dong luc hoc cong trinh - duong van thu

1

M

0

y P(t)

Hnh 1.1

K

NG LC HC CNG TRNH

Bin son: PGS. TS Dng Vn Th

CHNG 1: DAO NG CA H C MT BC T DO

1.1 MT S KHI NIM C BN V L THUYT DAO NG

1.1.1 Khi nim v chu k v tn s

Xt h trn hnh 1.1. H gm khi lng M c gn vo mt im c

nh nh l xo c cng K (l phn lc pht sinh trong l xo khi l xo bin dng

mt lng bng n v). Khi lng M chu tc ng ca mt lc ng P(t) c

phng theo phng ca chuyn ng (phng y), cn chiu v tr s thay i

theo thi gian.

Khi lng M chuyn ng, lc pht sinh trong l xo

thay i lm cho vt thc hin mt dao ng c hc.

Tu thuc vo quan h gia lc l xo v bin dng

ca l xo l tuyn tnh , hay phi tuyn, m ta c bi ton dao

ng tuyn tnh hay dao ng phi tuyn.

Dao ng ca vt thun ty do lc l xo sinh ra khi M

dch chuyn khi v tr cn bng ban u (do mt nguyn

nhn bt k no gy ra ri mt i) c gi l dao ng

t do hay l dao ng ring.

Dng chuyn v ca vt M c gi l dng dao ng ring. Nu trong qu

trnh dao ng lun lun tn ti lc ng P(t), ta c bi ton dao ng cng bc.

Lc ng P(t) cn c gi l lc kch thch.

S cc dao ng ton phn ca khi lng thc hin trong mt n v thi

gian, ch ph thuc vo cc c trng c hc ca h, gi l tn s dao ng ring

hay tn s dao ng t do, v c k hiu l f. Thi gian thc hin mt dao

ng ton phn c gi l chu k dao ng, v c k hiu l T. Nu T o bng

giy (s) (trong ng lc hc cng trnh thi gian thng c o bng giy), th

th nguyn ca f l 1/s. V tr s f v T l nghch o ca nhau.

2

1.1.2 Dao ng iu ho v vc t quay

Sau y ta xt mt dng dao ng quan trng c gi l dao ng iu

ha. y l dng dao ng c bn thng gp trong c hc, mt khc, cc dao

ng c chu k lun lun c th phn tch thnh cc dng dao ng iu ha n

gin ny.

Xt dao ng iu ha,

( ) sinS t A t (1-1)

C vn tc

( ) os tv t A c (1-2)

v gia tc

2( ) sina t A t (1-3)

Ta thy rng, c th miu t

chuyn ng ny nh chuyn dch

ca im mt vc t OA (c ln

bng A) ln mt trc S no khi

vc t ny quay quanh im c nh

O vi vn tc gc .(xem hnh 1.2).

Lc ny, tr s A c gi l

bin dao ng, cn vn tc gc

c gi l tn s vng ca dao ng

l s dao ng ton phn ca h

thc hin trong 2 giy.

Tht vy, theo nh ngha,

2T , nn 2 1

Tf

, do 2 f

Tm li, trong dao ng iu ha ta c cc quan h sau,

Acost

Asint

x

s

0

v

Hnh 1.2

A

a

t

3

22 f

T

(1-4)

1

2f

T

(1-5)

1 2

Tf

(1-6)

Sau ny trong tnh ton thc t, ngi ta hay dng hn f.

Kho st ba dao ng iu ha cng bin A v chu k T, nhng bin

t c cc thi im khc nhau; Cng c ngha l thi im bt u ca ba

dao ng ny l lch nhau. Ta ni ba dao ng lch pha nhau xem hnh 1.3;

Dao ng (c) bt u sm hn dao ng (b) mt khong thi gian t0; Ngha

l, sau khi vc t quay OA biu din dao ng (c) quay c mt gc = t0 th

dao ng (b) mi bt u. Ta ni t0 l lch pha, cn l gc lch pha (hay gc

pha). Tng t, dao ng (a) c gc pha l /2.

Cch biu din dao ng iu ha di dng vc t quay nh trn hnh 1.2,

gip ta thc hin thun tin vic hp cc dao ng iu ha. V d, xt hp ca hai

dao ng iu ha cng tn s (c th khc bin v lch pha).

1 1( ) sinS t A t (a)

2 2( ) sinS t A t (b)

Cc vc t quay biu din cc dao ng S1 v S2 ti thi im t no l

OA1 v OA2 nh trn hnh 1.4. Hp ca hai dao ng S1 v S2 chnh l hp ca hai

vc t OA1 v OA2 cho ta vc t OA c ln , theo qui tc hnh bnh hnh, l

T

A

A

b)

t

s

( ) Asin( t)S t

0

T

A

A

a) t0=

4

T

t

s

( ) Asin t-2

S t

0

T

A

A

c) 0

2t T

t

s

( ) Asin t-S t

0

Hnh 1.3

4

2 2

1 2 2os sinOA A A A c A (1-7)

v gc lch pha , m:

2

1 2

sin

os

Atg

A A c

(1-8)

Nh vy, hp ca hai dao ng iu ha cng tn s l mt dao ng iu

ha cng tn s, c bin A c tnh theo (1-7) v gc lch pha c tnh

theo (1-8)

1 2( ) ( ) ( ) Asin t+S t S t S t (c)

Ch rng, nu hai dao ng thnh

phn khc tn s, th hp ca chng khng

cn l dao ng iu ha na, m ch l dao

ng c chu k (chi tit c th xem

cc ti liu tham kho).

1.1.3 Lc cn v cc m hnh lc cn

Dao ng t do ca h do mt nguyn nhn tc dng tc thi no gy ra

ri mt i s khng tn ti mi, m s mt i sau mt khong thi gian. S d nh

vy l do trong qu trnh dao ng, h lun lun phi chu tc dng ca mt s lc

gy cn tr dao ng m ta gi l lc cn. Lc cn do nhiu nguyn nhn gy ra

nh : ma st gia cc mt tip xc m ta gi l lc cn ma st; sc cn ca mi

trng nh khng kh, cht lng hay lc ni ma st m ta gi chung l lc cn

nht.

Trong chuyn ng c hc, ngi ta thng chia lc cn thnh ba nhm

chnh:

1- Lc cn ma st c xc nh theo nh lut Culong

1.cR C N (1-9)

Trong : C1 l h s ma st,

A2 sin

A2 cos

A

A A2

A1

t

A

A

A x

A

s

A

Hnh 1.4

0

A

5

N l thnh phn php tuyn ca lc sinh ra gia hai mt tip xc khi

chuyn ng ( n ph thuc vo vn tc chuyn ng)

2- Lc cn nht tuyn tnh Newton t l bc nht vi vn tc chuyn ng

2.cR C v (1-10)

Trong : C2 l h s cn nht

v l vn tc chuyn ng, v = (t)

y l m hnh lc cn c dng nhiu trong thc t xy dng; v c

m t bng mt pt tng chuyn ng trong cht lng nht nh trn hnh 1.6d.

3- Lc cn t l bc cao vi vn tc (thng l bc hai). Lc cn ny

thng xy ra khi vt chuyn ng trong mi trng cht lng hay cht kh vi

vn tc tng i ln.

3.cR C v (1-11)

S thay i ca ba nhm lc cn ny trong dao ng iu ha c th

hin trn hnh 1.5;

1, Lc cn Culng

2, Lc cn nht tuyn tnh

3, Lc cn nht phi tuyn

1.2 PHNG TRNH VI PHN DAO NG NGANG TNG QUT

CA H MT BC T DO

Xt h mt bc t do gm dm n hi gi thit khng c khi lng, trn

c t khi lng tp trung M, chu tc dng ca ti trng ng P(t) t ti khi

lng v c phng theo phng chuyn ng ca khi lng (xem hnh 1.6a).

Trng hp ti trng khng t ti khi lng th phi chuyn tng ng v t

T

A

3

A 1

A

Rc

t

ng chuyn ng

Hnh 1.5: Lc cn trong dao ng iu ha

2

A

6

ti khi lng. Mt trong cc cch chuyn tng ng nh vy s c trnh by

chi tit mc 2-4. Kt cu c t trong h ta yz nh trn hnh v.

Khi trn h cha chu tc ng ca lc ng P(t), nhng do trng lng ca

khi lng M ,( G = Mg), h c bin dng v chuyn dch ti v tr 1 nh trn

hnh 1.6a; Trng thi tng ng vi v tr ny ca h ta gi l trng thi cn bng

tnh ban u ca h. Khi h chu tc dng ca ti trng ng P(t), h s dao ng

xung quanh v tr cn bng ny. Gi s, n thi im t no , h ang chuyn

ng hng xung v ti v tr 2 nh trn hnh 1.6a;

Do y ta ch xt nh hng ca lc ng P(t), ng thi do gi thit bin

dng b, nn trng thi cn bng tnh ban u c th coi gn ng nh trng hp

cha c bin dng (Hnh 1.6b). Tt nhin, khi xc nh mt i lng nghin cu

no , ta phi k ti gi tr do M gy ra theo nguyn l cng tc dng.

Xt h dao ng chu lc cn nht tuyn tnh Newton, th dao ng ca h

trn hnh 1.6b c th c m hnh ha nh trn hnh 1.6d; gm khi lng M

c treo vo l xo c cng K , v gn vo pt tng chuyn ng trong cht

lng nht c h s cn C.

Xt h thi im t no ang chuyn ng hng xung cng chiu vi

lc P(t). Khi h chu tc dng ca cc lc sau: lc ng P(t); lc n hi sinh ra

P=1

z

y

c)

M

Rh

( )cR t

( )z t

P(t)

f)

Hnh 1.6

P(t)

y(t) yt M

1

2

z

y

a)

P(t)

y(t) 2

z

y

b)

M

1K

M hnh tnh

c

d)

P(t)

7

trong l xo ph thuc dch chuyn y ca khi lng, Rh(y) = K.y(t), c chiu

hng ln; lc qun tnh Z(t) = -M (t) c chiu hng xung cng chiu vi

chuyn ng; v lc cn nht tuyn tnh Rc = C (t) c chiu hng ln ngc vi

chiu chuyn ng (xem hnh 1.6f). H trng thi cn bng ng, nn:

Rh + Rc Z(t) P(t) = 0

Hay ( ) ( ) ( ) ( )My t Cy t Ky t P t (1-12)

Phng trnh (1-12) l phng trnh vi phn (PTVP) dao ng ngang tng

qut ca h n hi tuyn tnh mt bc t do chu lc cn nht tuyn tnh. Trong

, C l h s cn c th nguyn l [ lc thi gian / chiu di]; K l cng ca

h, l gi tr lc t tnh ti khi lng lm cho khi lng dch chuyn mt lng

bng n v, v c th nguyn l [lc / chiu di ].

Phng trnh (1-12) cng c th c thit lp da vo biu thc chuyn

v. Tht vy, nu k hiu l chuyn v n v theo phng chuyn ng ti ni

t khi lng (hnh 1.6c) cn gi l mm ca h mt bc t do- th dch

chuyn y(t) ca khi lng ti thi im t do tt c cc lc tc dng trn h gy ra,

theo nguyn l cng tc dng s l:

( ) ( ) ( ) ( )y t P t My t Cy t

Hay ( ) ( ) ( ) ( )My t Cy t Ky t P t chnh l (1-12)

Trong 1

K

(1-13)

c gi l cng ca h.

