k i ij p=k N Nj

p=k j jj

k 1j k ij k jj k Nj

j v=1



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

]][][[2

1vKvU T= (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 la cac ma tran xac nh dng (Positive Definite), khong suy bien va nghch ao c. Thiet lap quan he gia [K] va [f], t (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 tran mem 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 trang thai (b) tren chuyen v trang thai (a)

[paT] [vb] = [pbT] [va] (3.23) hay [paT][f][pb] = {[pbT][f][pa]}T = [paT] [fT] [pb] suy ra: [f] = [fT] Ma tran oi xng (3.24) Mot cach tng t ta cung co ma tran cng oi xng: K = KT (3.25)

3.2.1.4 Thiet lap ma tran o cng bang phng phap phan t hu han (FEM)

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

a1 p

v a 1

a2 p

a 2 v

a3 p

v a 3

b 1

p b1

v b 2

p b2

v b 3

p b3

v

1 2 3 Trang thai (a)

Trang thai (b)



Xet phan t dam thang co 2 nut nh hnh ve: Co hai bac t do moi nut: bao gom chuyen v thang va goc xoay.

Ham dang i(x) ch chuyen v vi = 1 gay ra, con cac chuyen v nut khac eu bang 0. Ham i(x) phai thoa man ieu kien bien, nhng thng chon ham chuyen v trong dam co o cng EI = const do chuyen v nut vi = 1 gay ra. o la cac ham a thc Hermit bac ba nh sau:

1(x) = 1 - 3 2

L

x + 23

L

x (a)

3(x) = x(1- L

x )2 (b)

2(x) = 3 2

L

x - 23

L

x (c)

4(x) =

12

L

x

L

x (d) (3.26)

Dung bon ham noi suy nay, chuyen v cua dam xac 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 do chuyen v nut gay ra. e n gian ta xet phan t dam nh hnh ve. He so k13, tc la phan lc pa tren hnh ve c xac nh nh sau:

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

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

EI(x) L

x a b v(x)

1 v v 3 2 v

4 v

1

a v =v =1 1

=v =1 a 3

(x) 1

3 (x)

=v3 =1 1 3 (x) v = v1 a

k13 = pa =p

a

v(x)= (x) v1 1 (chuyen v kha d)



Cong kha d cua noi lc: WI = v1 dxxxxEIL

)()()(''

3

0

''

1

Cho WI = WE suy ra: k13 = dxxxxEIL

)()()(''

3

0

''

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 eu EI = const, ta co:

4

3

2

1

S

S

S

S

f

f

f

f

= 3

2

L

EI

22

22

233

233

3366

3366

LLLL

LLLL

LL

LL

4

3

2

1

v

v

v

v

(3.30)

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

He so o cng kij cua ket cau bang tong cac he so cng tng ng cua cac phan t noi vao nut. Chang han, neu cac phan t m, n, p cung noi vao nut i th he so cng cua ket cau tai nut i la:

kii = )(miik + )(n

iik + )( p

iik (3.31)

trong o )(miik , )(n

iik ,)( p

iik la he so cng cua phan t a bien oi sang he toa o chung(t toa a phng).



Th du:

Xet he nh hnh ve, gom 3 phan t noi tai 2 nut. Bo qua bien dang doc truc, he co 3 bac t do: v1, v2 va v3

Cac he so o cng cua he c xac nh bang cach lan lt cho cac chuyen v cng bc n v vi = 1 va cong lc nut ng vi cac phan t. Ma tran o cng ket cau:

)26(2311

xL

EIk = )3(2

321L

L

EIk = )3(2

331L

L

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 ==

=

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 bo thng oi hoi nhieu bac t do hn so vi bai toan tnh, do anh hng cua lc quan tnh. Tuy nhien, khi a chon cac

2L

L EI EI

4EI

v1 v2 v3

EI EI

4EI

k11 k21 k31

v1=1

EI EI

4EI

k12 k22 k32

v2=1



bac t do cho bai toan ong roi th viec xay dng ma tran cng giong nh trng hp bai toan tnh.

