dong luc hoc cong trinh - duong van thu
DESCRIPTION
jjjTRANSCRIPT
-
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 ;