Gii PTVP (1-12) s xc nh c phng trnh chuyn ng, vn tc, v

gia tc chuyn ng ca khi lng; T c th xc nh c cc i lng

nghin cu trong h. Sau y ta s gii bi ton trong mt s trng hp.

1.3 DAO NG T DO-TN S DAO NG T DO ( HAY TN S

DAO NG RING )

1.3.1 Dao ng t do khng c lc cn

y l trng hp l tng ha, v trong thc t lc cn lun tn ti. PTVP

dao ng lc ny c dng n gin [cho C v P(t) trong (1-12) bng khng].

( ) ( ) 0My t Ky t

8

Hay l 2( ) ( ) 0y t y t (1-14)

Trong 2( )

1M

t

K g g

M M G y

(1-15)

y, ta k hiu G = yt(M)

, v mt ngha, n l chuyn v tnh ca khi

lng M do trng lng ca khi lng, G , t tnh theo phng chuyn ng

gy ra (xem hnh 1.6a); cn g l gia tc trng trng. Phng trnh vi phn (1-14)

c nghim tng qut l:

1 2( ) os t+A siny t Ac t (a)

Cc hng s tch phn A1v A2 c xc nh t cc iu kin u: Ti thi im

bt u dao ng (t=0), gi s h c chuyn v ban u yo v vn tc ban u v0

0 00 0;t ty y v v (1-16)

Thay (1-16) vo (a) vi ch ; 1 2( ) ( ) sin os tv t y t A t A c , ta c:

A1 = y0 ; v A2 = v0 (b)

Thay (b) vo (a) ta c phng trnh dao ng t do khng c lc cn ca h mt

bc t do:

00

v( ) os t+ siny t y c t

(1-17)

Hay 00v

( ) sin t+ + sin2

y t y t

(1-17)

iu ny c ngha l, dao ng t do khng cn ca khi lng l hp ca

hai dao ng iu ha cng tn s v lch pha /2. S dng khi nim vc t

quay, theo (1-7) v (1-8) , phng trnh (1-17) c dng n gin:

( ) Asin t+y t (1-18)

Trong 2

2 00

vA y

v 0

0

yarctg

v

(1-19)

9

G=Mg

a)

4

l

3

4

l

b)

P=1

Hnh 1.8

c)

P=1

3

16m

M

C

Nh vy, dao ng t do ca h mt bc t do (BTD), khi khng c lc

cn, l mt dao ng iu ha, c tn s c tnh theo (1-15) , c bin v

gc lch pha c tnh theo (1-19), cn chu k dao ng c tnh theo (1-6).

Nhn vo (1-15) ta thy ch ph thuc yt(M),

cng tc l ph thuc hay

K, ngha l ch ph thuc vo n hi ca h. Nn tn s dao ng t do cn

c gi l tn s dao ng ring ca h; N l mt c trng ca h dao ng.

Dao ng t do khng cn c dng nh trn hnh 1-3; Ph thuc iu kin

ban u m c dng (hnh 1.3a, b, hay c). V d, khi khng c chuyn v ban u

(y0 = 0), th = 0, nn dng dao ng nh trn hnh 1.3b; Khi khng c vn tc

ban u (v0 = 0), th gc pha bng /2, dng dao ng nh trn hnh 1.3a; Cn

dng dao ng trn hnh 1.3c tng ng vi khi c y0 v v0 u khc khng.

Ch : Khi khi lng c lin kt bng nhiu l xo mc song song hay ni

tip nh trn hnh 1.7, khi cng tng cng c tnh nh sau:

V D 1.1:

Trn dm n gin hai u khp, t

ti C mt khi lng tp trung M c trng

lng G = 0,75 kN nh trn hnh 1.8a; Bit

E = 2,1.104 kN/cm

2;

4410

12J cm ; l=1m.

(1-20)

M

K1 K2

P(t)

i

i

k k

M

K1

K2

P(t)

1 1

i ik k

Hnh 1.7

M

K1 K2

P(t)

2 1

2sini ii

k k

10

Yu cu: Xc nh tn s vng v chu k dao ng ring ca h. B qua khi

lng dm, v ly g = 981 cm/s2.

Gii: Chuyn v n v tai C, theo phng chuyn ng, do lc P = 1 gy ra, theo

cng thc Maxwell Mohr l ( xem hnh 1.8b):

31 3 1 3 1 2 3 3

4 4 16 2 3 16 256

mm m m

EJ EJ

(a)

Chuyn v tnh ti ni t khi lng do trng lng ca khi lng gy ra l:

3 3

( ) 3 2,25. 0,75256 256

M

t

m kNmy G kN

EJ EJ (b)

Tn s dao ng ring ca h , theo (1-15) l:

4 4

1

3

256 2,1 10 4981 70,6

2,25 12 100s

(c)

Chu k dao ng ring tnh theo (1-6) l:

2 2 3,1416

0,08970,6

T s

(d)

V D 1.2:

Trn khung ba khp c t vt nng trng lng G (hnh 1.9a). B qua nh

hng ca khi lng khung, lc ct , v lc dc ti din dng. Hy xc nh tn

s dao ng ring theo phng ng v phng ngang ca h.

Gii: Chuyn v n v theo phng ng g, v phng ngang ng ti ni t

khi lng c tnh theo cng thc Maxwell Mohr. T cc biu m men

n v trn hnh 1.9b, v c, ta c:

g31 2 1

24 2 2 3 4 48

l l l l

EJ EJ

(a)

ng 3 2. 2 . 2 1

2 3 2 3 3

h h h l h h lh l

EJ EJ

(b)

Thay (a) v (b) vo (1-15) ta c tn s dao ng ring theo phng

ng v phng ngang l:

11

g = sGl

EJg

G

g

1483

; ng = slhhGEJg

G

g

ng

1323

1.3.2 Dao ng t do c lc cn

Khi coi lc cn t l vi vn tc, PTVP dao ng t do tng qut c dng:

( ) ( ) ( ) 0My t Cy t Ky t (1-21)

Hay 2( ) 2 ( ) ( ) 0y t y t y t (1-21)

y ta t 2c

M cng c gi l h s cn (1-22)

Phng trnh c trng ca PTVP (1-21) c nghim l:

2 21,2 (a)

nn nghim tng qut ca (1-21): 1 21 2( )t ty t Ae A e

s c dng:

2 2 2 2

1 2( )t t

ty t e Ae A e

(1-23)

Chuyn ng ca khi lng, theo (1-23), ph thuc vo h s . Phn

tch tng trng hp ta thy:

1- Khi 2 2; hay C 2 KM y(t)

t

0

Hnh 1.10

2l

2l

h

G

(EJ=hng s)

a)

P=1 h

2l

2l

c)

Hnh 1.9

P=1

4l

2l

2l

b)

12

Khi > ta gi l lc cn ln; cn khi = ta gi l lc cn trung bnh

(hay lc cn gii hn). Lc ny l mt s thc; Hn na, v nn 22 <

, (bng khng khi = ). Do c hai nghim tnh theo (a) u m. Nh vy,

chuyn ng ca khi lng khi lc cn ln v trung bnh , theo (1-23), l tng

ca hai hm s m m. H khng giao ng m chuyn ng tim cn dn ti v

tr cn bng nh trn hnh 1.10;

2- Khi 2 < 2:

Trng hp ny c gi l lc cn b. Lc ny nghim l phc.

t

2 2 21 (1-24)

Khi nghim ca phng trnh c trng (xem (a ) s l:

1,2 1i (b)

V phng trnh chuyn ng (1-23) tr thnh:

1 2

1 2( )t ti ity t e Ae A e

(1-23)

S dng cng thc Euller

cos sin

cos sin

i

i

e i

e i

(1-25)

thay vo (1-23) ta c:

1 2 1 1 2 1( ) cos sinty t e A A t i A A t

hay l, 1 1 2 1( ) cos sinty t e B t B t (1-23)

Trong , B1 = A1 + A2 ; B2 = i ( A1 A2 ) (c)

Cc hng s B1, B2 xc nh c t cc iu kin u (1-16)

B1 = y0 ; B2 = ( v0 + y0 ) / 1 (d)

13

Thay (d) vo (1-23), v li p dng khi nim vc t quay hp hai dao

ng iu ha trong du mc vung, ta c phng trnh dao ng t do ca h

mt bc t do khi lc cn b l:

1( ) sin( )ty t Ae t (1-26)

Trong , A =

2

1

002

0

yvy

(1-27)

v = arctg (00

10

yv

y

)

Dng dao ng trong trng hp ny c th hin trn hnh 1.11;

T (1-26), hay t hnh 1-11 ta thy, dao ng t do ca h mt bc t do

khi lc cn b, cng l mt dao ng iu ha c tn s vng 1 tnh theo (1-24),

v chu k T1 tnh theo (1-28)

T1 = 1

2 =

22

2

(1-28)

song bin dao ng gim dn theo lut hm s m m : Ae -t.

nghin cu tt dn ca dao ng, ta xt t s gia hai bin dao ng

lin k nhau (cch nhau mt chu k T1). K hiu bin t c ti thi im t

no l An, cn ti thi im ( t + T1) l A n+1, th t (1-26) ta c:

A

A

0

yn tAe

tAe

yn+1

t

y(t)

Hnh 1.11 : Dao ng t do khi lc cn b

T1

14

1

11

11

1

1 sin

sin TTt

t

Tt

t

n

n ee

e

TtAe

tAe

A

A

= hng s

Suy ra, T1 = ln (1n

n

A

A

) = (1-29)

Nh vy, t s gia hai bin lin k nhau l mt hng s; Cn logarit t

nhin ca t s ny, k hiu l , l mt i lng ph thuc vo h s cn v

ng nhin l c 1 ca h, dng nh gi tt dn ca dao ng , ngi ta

gi l h s cn logarit, hay l Dekremen logrit ca dao ng t do c cn b.

H s cn logarit ng vai tr quan trng trong thc t. N gip xc nh

h s cn nh th nghim o bin dao ng An v A n+1. Sau y l mt s kt

qu th nghim tm c cho mt s loi kt cu xy dng.

1, i vi cc kt cu thp

T1 = (0,016 0,08)2 0,1 0,15

2, i vi kt cu g --- = (0,005 0,022)2 0,03 0,15

3, i vi cc kt cu b tng ct thp

T1 = (0,016 0,032)2 0,08 0,2

4, i vi cu thp --- = (0,01 0,15 ); trung bnh 0,28

5, Vi cu b tng ct thp: --- = 0,31

6, Vi dm b tng ct thp: --- = (0,17 0,39 ); trung bnh 0,28

7, Vi khung b tng ct thp: --- = (0,08 0,16 ); trung bnh 0,12

So snh hai phng trnh dao ng t do khng cn (1-18) v c cn b (1-

26) ta thy, tn s ring khi c cn b 1< khi khng c cn, cn chu k T1 > T;

C ngha l, khi c cn b, dao ng chm hn so vi khng c lc cn. Tuy

nhin, s sai khc ny cng rt nh. Do trong xy dng, do ch yu l cn b,

ngi ta thng coi gn ng 1 , v T1 T trong tnh ton.

Tht vy, ta xt mt trng hp dao ng tt kh nhanh.

V d, An / A n+1 = 0,5.

Khi = ln(A n/A n+1) = ln0,5 = 0,693. suy ra,

= 0,693 / T1 = 0,6931 / 2 = 0,111 hay

15

1 = 22 = 21

2 0,11 = 0,994 .