3.2.2 Tnh chat khoi lng

3.2.2.1 Ma tran khoi lng thu gon

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

[M] =

Nm

m

m

00

0

0

00

2

1

L

OM

M

L

(3.32)

trong o: mij = 0 vi i j, v gia toc tai khoi lng nao 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 nh hnh ve. Dung cac ham noi suy i(x) nh trong ma tran cng.

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

3v&& = a&& = 1, gia toc chuyen ong ngang cua dam se la:

)()( 33 xvxv &&&& = (3.33)

1m m2 m3

1 2 3

L m(x)

v(x) v 1

a 3 v v 4 b

2 v x

v = v

=v =1 a 3

a

(chuyen v kha d) v(x)= (x) v

m =p

1

13 a

1 f (x)

1 1

.. ..



Lc quan tnh co tr so: )()()()()( 33 xvxmxvxmxf 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

Tong quat: mij = dxxxxm jL

i)()()(

0

Khoi lng suy rong (3.35)

v mij = mji, nen ma tran khoi lng 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

][ LM

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 chong chat t ma tran cua phan t, tng t nh ma tran cng.

Th du

m11 m21 m31 11 =v&&

=1

m12 m22 m32

12 =v&&

2L

L m m

v1 v2 v3

1.5m

v1 v2 v3 1.5m L

0.5m L

0.5m L 0.5m L

0.5m L

1.5m L

m11= 4m L m22 = m33 = 0



Thanh lap ma tran khoi lng cho ket cau nh hnh ve theo hai phng phap. Qua trnh tnh cac he so khoi lng c ch ro tren cac hnh ve.

Ma tran khoi lng thu gon:

[M] =

0

0

840

210

Lm

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

Ma tran khoi lng tng thch:

768210

25.1)2156(420

11

LmLxmx

Lmm =+= LLmLLmmm 11

210)22(

4203121 ===

222

3322 26210

)2(4420

25.14

420L

LmL

LxmL

Lmmm =+== 2232 )18(

210)2()3(

420

25.1L

LmLx

Lxmm ==

[M] =

22

22

261811

182611

1111786

210LLL

LLL

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 ng cheo. Cac he so cua [M] thu gon ng vi cac chuyen v xoay cung bang 0, cang lam cho bai toan n gian hn.

- Dung [M] thu gon co the loai bo cac chuyen v xoay, nhng dung [M] tng thch th khong the loai bo c.

3.2.3 Tnh chat can

He so can cua phan t c xac nh bi FEM, cho bi cong thc:

cij = dxxxxc jL

i )()()(0

He so can suy rong (3.37)

trong o: c(x) - tnh chat can phan bo cua phan t.

Ma tran can ket cau cung c chong chat t ma tran can cua phan t, tng t ma tran o cng hoac ma tran khoi lng.



Tuy nhien, e xac nh ham c(x) trong thc te th khong lam c. Thng tnh can cua ket cau xac nh bi thc nghiem bang t so can .

3.2.4 Tai trong

Neu tai trong tac dung tren phan t th phai thay the bang tai trong nut tng ng, dung khai niem lc suy rong. Co hai phng phap:

3.2.4.1 Tai trong nut tng ng tnh hoc

Xem nh tai trong at tren dam phu co mat truyen lc at tai nut. Lc truyen vao nut se thay the cho tai trong at tren phan t. Nh vay khong truyen mo men tap trung vao nut.

3.2.4.2 Tai trong nut tng thch

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

p1(t) = dxxtxpL

0

1 )(),(

Tong quat:

pi(t) = dxxtxpL

i0

)(),( Tai trong suy rong (3.38)

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

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

th lc nut suy rong tr thanh:

pi(t) = (t) dxxxL

i0

)()( (3.39)

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

pi(t) pj(t)

Lc nut tng ng

p

a 3 p

1

p 4

b

2 p L v(x)= (x) v

1 1 v = v a 1

p(x,t)

Tai trong suy rong



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 hoc

o cng hnh hoc the hien khuynh hng lam tang chuyen v uon cua lc nen N. He so cng hnh hoc chnh la lc nut do N tao ra. Gia thiet rang lc nen N do tai trong tnh gay ra la chu yeu; phan do lc ong gay ra co the bo qua c. V vay, coi N khong oi trong qua trnh dao ong. (Neu N(t) thay oi theo thi gian th [KG] cung thay oi theo thi gian. Bai toan tr nen phi tuyen).