Tr li trng hp lc cn trung bnh (cn gii hn) 2 = 2. Lc ny,

= T = .

2 = 2; Do :

1n

n

A

A

= e T

= e 2

= 529.

Ngha l bin dao ng sau mt chu k gim i 529 ln, hay ni cch

khc, khi h chu lc cn trung bnh, h gn nh khng dao ng m ch chuyn

ng tim cn dn ti v tr cn bng ban u. iu ny nht qun vi kt lun

c cp ti mc a.

1.4 DAO NG CNG BC CHU LC KCH THCH IU HO

P(t)=P0sinrt - H S NG

Phng trnh vi phn dao ng tng qut trong trng hp ny, theo (1-12) s l:

0( ) ( ) ( ) s inrtMy t Cy t Ky t P (1-30)

Hay l 2 0( ) 2 ( ) ( ) s inrtP

y t y t y tM

(1-30)

Trong , P0 v r ln lt l bin v tn s ca lc kch thch; Cn v

nh k hiu trc y. y l PTVP bc hai tuyn tnh chun c v phi l mt

hm iu ha. Nghim tng qut ca (1-30) bng nghim tng qut ca PTVP

thun nht k hiu l y0(t), cng vi mt nghim ring k hiu l y1(t).

y(t) = y0(t) + y1(t) (a)

1.4.1 Xt trng hp lc cn b:

Nghim y0(t) tnh theo (1-26), cn nghim ring y1(t) c th xc nh bng

nhiu cch, v d phng php bin thin hng s Lagrange.Song thun tin hn,

y ta gii bng phng php na ngc nh sau:

Gi thit nghim ring di dng tng qut sau

y1(t) = A1sinrt + A2cosrt

Hay l y1(t) = A0 sin(rt - ) (1-31)

Trong r l tn s lc kch thch bit, cn A0 v l bin v gc

lch pha cha bit. R rng l nu ta tm c mt A0, v mt (1-31) tha

16

mn phng trnh (1-30), th (1-31) l mt nghim ring ca (1-30). Tht vy, thay

y1(t) v cc o hm ca n

1 0( ) os(rt- )y t rA c v 2

1 0( ) sin( )y t r A rt (b)

vo phng trnh (1-30) ta c,

2 2 00 0 0sin( ) 2 os(rt- )+ sin( ) sinrtP

r A rt rA c A rtM

(c)

Khai trin sin(rt-) v cos(rt-), ri nhm cc s hng c cha sinrt v cosrt ta

c:

2 2 2 200 0 0 0 0 0

Psinrt -r os +2 rA sin os - osrt r sin 2 os - sin 0

MA c A c c A rA c A

(d)

Biu thc (d) phi bng khng vi mi t ty ; Mun vy, cc biu thc

h s ca sinrt v cosrt phi bng khng. T suy ra:

A0 = sin 2rcosrMP

22

0

(1-32)

tg =22 r

2r

(1-32)

Thay (1-32) v (1-32) vo (1-31) ta c nghim ring y1(t); Ri li thay

(1-26) v (1-31) vo (a) ta c nghim tng qut ca PTVP dao ng (1-30) l:

1 0( ) sin( ) sin( )y t A t A rt (1-33)

Trong : A, tnh theo (1-27) cha cc iu kin u y0 v v0.

A0, tnh theo (1-32) cha bin P0 v tn s r ca lc kch thch

iu ha. Phn tch (1-33) ta thy:

S hng th nht lin quan ti dao ng t do ca h. Trong thc t lun

lun tn ti lc cn. Nhng cho d lc cn l b, th phn dao ng t do ny, sm

hay mun, cng s mt i sau mt khong thi gian no . Dao ng ca h lc

ny c coi l n nh, v c biu din bng s hng th hai trong (1-33).

1 0( ) ( ) sin( )y t y t A rt (1-34)

Nh vy, dao ng cng bc - lc cn b - ca h mt bc t do chu lc

kch thch iu ha P0 sin rt, khi n nh, l mt dao ng iu ha c cng

17

tn s v chu k vi tn s v chu k ca lc kch thch, cn bin A0 v gc pha

c tnh theo (1-32).

Bin dao ng A0 cng thng c biu din dng khc tin li hn

nh sau:

T (1-32) ta c, 2r = [(2 r2)sin]/ cos, ri thay vo (1-32) c:

A0 = P0 cos / M(2-r

2) (f )

Thay tnh theo (1-32) vo (f ) vi ch : M = 2

1

v Cos(artg) =

21

1

(g)

Ta c,

A0 =

2222222

22

0

2

22

22

0

r

2rrrM

P

r

2r1

1

rM

P

hay

A0 =

4

222

2

2

0

22222

0

4r

r1

P

4rrM

P

K hiu: 0( )0.P

tP y l chuyn v tnh ti ni t khi lng do lc c

tr s bng bin lc ng P0 t tnh ti gy ra, v

K =

4

222

2

2

4r

r1

1

(1-35)

Th ta c 0( )0 .P

tA y K (1-32)

iu ny c ngha l, khi h chu tc dng ca ti trng ng iu ha

P0sinrt, th bin chuyn v ng A0 ln gp K ln so vi chuyn v khi P0 t

tnh gy ra. K c gi l h s ng.

18

H s ng cng c th c biu din qua h s cn c. c gi c th t

vit cng thc ny.

1.4.2 Xt trng hp khi khng c lc cn :

H s ng trong trng hp ny c dng n gin hn (cho = 0 trong

cng thc 1-35)

K =

2

2

r1

1 (1-36)

Kt qu ny cng c th tm c nh gii trc tip PTVP dao ng cng

bc khng c lc cn. c gi c th t thc hin iu ny.

1.4.3 Phn tch h s ng Hin tng cng hng

Nhn vo cng thc (1-35) v (1-36) ta thy, h s ng ph thuc vo t s r/.

a) Xt trng hp khng c lc cn:

th quan h gia h s ng v t s r/ v c nh trn hnh (1.12a) vi

ch l h s ng ch ly gi tr dng

.Ta thy rng,

Khi t s

r 0 th K 1

r th K 0

r 1 th K

Ngha l, khi tn s lc kch thch ln hn nhiu tn s ring ca h, h s

ng c gi tr nh, thm ch bin dao ng cn nh hn c chuyn v tnh do

Po gy ra. C th l gii iu ny l do khi r>, K c tr s m, v mt ngha,

19

iu ny c ngha l dao ng ca khi lng ngc pha vi lc kch thch (chiu

chuyn ng ngc vi chiu ca lc kch thch), nn lc kch thch chng li

chuyn ng.

Khi r

20

P(t)

t 0

Hnh 1.13: Ti trng kch ngng

f(t) P0

h s K lun lun nh hn mt. Trng hp ring khi h s cn ly du bng

trong cng thc (1-37) c gi l h s cn l tng; v c ngha quan trng

khi ch to cc thit b o dao ng.

b2- Khc vi trng hp khng cn, khi c lc cn, h s ng c gi tr

ln nht khng phi khi r/ bng mt, m khi t s ny nh hn mt. Tht vy,

kho st biu thc K theo t s r/, t (1-35) hay (1-35) ta c K t cc tr khi :

rd

dK = 0 suy ra

r

22

2

2

2

2M

c1

21 < 1 (1-37)

(B qua bin i chi tit)

Tuy nhin s sai khc ny l nh, nn thc t vn coi gn ng K t gi

tr ln nht khi r/ 1.

1.5 H MT BC T DO CHU TI TRNG KCH NG

HM NG LC V TCH PHN DUHAMEL

Nh trnh by trong phn m u, ti trng kch ng l ti trng tc

dng vo cng trnh mt cch t ngt vi cng ln, ri gim nhanh sau mt

khong thi gian tng i ngn. Tuy thi gian

cht ti ngn, nhng ta cng khng th b qua yu

t thi gian ny trong tnh ton.

K hiu P0 l gi tr ln nht m ti trng

t c, f(t) l hm biu din lut bin i ca ti

trng theo thi gian, cn gi l hm cht ti. Khi

c th biu din ti trng kch ng di dng

tng qut nh sau (hnh 1.13).

P(t) = P0f(t) (1-38)

Do chu ti kch ng, nn trng thi nguy him ca kt cu xy ra kh

nhanh sau khi chu ti. Bi vy, trong trng hp ny ngi ta thng b qua nh

hng ca lc cn. PTVP dao ng tng qut c dng:

0( ) ( ) ( )My t Ky t P f t (1-39)

hay 2 0( ) ( ) ( )P

y t y t f tM

(1-39)

21

C th gii phng trnh ny bng nhiu cch. y ta gii theo cch h

dn bc o hm bng cc php bin i tng ng nh sau .

Trc ht nhn hai v ca (1-39) vi sint, cng v tr vo v tri hm

( ) os( t)y t c ta c:

2 0sin os t sin os t ( )sinP

y t yc y t yc f t tM

Hay 0sin os t ( )sinPd d

y t y c f t tdt dt M

(a)

Tch phn hai v ca (a) theo cn t t0 ti t ta c:

0 0

0

0sin os ( )sin

tt t

t tt

Py y c f d

M (b)

Trong l mt thi im no trong khong t t0 ti t (do cn tch

phn l t nn bin tch phn phi l )

S dng iu kin u: 0

0( ) ty y ; 0 0( )

ty v

(c)

th phng trnh (b) tr thnh:

0

00 0 0 0sin sin os t+y os t ( )sin( )

t

t

Py t v t y c c f d

M (1-40)

Tip theo, ta li thc hin cc php tnh theo ng th t nh trn nhng

nhn hai v ca (1-39) vi cost; Sau cng v tr vo v tri hm ( sin )y t , ri

tch phn hai v vi cn t t0 ti t, v s dng iu kin u (c); Ta li c mt

biu thc c dng tng t (1-40):

0

00 0 0 0os os sin t-y sin t ( ) os( )

t

t

Pyc t v c t y f c d

M (1-40)

Cc phng trnh (1-40) v (1-40) ch l dng khc ca (1-39) nh cc

bin i tng ng. By gi ta li nhn hai v ca (1-40) vi cost, v vi

(1-40) l sint; ri tr hai phng trnh cho nhau, vi ch cc quan h lng

gic sau:

sin(a-b) = sina cosb - cosa sinb

cos(a-b) = cosa cosb + sina sinb (d)

22

Ta c

0

00 0 0 0( ) sin ( ) os (t-t ) ( )sin ( )

t

t

Py t v t t y c f t d

M

Suy ra

0

0 00 0 0( ) sin ( ) os (t-t ) ( )sin ( )

t

t

v Py t t t y c f t d

M

Hay

0

0

00 0 0( ) os (t-t ) sin ( ) ( )sin ( )

t

P

t

t

vy t y c t t y f t d

(1-41)

Trong , 0( )0

P

ty P l chuyn v tnh ca khi lng do lc c tr s bng

P0 t tnh gy ra.

(1-41) l nghim tng qut ca PTVP (1-39), trong c cha tch phn

0

( ) ( )sin ( )

t

t

K t f t d (1-42)

c gi l tch phn Duhamel.

Nh vy, phng trnh chuyn ng ca h mt bc t do, chu tc dng

ca lc kch ng vit di dng (1-38), l hon ton xc nh nu bit cc iu

kin u (y0,v0) v hm cht ti f(t). Khi khng c ti trng tc dng, phng trnh

(1-41) tr v phng trnh (1-18) l phng trnh vi phn dao ng t do ca h

khi khng c lc cn.

Nu iu kin u y0 =0, v v0 =0; th phng trnh chuyn ng ch cn

li s hng th ba trong (1-41).

0( )( ) ( )Pty t y K t (1-43)

Ch : Li gii (1-41), hay (1-43) l li gii tng qut khng nhng cho trng

hp ti trng kch ng nh trnh by trn, m cho ti trng ng bt k c th

biu din c dng (1-38).

Hm K(t) ng vai tr nh hng ca tc dng ng, n l hm ca thi

gian, c gi l hm nhn t ng hay l hm ng lc. Gi tr ln nht ca K(t)

chnh l h s ng. Trong thc t tnh ton, ta cn xc nh gi tr ln nht ny.

23

Sau y ta xt mt s dng ti trng kch ng thng gp, vi gi thit ban

u h trng thi tnh, ngha l y0 = 0, v v0 = 0. Lc ny phng trnh chuyn

ng ca h l (1-43).

1) Lc khng i tc ng

t ngt vo khi lng.

th hm cht ti nh

trn hnh 1.14a; Lc ny c:

P = P0

f(t) = 1 (t 0) (a)

Nn,

K(t) = t

0

sin(t-) d

= 1 cost (b)

th hm K(t) ny nh trn hnh 1.14b, v ta c

K = max K(t) = 2

2- Ti trng kch ng dng ch nht (nh trn hnh 1.15a)

Khi 0 t t1, c P = P0, v f(t) = 1; nn theo (b) ta c:

K(t) = 1 cost (c1)

Khi t1 t , c P = 0 , v f(t) = 0; nn theo (1-42) ta c:

K(t) =2sin(2

t1 ) sin(t-2

t1 ) (c2)

Trong t1 l thi gian cht ti.

Trong trng hp ny, s bin i ca hm ng lc , cng nh gi tr ln

nht ca n (K) ph thuc t1. S bin i ca K(t) theo thi gian, ng vi cc t1

khc nhau, c th hin trn hnh 1-15b; Cn quan h gia maxK(t) = K vi t

s T

t1 c th hin trn hnh 1.15c. R rng l, khi t1 cng ln, trng hp ny s

tr v trng hp (1). V trong thc t, khi t1 2

T l c th coi nh trng hp

(1) xem hnh 1.5c; Lc ny K 2. Cn t1 cng ln th tn s cng ln. y, T

l chu k dao ng t do.

K(t)

P t

P(t)

0

t

2T

2T

0

1

2

Hnh 1.14: Lc tc ng t ngt

24

3- Ti trng tng tuyn tnh ri sau khng i (nh trn hnh 1.16a.)

Khi 0 t t1, c P = P0(1t

t); Cn f(t) =

1t

t; Thay vo (1-42) ta c

hm ng lc trong trng hp ny l:

K(t) = 1t

t -

1t

tsin

=

1t

t (

1 t2

T

)sint (d1)

Khi t1 t, c P = P0; Cn f(t) = 1; Nn trong trng hp ny

K(t) = 1 + (1 t2

T

)[sin(t-t1) sint] (d2)

Trong , T=

2 l chu k dao ng t do.

th bin i ca K(t) theo thi gian, ng vi cc t1 khc nhau, nh trn

hnh 1.16b; Cn quan h gia maxK(t) = K vi t s T

t1 nh trn hnh 1.16c. Ta

thy, khi t1 cng nh (t1 0) , n tin dn ti trng hp (1): K 2.

1

5

4t T

110

Tt

5t1 4t1 t1

0

1

2

k(t)

t

b)

Hnh 1.15

0,6 0,4 0,2

0

1

2

max k(t)

1t

T

0,8 c)

P(t)

P

t1

t

a)

0

Dng cht ti Bin i ca K(t) ng vi cc t1 khc nhau

Quan h gia K vi 1t

T

25

4, Ti trng kch ng dng tam gic (nh trn hnh 1.17a.)

Khi 0 t 2

t1 , c P = 2(1t

t)P0; Cn f(t) =

1t

2t; Nn theo (1-42) ta c:

K(t) = 1t

2t (

1 t

T

)sint (f1)

Khi 2

t1 t t1, c P = (2-1t

2t)P0; Cn f(t) = (2-

1t

2t); Nn ta c:

K(t) = 2 1t

2t + (

1 t

T

)[2sin(t-

2

t1 ) - sint (f2)

Khi t1 t; c P = 0; Cn f(t) = 0; Nn lc ny ta c:

K(t) = (1 t

T

)[- sin(t-t1) + 2sin(t-

2

t1 ) sint] (f3)

S bin i ca K(t) ng vi cc t1 khc nhau nh trn hnh 1.7b; Cn quan

h gia maxK(t) = K vi T

t1 nh trn hnh 1.17c. V ta thy K lun lun nh hn

hai.

Qua cc v d trn, ta c th rt ra mt s nhn xt quan trng.

3 2 1

0

1

2

max k(t)

1t

T

4 c)

3t1 2t1 t1

0

1

2

k(t)

4t1 b)

t

14

Tt 1

10

3

Tt

Hnh 1.16

P(t)

P

t1

t

a)

0

3 2 1

0

1

2

max k(t)

1t

T

4 c)

P(t)

P

t1

t

a)

0

3t1 2t1 t1

0

1

2

k(t)

4t1 b)

t

1

5

4

Tt 1

4

Tt

Hnh 1.17

26

a, Khi chu tc dng ca ti trng kch ng, h s ng c gi tr nh hn ,

hoc bng hai.

b, Khi thi gian cht ti kch ng t1 l nh so vi chu k dao ng ring, ta c

th gii gn ng bi ton vi gi thit: khi lng ch bt u chuyn ng sau

thi gian t1. Nh vy, da vo nguyn l ng lng ta c:

0

0

( )

t

J P t dt Mv ; Suy ra 0J

vM

(g)

Ngha l, c th thay bi ton h chu ti kch ng c t1 nh, bng bi ton

h chuyn ng c vn tc ban u v0 gii n gin hn nhiu. Li gii loi bi

ton ny c th tm thy trong cc ti liu.

***********

27

CHNG 2: DAO NG CA H C NHIU BC T DO

2.1 KHI NIM BAN U

Nh trnh by chng 1; h mt BTD c c trng bng mt dng

dao ng ring vi tn s . Tng t nh vy, dao ng t do ca h nhiu bc

t do cng c c trng bng cc tn s dao ng ring, v ng vi mi tn s

ring h c mt dng dao ng ring tng ng. Hay ni cch khc nh sau ny s

chng minh, h c bao nhiu bc t do s c by nhiu tn s dao ng ring, v

trong cc iu kin nht nh, ta c th lm cho tt c cc khi lng ti mt

thi im no - ch thc hin dao ng tng ng vi mt tn s no trong

s cc tn s ring. Nhng dng dao ng nh vy c gi l nhng dng dao

ng ring chnh, hay dng dao ng chun. Tt nhin dao ng t do ca h l

tng hp ca tt c cc dng dao ng ring ny.

Vic nghin cu cc dng dao ng ring chnh l rt quan trng v n n

gin (nh h mt bc t do); sau hp cc dao ng ny s cho dao ng tng

cng. Trong thc t ta cng gp nhiu bi ton c s BTD hu hn, bi v ngi

ta thng chuyn bi ton c v hn BTD ( gii phc tp)v bi ton c s BTD

hu hn gii n gin hn.

2.2 PHNG TRNH VI PHN DAO NG NGANG TNG QUT

CA H C n BC T DO

Xt h c n BTD, n khi lng tp trung M1,M2,...,Mn, nh trn hnh

2.1(b qua khi lng kt cu). H dao ng di tc dng ca h lc ng P1(t),

P2(t), ...,Pn(t), trong trng hp tng qut, gi thit t ti tt c cc khi lng, v

c phng theo phng chuyn ng. Trng hp c cc ti trng khng t ti

khi lng, th ta phi chuyn tng ng v t ti khi lng (xem mc 2.4).

P1(t)

y1(t)

z

y

P2(t) Pn(t)

y2(t) yn(t)

M1 M2 Mn .....

.

.....

.

Hnh 2.1

28

Khi dao ng, ti mi khi lng u chu tc dng ca cc ngoi lc nh

sau,

+ Ngai lc ng (nu c) Pk(t);

+ Lc qun tnh Zk(t) = - Mkk(t)

+ Lc cn Rk(t)

y, k l khi lng th k;( k = 1, 2, ,n); Cn lc n hi Rh(t) khng

phi l ngoi lc. Hp ca cc ngoi lc ny , k hiu l Fk(t), th:

( ) ( ) ( ) ( )k k k kF t Z t R t P t (a)

Gi s ti thi im t ang xt, khi lng th k chuyn ng hng xung

cng chiu vi lc P(t), th nh phn tch mc 1.2, biu thc (a) c dng:

( ) ( ) ( ) ( )k k k k kF t M y t R t P t (b)

Di tc ng ca h lc ny, dm s thc hin dao ng . PTVP dao ng

ngang tng qut ca h cng c th thit lp c t iu kin cn bng ng vit

ti tng khi lng.

Rhk(t) Fk(t) = 0 (c)

( k = 1, 2, ,n)

Song trong trng hp ny, s dng biu thc chuyn v t ra thun tin hn.

Chuyn v ca cc khi lng ti thi im no , gi s xt khi lng

th k,

yk(t) = k1 F1(t) + k2 F2(t) +..+ kn Fn(t) (d)

Cho k bin thin t ( k = 1, 2, , n); ta c h n PTVP chuyn ng ca

n khi lng ti thi im t l:

1 11 1 12 2 1

2 21 1 22 2 2

1 1 2 2

( ) ( ) ( ) .... ( )

( ) ( ) ( ) .... ( )

................................................................

( ) ( ) ( ) .... ( )

n n

n n

n n n nn n

y t F t F t F t

y t F t F t F t

y t F t F t F t

(f)

29

Hay dng ma trn,

1 11 12 1 1

2 21 22 2 2

1 2

( ) .... ( )

( ) .... ( )

.... .... .... .... .... ....

( ) .... ( )

n

n

n n n nn n

y t F t

y t F t

y t F t

(f)

Trong , kj, (k, j = 1,2,,n) l chuyn v n v ti khi lng th k do

lc bng n v t ti khi lng th j gy ra.

K hiu cc ma trn v cc vc t nh sau:

11 12 1

21 22 2

1 2

....

....

.... .... .... ....

....

n

n

n n nn

N

;

1

2

.... .... .... ....

n

M

MM

M

;

11 12 1

21 22 2

1 2

....

....

.... .... .... ....

....

n

n

n n nn

C C C

C C CC

C C C

(2-1)

1 1 1 1

2 2 2 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ; ( ) ; ( ) ; ( )

.... .... .... ....

( ) ( ) ( ) ( )n n n n

y t y t y t P t

y t y t y t P ty t y t y t P t

y t y t y t P t

( 2-2)

Trong , [N] l ma trn i xng, v c gi l ma trn mm ca h, gm

c cc phn t l cc chuyn v n v ti ni t cc khi lng , theo phng

chuyn ng.

[M] l ma trn khi lng , l ma trn ng cho. Cc phn t trn

ng cho chnh ln lt l cc khi lng tp trung t trn h.