Xap x tuyen tnh: 1 BTD/nut

Gia s lc doc trong phan t i la Ni. Coi phan t i thang th lc nut fGi va fGj c xac nh theo lc nen Ni 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

N

l

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)

+ o cng hnh hoc tng thch:

Dung khai niem phan t hu han (FE), ta thu c cong thc:

i v j v

i j x

v

N O

N

N

i

i

Li

i v j v i

Gi f = v - i v j L i

i N

i N Gj f = L i v - j v i

Bieu o N(x)

PG2

b PG4

PG1

PG3 a



( ) ( ) ( )dxxxxNk jiL

oGij

'' = (3.43)

Neu phan t co lc doc N(x) = N = const, dung cac ham noi suy trc ay, ta thu c ma tran cng 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

NK eG

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

Ma tran [KG] cua ket cau suy ra t [ eGK ] tng t nh [K], [M].

3.2.6 La chon cach thiet lap ma tran tnh chat ket cau Co 2 cach tnh gan ung cac ma tran khoi lng, o cng hnh hoc, tai trong: - Phng phap s cap ch xet en chuyen v thang. - Phng phap tng thch xet ca chuyen v thang chuyen v xoay. Ve nguyen tac, phng phap tng thch cho o chnh xac cao hn, v xet ay

u va he thong hn cac phan nang lng lien quan en s lam viec ong cua ket cau. Tuy nhien, trong thc te th o chnh xac cua phng phap tng thch khong troi bao nhieu so vi phng phap s cap, nhng khoi lng tnh toan th ln hn nhieu, v bac t do xoay ong vai tro kem quan trong so vi chuyen v thang.

Phng phap s cap de dang hn, v cac ma tran xuat phat de tnh hn va so bac t do phai xet cung t hn.

Neu phng phap thu gon khoi lng c dung vi ma tran cng thiet lap bang FEM (tc la ke en bac t do chuyen v xoay) th co the loai tr cac chuyen v xoay nay trong phng trnh chuyen ong. Khi o ma tran cng cung c rut gon lai, goi la Static Condensation (kch thc ma tran cng thu nho lai). e minh hoa, ta viet lai phng trnh (3.2) trong o a sap xep lai cac chuyen v thanh 2 nhom: vt la thanh phan chuyen v thang va vo la thanh phan chuyen v xoay.

Phng trnh chuyen ong c viet lai dang ma 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 momen nut an hoi bang 0, neu tac ong tren he ch la lc ch khong co momen tap trung at ngay tai nut.

Trong (3.45) co the bieu dien cac chuyen v xoay { }v theo chuyen v thang { }tv :

{ } { }tt vKKv ][][ 1 = (3.46)

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

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

[ ]{ } { }Stttttt fvKKKK = ][]][[][ 1 hay { } { }Sttt fvK =][ (3.47)

trong o [ ]][]][[][][ 1 ttttt 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 FEM co the loai tr va so bac t do thc s phai giai quyet giam 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

Ta co:

][ K = 32

L

EI

22

22

62

26

LL

LL = L

EI4

31

13

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 = 32

L

EI [ ]

L

LLL

8

38

3

3312 = 3

2

L

EI

4

39

2L

L EI EI

4EI

v1 v2 v3



Bai tap: 11 8 page 175

3.3 DAO ONG T DO KHONG CAN

3.3.1 Phan tch tan so dao ong

T phng trnh (3.8), phan tch dao ong t do nen vect tai trong ngoai p(t) = 0, ta co:

{ } { } { } { }0)(][)(][)(][ =++ tvKtvCtvM &&&

Bo qua thanh phan lc can [C]= [0]

{ } { } { }0)(][)(][ =+ tvKtvM && (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 - la bien o dao ong.