[C] l ma trn cn. Vic xc nh cc phn t ca [C] kh phc tp.

Trong tnh ton thc t, ngi ta thng coi gn ng [C] t l vi ma trn cng

[K].

30

( ) ; ( ) ; ( )y t y t y t , ln lt l vc t chuyn v, vc t vn tc, v

vc t gia tc chuyn ng ca h, m cc phn t ca n, ln lt l chuyn v,

vn tc, v gia tc chuyn ng ca cc khi lng.

{P(t)} l vc t ngoi lc ng, c cc phn t l cc ngoi lc ng

tc dng ti cc khi lng..

Cn

1 2( ) ( ) ( ) .... ( ) ( )T

c nR t R t R t R t C y t (2-3)

l vc t lc cn nht tuyn tnh (t l bc mt vi vn tc ). Thay (b) kt hp vi

(2-3) vo (f) v chuyn v, ta c:

( ) ( ) ( ) ( )N M y t N C y t E y t N P t (f)

y, [E] l ma trn n v cp n.

Nhn bn tri (f) vi

11 12 1

1 21 22 2

1 2

....

....

.... .... .... ....

....

n

n

n n nn

k k k

k k kK N

k k k

(2-4)

ta c PTVP dao ng ngang tng qut ca h c n BTD , cn nht tuyn

tnh,di dng ma trn nh sau:

( ) ( ) ( ) ( )M y t C y t K y t P t (2-5)

Ma trn [K] i xng, v c gi l ma trn cng ca h.

So snh hai phng trnh (2-5) v (1-12) ta thy chng hon ton ging

nhau v hnh thc, cho nn cch gii cng c phn tng t nhau. Tuy nhin, gii

h phng trnh (2-5) phc tp hn rt nhiu, v [M], [C], [K] l cc ma trn ch

khng phi l cc con s nh trong (1-12), cn ( ) ; ( ) ; ( )y t y t y t l cc vc t

hm. Sau y ta s gii mt s trng hp ring.

2.3 DAO NG T DO CA H C n BC T DO PHNG TRNH

TN S

2.3.1 Tn s v phng trnh tn s

Khi nghin cu dao ng h mt bc t do ta thy rng, khi lc cn b, tn

s ring 1 ; Bi vy, i vi h nhiu bc t do, khi nghin cu dao ng t

do, ta quan tm ch yu ti trng hp gi thit khng c lc cn.

31

Phng trnh vi phn dao ng t do lc ny c dng n gin:

( ) ( ) 0M y t K y t (2-6)

Gi thit dao ng t do l iu ha, nn phng trnh dao ng t do

khng lc cn ca khi lng th k, cng nh (1-18), c dng:

( ) sin( )k ky t A t (2-7)

v gia tc 2( ) sin( )k ky t A t (2-8)

Trong , Ak l bin dao ng ca khi lng th k, v l tn s v

gc lch pha ca dao ng.

Thay (2-8) v (2-7) vo (2-6), ri khai trin vi (k = 1, 2,,n); v t

sin(t+) lm tha s chung, ta c:

2 sin( ) 0M A K A t

Do phi tn ti dao ng, sin(t+) 0;

nn, 2 2 0K A M A K M A (2-9)

Trong , {A} = {A1, A2, ,An}T l vc t ct cha cc bin dao ng

ca cc khi lng th nht, th hai, ..., th n, v c gi l vc t bin dao

ng t do ca h. Do phi tn ti dao ng, ngha l {A} {0}. T suy ra

nh thc

2 0D K M (2-10)

Hay dng khai trin:

2

11 1 12 1

2

21 22 2 2

2

1 2

....

....0

.... .... .... ....

....

n

n

n n nn n

k M k k

k k M kD

k k k M

(2-10)

(2-10) l phng trnh bc n i vi 2. Do [K] v [M] l cc ma trn i

xng, v xc nh dng, nn gii (2-10), ta s xc nh c n nghim thc v

32

dng: 12, 2

2,,n2 ; cng c ngha l ta c n tn s dao ng ring vi qui

c k hiu, 1 < 2 < < n ; (cc gi tr m ca php khai cn khng c

ngha vt l nn b i). iu ny trng vi kt lun nu ra mc 2.1 ca

chng: h c bao nhiu bc t do s c by nhiu tn s ring. Phng trnh

(2-10) c gi l phng trnh tn s.(hay cn gi l phng trnh th k). Tn

s ring b nht 1 c gi l tn s c bn, v c vai tr quan trng trong tnh

ton kt cu khi chu ti trng ng. iu ny s c sng t cc phn sau.

Phng trnh tn s (2-10) cng c th biu din qua ma trn mm.

Mun vy, ta nhn bn tri hai v ca (2-9) vi [N]( 2

1) c:

22 21 1

0N K N M A

Hay 21

0E N M A

(2-9)

y, [E] l ma trn n v cp n, v k hiu: 2

1u

(2-11)

Th t (2-9) ta suy ra phng trnh tn s biu din qua ma trn mm l,

0D u E N M

Hay 0D N M u E (2-11)

hay dng khai trin:

1 11 2 12 1

1 21 2 22 2

1 1 2 2

....

....0

.... .... .... ....

....

n n

n n

n n n nn

M u M M

M M u MD

M M M u

(2-11)

Nh vy, phng trnh tn s c th biu din qua ma trn cng hoc qua

ma trn mm. Tuy nhin trong thc t, ngi ta hay dng ma trn mm hn, v cc

phn t ca n c xc nh d dng hn nh cng thc tnh chuyn v

Maxwell- Mohr quen thuc.

V D 2.1

Xc nh cc tn s dao ng ring ca dm cho trn hnh 2-2a. Bit dm c

E J = hng s, M1 =M2 =M, khi tnh b qua khi lng dm.

33

Bi gii

Bi ton ny c hai BTD, nn phng trnh

tn s l nh thc cp hai sau:

uMMMuM

222121

212111

= 0 (a)

Dm cho l siu tnh, cng thc

Maxwell- Mohr tnh chuyn v l

(Xem gio trnh c hc kt cu)

ik = k0i0ki MMMM (b) y, biu m men n v c

thm ch s 0 l trn h tnh nh (ng vi

trng thi gi to).

Cc biu m men n v v c

nh trn hnh 2-2b,c,d,e; Thc hin nhn cc

biu ta c:

3 3

11 22 12 21

23 3;

1536 512

l l

EJ EJ

(c)

Thay (c) vo (a) v gii phng trnh

bc hai ny i vi u ta c:

3

148

Mlu

EJ

suy ra 1 3 3

1

1 486,9282

EJ EJ

u Ml Ml

v 3

2

7

768

Mlu

EJ

suy ra 2 3 3

2

1 109,7210,4745

EJ EJ

u Ml Ml

Dng dao ng ng vi 1 c dng phn i xng (px) nh trn hnh 2-2f;

cn ng vi 2 l dng dao ng i xng (x) nh trn hnh 2-2g.

z

y

M1 M2

2l

2l

2l

2l

a)

1M

P1=1

b)

4l

332

l

2M

P2=1

d)

4l

332

l

o

kM

Pk=1

c)

4l

o

kM

Pk=1

e)

4l

px f)

y

x g)

y

Hnh 2.2

34

2.3.2 Dng dao ng ring v tnh cht trc giao ca cc dng

dao ng ring

A- Dng dao ng ring

Nu ta thay ln lt cc tn s dao ng ring 1, 2, ...., nvo phng

trnh (2-9), s xc nh c n vc t t s bin dao ng k hiu l{a1}, {a2},

....,{an} ng vi tng tn s ring. V d, ng vi tn s ring th i ta c vc t

bin dao ng {ai} c cc phn t k hiu l (a1i, a2i, ....aki, ....ani); l bin

dao ng ca cc khi lng th (1, 2, ...,k, ...,n) ng vi tn s ring i:

1 2 .... ....T

i i i ki nia a a a a (2-12)

Cc aki (k = 1, 2, , n) l nghim ca phng trnh (2-9) sau y,

([K]-[M]i2){ai} = {0} (2-9)

Cn ch rng, y ta ch xc nh c dng ca cc dao ng ring,

hay ni cch khc, ch xc nh c t s (quan h) gia cc bin dao ng

ca cc khi lng ng vi mt tn s c th. S d nh vy l v, (2-9) l

phng trnh i s tuyn tnh thun nht, s c v s nghim. Mun xc nh mt

h nghim no , ta phi gi thit trc mt bin aki no lm bin c s; Sau

s gii nt (n-1) bin cn li qua bin c s aki ny. R rng, khi cho bin c

s cc tr khc nhau ta s c cc vc t {ai} khc nhau. Tuy vy, t s gia cc

phn t trong vc t ny vi bin c s chn trc lun khng i.

Nu chn n c s ban u aki = 1, th cc t s ny chnh l cc phn t

trong vc t (2-12). Trong thc t, ngi ta thng chn n c s ban u l

a1i = 1, khi vc t bin dao ng ng vi tn s ring i s l:

1 21 .... ....T

i i i ki nia a a a a (2-12)

Trong , cc aki (k = 2, 3, ...n) l nghim ca phng trnh (2-9) ng vi a1i =1.

Cc phn t ca vc t bin (2-12) cho ta dng dao ng ca h ng vi

tn s ring th i c gi l dng dao ng ring th i (hay dng dao ng chnh

th i). Nh vy, h c bao nhiu bc t do s c by nhiu dng dao ng ring.

Nu ta t tt c cc vc t biu din cc dng dao ng ring vo trong

mt ma trn vung, k hiu l [A], th [A] c gi l ma trn cc dng dao ng

ring ca h.

35

11 12 121 22 2

1 2

1 2

1 1 .... 1

........

.... .... .... ....

....

n

n

n

n n nn

a a a

a a aA a a a

a a a

(2-13)

Cng cn phi ni thm rng, (2-9) hay (2-9)l bi ton tr ring in hnh,

nn vic gii phng trnh (2-9) xc nh cc tn s dao ng ring v cc

dng dao ng ring tng ng nh trnh by trn, thc cht l xc nh cc

gi tr ring v cc vc t ring tng ng ca bi ton tr ring ny nh ta quen

thuc trong i s hc.

B- Tnh cht trc giao gia cc dng dao ng ring

Cc dng dao ng ring ca h nhiu bc t do c tnh cht trc giao.

Tht vy, xt hai dng dao ng th i v th k. Thay i v k vo (2-9) ri chuyn

v, ta c:

Vi i c: 2i i iK a M a (a)

Vi k c: 2k k kK a M a (b)

Chuyn tr (a) v ch rng, [M]T = [M]; [K]T = [K] th (a) tr thnh,

2T

i i ia K a M (c)

Nhn bn phi vc t {ak} vo(c), nhn bn tri vc t {ai}T vo (b) ta c:

2T T

i k i i ka K a a M a (c)

2T T

i k k i ka K a a M a (b)

Tr hai phng trnh cho nhau: 2 2( ) ' ( ) ' 0Ti k i kc b a M a

V i k, ta suy ra: 0T

i ka M a (2-14)

V mt ton hc, (2-14) l iu kin trc giao ca hai vc t {ai} v {ak},

cng tc l ca hai dng dao ng ring th i v th k. y l iu phi chng

minh.