{ } { } )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 N phai triet tieu:

0]][][[det 2 = MK (3.52)

ay la phng trnh ai so bac N, do o co N nghiem 21 , 22 , ..., 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

M

2

1

(3.53)

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



T = 2/ va f = T

1

Th du (E12-1)

Tnh tan so rieng cua khung san cng: khoi lng va o cng nh hnh ve (a). Cac he so cng tnh tren hnh ve (b).

Cac ma tran cua khung:

=

0,20

5,1

00,1

][M (kip.s2/in)

=

520

231

011

600][K (kip.s/in)

Phng trnh ac trng (3.52):

0

2520

25,131

011

600

0,20

5,1

00,1

520

231

011

600][][ 22

=

=

=

B

B

B

MK

vi 600

2=B

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

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

1,0 kip.s2/in

1,5

2,0

v1

k in 600

1200

1800

v1 =1 K = 600 11

K = - 600 21

K =0 31

K = -600 12

K = 1800 22

K = -1200 32

K = 0 13

K = -1200 23

K = 3000 33

v =1 2

V =1 3

(a) (b)

v2

v3



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 so n ta co mot vect rieng { }nv . Nhng v nh thc (3.52) triet tieu, nen hang cua ma tran ch con N-1, do o ch co N-1 thanh phan cua { }v oc lap. Thng chon thanh phan au tien { } 11 =nv , khi o vect chuyen v tr thanh:

{ }

=

=

Nn

n

Nn

n

n

n

v

v

v

v

v

v

1

22

1

MM

at: ][][][ 2)( MKE nn = (3.53)

Phng trnh (3.51) c viet lai:

=

0

0

0

1

2

)()(

2

)(

1

)(

2

)(

22

)(

21

)(

1

)(

22

)(

11

MM

L

MLMM

L

L

Nn

n

n

NN

n

N

n

N

n

N

nn

n

N

nn

v

v

eee

eee

eee

(3.54)

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

][][

][)(

00

)(

01

)(

10

)(

11

nn

nn

EE

Ee

{ }

nv0

1 =

{ }

0

0

Tng ng vi 2 phng trnh:

{ }0]][[

0]][[][

0

)(

10

)(

11

0

)(

00

)(

01

=+

=+

n

nn

n

nn

vEe

vEE (a)

Giai he phng trnh (a) tren ta c:

{ } ][][ )(011)(00 nnon EEv = (3.55)



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

=

=

Nn

n

kn

Nn

n

n

n

v

v

v

1

1][

22

1

MM

(3.56)

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

Ma tran dang dao ong (Mode shape matric) la tap hp cua N vect dang dao ong:

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

NNNN

N

N

L

MLMM

L

L

21

22221

11211

(3.57)

Nh vay khi xac nh c [i] ta se biet c hnh dang 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. Hai chuyen v tang di cua mode n c tm theo (3.55):

n

n

3

2

= -

n

n

B

B

252

25,13

0

1

vi Bn =600

2

n

Ket qua nh hnh ve.

3.3.3 Phan tch tan so theo ma tran mem

Nhieu bai toan dung ma tran mem [f] tien hn ma tran cng [K]. Khi o can xac nh tan so rieng theo [f].

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

{ } { }0]][][[ 2 = vMK (3.51)

1.000 1.000 1.000

-2.570

2.470

-0.601

-0.676

0.644

0.300

Mode 1 Mode 2 Mode 3 =14.5 1 =31.1 2 =46.1 3



Nhan 2 ve vi [f] ta c: { } { }0]]][[]][[1[2

= vMfKf

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

{ } { }0]]][[][1[2

= vMfI

(3.58)

do { } 0 v , nen ta co phng trnh tan so: 0]]][[][1[det2

= MfI

(3.59)

3.3.4 Anh hng cua lc hoc

3.3.4.1 Dao ong t do

Phng trnh dao ong (3.49) ke en o cng hnh hoc co dang:

{ } { } { } { }0)(][)(][)(][ =+ tvKtvKtvM G&& (3.60)

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

Phng trnh tan so:

0]][][[det 2 = MK (3.61)

Lc doc lam cho ket cau b mem hn, nen cac tan so rieng cung thap hn. Ket cau thng lam viec bat li hn di tac dung cua tai trong ong. Tng ng, cac dang dao ong chnh (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 tn

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