Thc hin php nhn ma trn, iu kin (2-14) c th vit dng khai trin

nh sau:

36

1 1

2 2

1 2 1 1 1 2 2 2..... ........ .... .... .... ....

k

k

i i ni i k i k ni n nk

n nk

M a

M aa a a a M a a M a a M a

M a

1

0n

ji j jk

j

a M a

(2-14)

C- Chun ha cc dng dao ng ring

Nu ta thay vc t dng dao ng ring th i, {ai} bng vc t {bi} tha

mn iu kin

{bi}T[M] {bi} = 1 (2-15)

th vc t {bi} c gi l vc t biu din dng dao ng ring th i c

chun ha, hay gi ngn gn l vc t chun ha dng dao ng ring th i.

Nu t cc vc t {bi} vo trong mt ma trn vung, k hiu l [B],

11 12 1

21 22 2

1 2

1 2

....

........

.... .... .... ....

....

n

n

n

n n nn

b b b

b b bB b b b

b b b

(2-13)

th [B] c gi l ma trn chun ha cc dng dao ng ring ca h. Lc ny

theo (2-15) ta c:

T

B M B E (2-15)

Trong [E] l ma trn n v cp n.

S dng dng chun ha ca cc dng dao ng ring kt hp vi h ta

chnh s cho php ta chuyn vic gii bi ton c n BTD v gii n bi ton c mt

BTD n gin hn nhiu c trnh by chi tit trong chng 1.

Tht vy, xc nh {bi}, ta t

1

i i

i

b ad

(2-16)

Trong di l mt h s.

Thay (2-16) vo (2-15) ta rt ra: 2T

i i id a M a (2-16)

37

K hiu ma trn

2

1

2

2

2

.... .... .... ....

n

(2-17)

c gi l ma trn cc tn s dao ng ring, hay ma trn tn s.

Thay (2-16) vo ma trn [A] (2-13), ri thay vo (2-9) ta c:

K B M B (a)

Nhn bn tri c hai v ca (a) vi [B]T v ch ti (2-15) ta c:

T

B K B (2-18)

K hiu vc t {q(t)} nh sau c gi l h ta chnh,

( ) ( )y t B q t (2-19)

Thay (2-19) vo PTVP dao ng t do (2-6) ta c,

( ) ( ) 0M B q t K B q t (b)

Nhn bn tri (b) bi [B]T, kt hp vi (2-15) v (2-18) ta c:

( ) ( ) 0q t q t (2-20)

Phng trnh (2-20) l PTVP dao ng t do khng c lc cn ca h c n

BTD c vit trong h ta chnh di dng ma trn . (2-20) l mt h phng

trnh gm n phng trnh c lp (v l ma trn ng cho) m trong mi

phng trnh ch cha mt hm n. Hay ni cch khc, (2-20) l mt h gm n

phng trnh c lp c dng sau y- l dng PTVP dao ng ca h mt BTD

khng c lc cn(1-14):

2( ) ( ) 0i i iq t q t (2-21)

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

Gii (2-21) (Xem chng 1- h mt BTD) ta c cc nghim qi(t), ri thay

vo (2-19) ta c li gii ca bi ton.

38

V D 2.2 Xc nh cc tn s dao ng ring v cc dng dao ng tng

ng ca dm conson trn c t hai khi lng tp trung nh trn hnh 2-3a.

Dm c EJ khng i v b qua khi lng dm khi tnh. Cho M = 4

ml ( m l

cng khi lng phn b)

Bi gii:

H c hai BTD. Cc chuyn v n v tnh c theo cng thc Maxwell-

Mohr v cho kt qu nh sau:

EJ

l

24

3

11 ; EJ

l

3

3

22 ; EJ

l

48

5 3

2112 (a)

1, Xc nh cc tn s dao ng ring

Cng nh v d 2-1, thay (a) vo phng trnh tn s (2-11) ta c mt

phng trnh bc hai i vi u, gii phng rnh ny ta c (b qua tnh ton chi

tit):

4

1 0,1004ml

uEJ

, suy ra 1 23,156 EJ

l m (b)

v 4

2 0,0043ml

uEJ

, suy ra 2 216,258 EJ

l m

3,05472

3,05472

y 1

b)

0,65472

0,65472

y 1

c)

Hnh 2-3

M1=2M M2=M

y

a)

2l 2

l

39

2, Xc nh cc dng dao ng ring

Thay ln lt 1,2 (hay u1,u2) vo h phng trnh (2-9) (l h hai

phng trnh hai n) , ri gi thit trc n th nht bng 1, ta s gii ra n th hai

l cc bin chuyn ng ca khi lng th nht v th hai, cc dch chuyn

ny cho ta dng dao ng tng ng.C th:

Dng dao ng th nht: Thay u1 vo phng trnh th nht (hoc th hai)

ca (2-9) v cho a11 =1 ta c mt phng trnh cha mt bin a21 nh sau,

1 11 1 11 2 12 211 0M u a M a

Thay M1, M2, 11, 12, u1 vo ri gii ta c, a21 = 3,05472; Vc t bin dao

ng cho ta dng dao ng ring th nht l:

1 11 21 1,0 3,05472T T

a a a

Dng dao ng ny nh trn hnh 2-3b.

Dng dao ng ring th hai hon ton tng t, thay u2 vo (2-9) ri cho

a12 = 1, ta s gii c a22 = -0,655. Do vc t bin cho ta dng dao ng

ring th hai l:

2 12 22 1,0 0,65472T T

a a a

Dng dao ng ring th hai nh trn hnh 2-3c.

Ma trn cc dng dao ng ring ca bi ton ny l:

[A] =

0,654723,05472

1,01,0

aa

aa

2221

1211 (c)

3, Chun ha cc dng dao ng ring

xc nh ma trn chun ha cc dng dao ng ring [B] ta phi tnh cc

h s di. Theo (2-16)ta c:

d12 = {a1}

T[M]{a1} = {1,0 3,05472}

3,05472

1,0M

10

02 = 11,33133M

Suy ra d1 = 3,3662 M

d22 = {1,0 -0,65472}

0,65472-

1,0M

10

02 = 2,42866M, suy ra d2 = 1,55842 M

40

By gi li thay d1,d2 vo (2-16) s c ma trn chun ha [B] nh sau:

{b1} = 1d

1{a1} =

M

1

0,90747

0,2971

3,05472

1,0

M3,3662

1

{b2} = 2d

1{a2} =

M

1

0,42012-

0,64168

0,65472-

1,0

M1,55842

1

Ghp hai vc t b1 v b2 , ta c ma trn chun ha cc dng dao ng ring ca

h:

[B] = M

1

0,420120,90747

0,6416770,29707

(d)

2.3.3 Phn tch ti trng theo cc dng dao ng ring

Xt h c n bc t do, n khi lng M1, M2, ...., Mn; Trn c h ti trng

ng tc dng ti cc khi lng lp thnh vc t ti trng ng nh trong (2-2):

1

2

( )

( )

....

....( )

( )

....

....

( )

k

n

P t

P t

P tP t

P t

(a)

Ta s phn tch h ti trng ny thnh cc thnh phn t ti tt c cc khi

lng tng ng vi cc dng dao ng ring, ngha l:

' ' ' '1 11 12 1 1

' ' ' '2 21 22 2 2

( ) ( ) ( ) .... ( ) .... ( )

( ) ( ) ( ) .... ( ) .... ( )

.... .......................................................

....

( )

....

....

( )

i n

i n

k

n

P t P t P t P t P t

P t P t P t P t P t

P t

P t

' ' ' '

1 2

...

..........................................................

( ) ( ) .... ( ) .... ( )

...........................................................

.......................................

k k ki knP t P t P t P t

' ' ' '

1 2

....................

( ) ( ) .... ( ) .... ( )n n ni nnP t P t P t P t

(b)

41

hay vit dng vc t

' ' ' '1 2( ) ( ) ( ) ..... ( ) ..... ( )i nP t P t P t P t P t (2-22)

Trong , Pki(t) l thnh phn lc tc dng ti khi lng th k tng ng vi tn

s i (dng chnh th i). Nh vy theo (2-19), ta phn tch vc t ti trng

{P(t)} thnh tng ca n vc t ti trng tng ng vi n dng dao ng ring .

Vc t ' ' ' ' '1 11 21 1 1( ) ( ) ( ) .... ( ) .... ( )T

k nP t P t P t P t P t l h ti trng tc

dng ti n khi lng tng ng vi dng chnh th nht.

Vc t ' ' ' ' '2 12 22 2 2( ) ( ) ( ) .... ( ) .... ( )T

k nP t P t P t P t P t l h ti trng tc

dng ti n khi lng tng ng vi dng chnh th hai.

Tng t, ta c cc vc t ti trng tc dng ti cc khi lng tng ng

vi cc dng chnh th ba, th t, v.v. , th k, v.v., th n:

' ' ' ' '1 2( ) ( ) ( ) .... ( ) .... ( )T

n n n kn nnP t P t P t P t P t

xc nh n vc t ny, ta phi xc nh (nn) thnh phn Pki(t)

(k, i = 1,2,,n). V Pki(t) lin quan ti khi lng th k, v dng dao ng th i,

nn ta t:

' ( ) ( )ki k ki iP t M a H t (2-23)

Trong :

Mk l khi lng th k

aki l bin dao ng ca khi lng th k tng ng vi dng dao ng

th i [xem (2-12)]

Hi(t) l hm tng ng vi dng chnh th i, cha bit m ta phi xc nh.

p dng (2-23) cho tt c cc khi lng, ta c vc t ngoi lc ng tc

dng ti cc khi lng tng ng vi dng dao ng th i.

42

'

1 1 1

'

2 2 2

'

'

( ) ( )

( ) ( )

..........

..........

( ) ( )

..........

...........

( ) ( )

i i i

i i i

ki k ki i

ni n ni i

P t M a H t

P t M a H t

P t M a H t

P t M a H t

hay dng ma trn

'1 11

'2 22

'

'

'

( )

( )

.... .......( ) ( )

( )

.... .......

( )

ii

ii

i i

k kiki

n nini

M aP t

M aP t

P t H tM aP t

M aP t

Hay thu gn,

'( ) ( )i i iP t M a H t (2-24)

y, [M] l ma trn khi lng, cn {ai} l vc t (2-12) cho ta dng

dao ng th i . Thay (2-24) vo (2-22), vi (i = 1, 2, , n); ta c:

1 1 2 2( ) ( ) ( ) .... ( ) .... ( )i i n nP t M a H t M a H t M a H t M a H t (c)

Nhn bn tri hai v ca (c) vi vc t {ai}T, ta c:

1 1 2 2( ) ( ) ( ) .... ( )T T T T

i i i i i ia P t a M a H t a M a H t a M a H t

.... ( )T

i n na M a H t (d)

Do tnh cht trc giao (2-14), nn cc s hng th 1, 2, ...,(i-1), (i+1),

(i+2),....,n trong (d) u bng khng; ch c s hng th i l khc khng.

T suy ra:

( ) ( )T T

i i i ia P t a M a H t

hay

( )( )

T

i

i T

i i

a P tH t

a M a (2-25)

43

Thay (2-25) vo (2-24) ta c vc t ti trng tng ng vi dng dao ng th i

'

( )( )

T

i

i i T

i i

a P tP t M a

a M a

(2-24)

Ch :

1- Khi s dng cng thc (2-24) cn lu : (2-24) c hai tha s, (mi

tha s c t trong du ngoc n), v phi tnh ring tng tha s; Tha s

th nht l mt vc t c n phn t; tha s th hai cho ta mt con s; Tch hai

tha s ny l mt vc t c n phn t chnh l vc t ti trng tng ng vi dng

dao ng th i.

Ln lt cho ( i= 1, 2, ..., n) vo (2-24); ta xc nh c n vc t ti trng

tng ng vi n dng dao ng ring ca h, c tch ra t h ti trng cho

ban u. C th kim tra s ng n ca php phn tch t cng thc (b), hoc

(2-22).

2- Bng cch tng t, ta cng c th phn tch chuyn v theo cc dng

dao ng ring.

Nh phn tch h ti trng cho theo cc dng dao ng ring, v nhiu

khi c chuyn v, m sau ny, khi nghin cu dao ng cng bc ca h nhiu

bc t do, ta cng c th chuyn vic gii bi ton h nhiu bc t do phc tp v

gii nhiu bi ton nh h mt bc t do n gin c nghin cu k

chng 1.

2.4 CCH CHUYN TNG NG CC TI TRNG NG T

TI CC V TR BT K TRN KT CU V T TI CC KHI

LNG

Khi trn kt cu c cc lc khng t ti cc khi lng, k hiu l P*k(t)

tc dng, gi s c m lc, lp thnh vc t:

* * * *1 2( ) ( ) ( ) .... ( )T

mP t P t P t P t (2-26)

Lc ny, c th p dng cc kt qu trnh by phn trn, ta phi

thay th tng ng (chuyn tng ng) h lc ny thnh h lc t ti cc

khi lng

C nhiu cch chuyn tng ng nh vy, song ch l gn ng. Sau y

l mt trong cc cch chuyn tng ng nh vy da trn gi thit gn ng

44

cho rng: Hai h lc tng ng l hai h lc gy ra chuyn v tnh ti cc khi

lng bng nhau. K hiu vc t h lc thay th t ti n khi lng l ,

1 2( ) ( ) ( ) .... ( )T

nP t P t P t P t (2-27)

.

v *ik l chuyn v n v ti khi lng th i do lc bng n v t ti lc

P*k(t) gy ra;

ij l chuyn v n v ti khi lng th i do lc bng n v t ti khi

lng th j gy ra. Khi , chuyn v ti cc khi lng th (1, 2, , n ) do h lc

{P*(t)} gy ra, bng chuyn v ny, do h lc thay th {P(t)} gy ra, ngha l:

* * * * * *1 11 1 12 2 1

* * * * * *

1 1 2 2

( ) ( ) ( ) ... ( )

.... ...................................................

( ) ( ) ( ) ... ( )

.... ............................................

( )

m m

i i i im m

n

y t P t P t P t

y t P t P t P t

y t

11 1 12 2 1

1 1 2 2

* * * * * *

1 1 2 2

( ) ( ) ... ( )

................................................

( ) ( ) ... ( )

.......................................................

( ) ( ) ... ( )

n n

i i in n

n n nm m

P t P t P t

P t P t P t

P t P t P t

1 1 2 2

( )

.

( ) ( ) ... ( )n n nn n

c

P t P t P t

K hiu ma trn sau,

* * *

11 12 1

* * *

* 21 22 2

* * *

1 2

....

....

.... .... .... ....

....

m

m

n n nm

N

(2-28)

Lc ny c th biu din h n ng thc (c) di dng ma trn:

* *( ) ( )N P t N P t (c)

*

1 ( )P t

M1 M2 Mn .....

.

.....

.

*

2 ( )P t *( )mP t

Hnh 2.4

1( )P t

M1 M2 Mn .....

.

.....

.

2 ( )P t ( )nP t

45

Suy ra, 1 * *( ) ( )P t N N P t (2-29)

Trong , [N] l ma trn mm ca h c tnh theo (2-1).

2.5 DAO NG CNG BC CA H NHIU BC T DO, KHNG

LC CN CHU LC KCH THCH IU HO: P(t)=P0sinrt

2.5.1 Biu thc ni lc ng v chuyn v ng

Xt h nhiu bc t do chu tc dng ca cc lc kch thch iu ha cng

tn s. Cng nh h mt bc t do, trong thc t lun tn ti lc cn; nn d lc

cn rt nh, th sau mt khong thi gian no , dao ng t do cng s mt i.

Dao ng ca h lc ny hon ton ph thuc lc kch thch iu ha, nn ni lc,

ng sut, v.v. cng thay i iu ha cng chu k vi chu k ca lc kch thch.

Khi h c n BTD, s c n tn s dao ng ring. Khi mt trong s cc tn

s ring xp x bng tn s lc kch thch s xut hin hin tng cng hng.

Thc t, tn s lc kch thch thng nh hn nhiu so vi tn s dao ng ring

(r

46

V dao ng n nh, nn cc i lng nghin cu u bin i iu

ha theo cng mt tn s vi tn s ca lc kch thch. Bi vy, khi ti trng t

bin , th cc i lng nghin cu cng t bin , ngha l:

01 1 2 2 ...

P

K K K K Kn nS S S Z S Z S Z (2-30)

Trong ,

0PKS l tr s ca SK do bin lc ng P0 t tnh gy ra, v c xc

nh bng cc phng php c trnh by trong gio trnh c hc kt cu.

S ki l gi tr SK do lc qun tnh Zi = 1 t tnh gy ra. (i = 1, 2, ,n)

Zi l bin ca lc qun tnh Zi(t)

Nh vy, xc nh c Sk ta phi xc nh c cc bin lc qun tnh Zi.

2.5.2 Xc nh bin ca cc lc qun tnh

Khi b qua lc cn, phng trnh chuyn ng ca khi lng th i, theo

nguyn l cng tc dng , c dng:

1 1 2 2( ) ( ) ( ) ... ( ) ( )i i i in n iPy t Z t Z t Z t t (c)

Trong , iP(t) l chuyn v ca khi lng th i do cc lc ng P(t) gy ra.

Sau khi dao ng n nh, c yi(t), Zi(t), v iP(t) u bin i iu ha vi tn

s r ca lc kch thch. Ngha l:

( ) sinrti iy t v 0( ) sinrtiP iPt (d)

ng thi, 2( ) sinrti iy t r

Trong , i l bin chuyn v ng ca khi lng th i ta ang tnh.

0iP

l chuyn v ca khi lng th i do bin P0 ca cc lc

ng P(t) t tnh gy ra.

Mt khc, lc qun tnh ( ) ( )i i iZ t M y t (f)

Thay (d) vo (f) ta c: 2 2i( ) sinrt=M ( )i i i iZ t M r r y t

Hay rt ra: 2

1( ) ( )i i

i

y t Z tM r

(2-31)

47

Thay (a), (d), v (2-31) vo (c), ri chuyn v v t sinrt lm tha s

chung, ta c:

01 1 2 2 2

1... ... sin 0i i ii i in n iP

i

Z Z Z Z rtM r

(g)

V sinrt khc khng (do tn ti dao ng), nn t (g) ta rt ra c h

phng trnh dng xc nh bin ca cc lc qun tnh, khi h chu tc dng

ca cc lc kch thch iu ha v khng c lc cn, nh sau:

0

*

1 1 2 2 ... ... 0i i ii i in n iPZ Z Z Z (2-32)

( i = 1, 2, , n)

Trong ta k hiu, *2

1ii ii

iM r

(2-33)

( i = 1,2,,n)

Gii h phng trnh (2-32) ta c bin ca cc lc qun tnh. Nu kt

qu tnh ra dng, th chiu gi thit ban u ca lc l ng; Nu kt qu tnh ra

m, th chiu gi thit l sai v phi i ngc li. t cc lc qun tnh v cc lc

kch thch theo ng chiu , v c tr s bng bin ca chng; ta s xc nh

c i lng cn tm bng l thuyt tnh hc c trnh by trong gio trnh

c hc kt cu.

V D 2. 3

Cho dm di l = 6m, trn t hai m t, trng lng mi m t l G = 10

kN. Khi m t th nht quay vi vn tc n = 450 v/pht to ra lc ly tm P0 = 5

kN. Xem hnh 2.5a.

Yu cu : Xc nh cc tn s dao ng ring, v v biu bin m men

ng, v biu m men tng cng ca dm. Bit dm c: J = 8880 cm4;

E = 2,1. 104 kN/cm

2; Ly g = 981cm/s2 v b qua khi lng v trng lng ca

dm trong tnh ton.

48

Gii: 1, Xc nh cc tn s dao ng ring

H c hai bc t do, nn phng trnh tn s, theo (2-11) c dng

1 11 2 12

1 21 2 22

0M u M

DM M u

(a)

T cc biu m men n v trn hnh 2.5c v d; ta tnh c theo

Maxwell-Mohr:

3 3

11 22 12 21

4 7;

243 486

l l

EJ EJ (b)

Thay (b) vo (a) v gii, ta c:

3 3

1 2

5;

486 162

Ml Mlu u

EJ EJ

Nn 2 21 23 32 1

1 162 1 486;

5

EJ EJ

u Ml u Ml (c)

Thay 2 210

1,02 ; 69,81

G kNs kNsM l m

g m m v EJ vo (c), ta c

11 52,5s v 12 203s

P0

oPM b)

0

2

9P l

P1=1

1M c)

2

9l

P2=1

2M d)

2

9l

Hnh 2.5

d

kNm

M

58.41 54,855

( )

kNm

M

tM

20 20

tc

kNm

M

78.41 74,855

P(t)=P0sinrt

a) M1 M2

G2 G1

3l

3l

3l

49

2, Xc nh bin cc lc qun tnh

H phng trnh xc nh bin hai lc qun tnh Z1 v Z2, theo (2-32), trong

trng hp ny l,

0

0

*

11 1 12 2 1

*

21 1 22 2 2

0

0

P

P

Z Z

Z Z

(d)

Tn s lc kch thch r = 2n/60 = 50 s-1; ng thi thay l, E, J vo (b) ta

tnh c:

11 = 22 = 1,908.10-4

m

kN; 12 = 21 = 1,67. 10

-4 m

kN; (b)

Thay M, r, 11 vo (2-25) c: *11 = *22 = - 2,013.10-4 m

kN (f)

(n v ca ik trong (b) v (f) c gii thch v d 4- 1)

Cn 01P

, v 02P

c th tnh c t cc biu m men n v trn cc

hnh 2.5c v d, v biu 0P

M trn hnh 2.5b; Tuy nhin y c th tnh n

gin hn, bi v:

0

4

1 0 11. 5 .1,908 9,54.10Pm

P kN mkN

0

4

2 0 21. 5 .1,67 8,35.10Pm

P kN mkN

(g)

Thay (f), (g) vo (d) v gii h phng trnh ny ta c:

Z1 = 25,98 kN; Z2 = 25,65 kN (h)

3, V cc biu ni lc yu cu

C hai cch v biu bin m men ng

a, Cch v trc tip:

Theo cch ny, ta t vo h cc lc c tr s bng bin cc ngoi lc

ng v bin cc lc qun tnh, ri tnh ton nh vi bi ton tnh di tc

dng ca cc lc ny.

50

b, Cch v theo nguyn l cng tc dng:

Theo cch ny , biu bin m men ng c v theo cng thc (2-30)

M01 1 2 2

. . PM Z M Z M (i)

Kt qu cho trn hnh 2.5e;

So snh hai biu : M v 00

( )P

t PM M ta thy rng, trong trng hp

tng qut, h s K ti cc tit din khc nhau l khc nhau. Nh vy, khc vi h

mt bc t do, i vi h nhiu bc t do, ta khng c mt h s K chung cho tt

c cc tit din ; cng nh khng c mt h s K chung cho tt c cc i lng

nghin cu. Tuy nhin nh sau ny s thy, n gin trong tnh ton , ng thi

thin v an ton, i vi mt i lng nghin cu ta cng c th dng mt h s

K chung, l K ca tit din c tr s K ln nht. V d trng hp ang xt,

h s ng c tr s ln nht i vi m men l ti tit din t khi lng M2:

Max K = 3,335

58,855 = 16,4.

Thm ch, nhiu khi ngi ta cn dng mt h s K chung cho tt c cc

i lng. Lc ny h s ng c tnh theo cng thc (1-36) i vi h mt bc

t do, m ta ly tn s ring b nht 1 tnh ton.

c, Biu m men tng cng:

Trc khi ng c t ti khi lng th nht hot ng, trong dm c

ni lc do trng lng ca hai ng c gy ra. Biu m men ny v c nh

trn hnh 2.6f. (k hiu l Mt(M)

). R rng, khi ng c lm vic, m men ln nht

xut hin trong dm s l tng ca hai biu ny, v ta gi n l biu m men

tng cng.

Mtc = M + Mt(M)

(k)

Kt qu nh trn hnh 2.5g.

Ch :

Ta cng c th gii bi ton bng cch phn tch ti trng iu ha cho

ra n h ti trng tng ng vi cc dng dao ng ring (vi bi ton ang xt

n=2) ri gii bi ton nh h mt bc t do. Cch lm ny s c trnh by mc

tip theo y.

51

2.6 DAO NG CNG BC CA H NHIU BC T DO, KHNG

LC CN, CHU LC KCH THCH BT K P(t)

Trong trng hp ny, ta c th gii bi ton bng nhiu cch.

Cch 1: Ta phn tch ti trng bt k thnh cc hm iu ha di dng chui

lng gic, ri gii bi ton nh c trnh by mc 2.5. Khi ti trng ng

P(t) c chu k T, th c th phn tch thnh chui lng gic nh sau:

0 1 2 k 1 2 k( ) sinrt+a sin2rt+...+a sin .... osrt+b os2rt+...+b oskrt+...P t a a krt b c c c (2-34)

Trong , r = T

(c gi l tn s c bn ca lc kch thch)

a0 = dttPT

1T

0

ak = T

0

tPT

k sinkrt dt (2-34)

bk = T

0

tPT

kcoskrt dt

Cch 2: S dng h ta chnh a h n BTD v n bi ton h mt BTD.

Xt trng hp khng lc cn, PTVP dao ng tng qut c dng:

( ) ( ) ( )M y t K y t P t (2-35)

Nhn bn tri (2-35) vi [B]T , v ch ti (2-19) ta c:

( ) ( ) ( )T T T

B M B q t B K B q t B P t

Hay ( ) ( ) ( )T

q t q t B P t (2-35)

Phng trnh (2-35) l mt h gm n phng trnh c lp c dng nh

dng PTVP dao ng ca h mt BTD (1-39);

[iu ny cng c ni ti khi nghin cu dao ng t do im (c) ca mc

2.3.2]

2 1 1 2 2( ) ( ) ( ) ( ) .... ( )i i i i i in nq t q t B P t B P t B P t (2-36)

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

52

Trong , Bij (i, j = 1,2,....,n) l cc phn t ca ma trn chun ha B .

Nghim ca phng trnh (2-36) c biu din qua tch phn Duhamel

(1- 41) nh bit mc (1- 5). Thay cc nghim qi(t) vo (2-19) ta c li gii

ca bi ton.

Cch 3: Da vo (2-24) ta phn tch vc t ti trng {P(t)} theo cc dng chnh,

sau gii n bi ton nh h mt BTD c trnh by chng mt.C th l:

( B qua cc bin i chi tit):

Phng trnh chuyn ng ca khi lng th k di tc dng ca h lc

ng {P(t)} l:

1

( ) ( )n

k ki i

i

y t a S t

(2-37)

Trong Si(t) l nghim ca PTVP sau:

2( ) ( ) ( )i i i iS t S t H t (2-38)

y Hi(t) c tnh theo cng thc (2- 25), cn aki l thnh phn th k ca

vc t bin dao ng ca dng chnh th i. (xem 2-12); phng trnh vi phn

(2-38) c dng nh ca h mt BTD, nghim tng qut ca n c biu din qua

tch phn Duhamel (1-41) nh c trnh by trong mc 1-5.

************

53

Z(z,t)

b)

y

z a)

q(z,t)

c)

Rc(z,t)

Hnh 3.1

Chng 3 DAO NG NGANG CA THANH THNG

C V HN BC T DO

3.1 PHNG TRNH VI PHN TNG QUT DAO NG NGANG

CA THANH THNG

Mt h kt cu thc t lun lun c v hn bc t do. Xt on thanh thng

c t trong h ta (yz). Xt trng hp tng qut thanh c tit din thay i

vi khi lng phn b cng m(z), chu tc dng ca h lc ngang phn b

cng q(z,t) nh trn hnh 3.1a..

Dao ng ngang ca h ti thi

im no , chnh l v tr ng n

hi ca n ti thi im xt. Phng

trnh ng n hi khi h chu tc

dng ca ti trng ng, ph thuc

hai bin l z v t, ngha l:

y = y(z,t) (a)

Mi quan h gia ng n

hi ca trc thanh c tit din thay

i vi ti trng ngang phn b trn

thanh, trng hp ti trng tnh,

c nghin cu trong gio trnh Sc

bn vt liu:

zqzydz

dzEJ

dz

d2

2

2

2

(b)

Vi qui c trc y hng xung l dng, cn ti trng hng ln l dng.

Trng hp ti trng ng th:

2

2 2( ) ( , ) ( , )EJ z y z t p z t

z z

(3-1)

y, p(z,t) l tng ti trng ngang tc dng trn dm (chiu hng ln l

dng). Khi dao ng, gi s ti thi im t h ang chuyn ng hng xung

54

cng chiu vi trc y, ngoi lc kch thch q(z,t), thanh cn chu tc dng ca h

lc qun tnh phn b :

2

2( , ) ( ) ( , )Z z t m z y z t

t

(c)

V lc cn phn b: R(z,t) (ngc chiu chuyn ng) (d)

Do ta c: ( , ) ( , ) ( , ) ( , )P z t q z t Z z t R z t

hay l 2

2( , ) ( , ) ( ) ( , ) ( , )P z t q z t m z y z t R z t

t

(3-2)

Thay (3-2) vo (3-1) ri chuyn v, ta c PTVP dao ng ngang tng

qut ca thanh thng c tit din thay i l:

2 2 2

2 2 2( ) ( , ) ( ) ( , ) ( , ) ( , )EJ z y z t m z y z t R z t q z t

z z t

(3-3)

Trng hp ring, khi tit din thanh l hng s, th phng trnh (3-3) c

dng n gin hn:

),(),(),(),(2

2

2

2

2

2

tzqtzRtzyt

mtzyz

EJz

(3-3)

Trng hp dao ng t do th v phi ca (3-3) hay (3-3) bng khng.

Sau y ta gii PTVP (3-3) trong mt s trng hp ring.

3.2 DAO NG T DO KHNG C LC CN CA THANH THNG

TIT DIN HNG S - TNH CHT TRC GIAO CA CC DNG

DAO NG RING

3.2.1 Phng trnh vi phn dao ng t do khng c lc cn

Phng trnh vi phn dao ng trong trng hp ny, theo (3-3) l:

4 2

4 2( , ) ( , ) 0

my z t y z t

z EJ t

(3-4)

y l PTVP o hm ring cp bn thun nht, nghim ca n c th c

biu din di dng tch bin nh sau:

( , ) ( ) ( )y z t y z s t (3-5)

Thay (3-5) vo (3-4) ta c:

55

0zydt

tSdtS

dz

zyd

m

EJ2

2

4

4

Hay chuyn v c:

2

2

4

4

dt

tSd

tS

1-

dz

zyd1

m

EJ

zy (f)

Hai v ca (f) ph thuc hai bin khc nhau nn chng ch bng nhau khi c

hai v cng c gi tr bng mt hng s no , gi s k hiu l 2. Nh vy, t

(f) ta c th biu din PTVP o hm ring cp bn (3-4) bng hai PTVP thng

(ch ph thuc mt bin).

2

2

2

( )( ) 0

d S tS t

dt (3-6)

v

0zyEJ

m

dz

zyd 24

4

(3-7)

Nh , thay cho gii mt PTVP o hm ring (3-4) phc tp, ta gii hai

PTVP thng (3-6) v (3-7) n gin hn nhiu.

3.2.2 Gii PTVP (3-6)-Xc nh quy lut dao ng t do

Phng trnh vi phn (3-6) chnh l PTVP dao ng t do, khng lc cn,

ca h mt bc t do (1-14) c trnh by trong chng 1, nn nghim tng

qut ca n theo (1-18) s l:

( ) Asin( t+ )s t

Hay ( ) sin( t+ )s t (3-8)

y ta cho A = 1; S d lm c nh vy, bi v t (3-5) ta thy bin

dao ng chnh l hm y(z). Bi vy sau ny ta gp A nm trong y(z) lun

[xem(3-5)]. Theo (3-8), dao ng t do ca h c v hn BTD cng l dao ng

iu ha.

3.2.3 Gii PTVP (3-7) Xc nh tn s dao ng ring v dng

dao ng ring

Nghim ca (3-7) l hm y(z) s cho ta bin dao ng, cng chnh l

dng dao ng ring ca h. Do thanh c tit din khng i, nn (3-7) l PTVP

thng cp bn c h s l gng s; Nghim tng qut c dng:

56

31 2 41 2 3 4( )

zz z zy z a e a e a e a e

(g)

Trong a1, a2, a3, a4 l cc hng tch phn, cn 1, 2, 3, 4 l nghim ca

phng trnh c trng ca PTVP (3-7) nh sau:

4 k4 = 0, vi k hiu k4 = 2EJ

m (3-9)

Nn ta c: 1,2 = k; v 3,4 = ik (h)

Thay (h) vo (g) ta c nghim ca PTVP (3-7) l:

1 2 3 4( )

kz kz ikz ikzy z a e a e a e a e (i)

S dng quan h

1

2

x xChx e e v 1

2

x xShx e e (k)

Th (i) tr thnh:

( ) oskz+Dsinkzy z Achkz Bshkz Cc (3-10)

Trong , A, B, C, D, l cc hng tch phn, c xc nh t cc iu kin

bin nh sau:

a, Ti gi ta khp c: vng y(z) = 0; v

m men M(z) = 0 => 2

2

dz

zyd = 0

b, Ti ngm cng c: y(z) = 0 v gc xoay (z) =

dz

zyd = 0 (3-11)

c, Ti u t do c: M men M(z) = 0 => 2

2

dz

zyd = 0; v

Lc ct Q(z) = 0 => 3

3

dz

zyd = 0

thun tin cho tnh ton sau ny , ta t:

A = 2

CC 31 ; B = 2

CC 42 ; C = 2

CC 31 ; D = 2

CC 42 (3-12)

Khi nhim (3-10) c dng:

57

1 2 3 4( ) kz kz kz kzy z C A C B C C C D (3-13)

Trong ta k hiu cc hm nh sau:

Akz = 2

cosch kzkz ;