Tính toán xác định hệ số động của dầm cầu dưới tác động của tải di động : Luận án PTS. Cơ học: 2 02 02
Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (15.01 MB, 51 trang )
MUC LUC
Trang
Muc luc
L6i cam on
1
M6 DAU
Chuong 1
PHUONG TRINH DAO DONG CUA DAM DAN HOI
DU6l
TAC DUNG
3
CUA VAT THE DI DONG
1.1. Phirang trinh tong quat dao dong uon cua dam dan hoi
3
1.2. Phuong trinh dao dong cua dam voi tai h^ng so di dong
5
1.3. Phuong trinh dao dong cua dam v6i vat the di dong
6
1.3.1 Phuong trinh dao dong cua vat vdi nen dao dong
6
1.3.2 Phuong trinh dao dong cua dim du6i tac dung cua vat di
7
dong
1.3.3 He phuong trinh dao dong cua dSm voi vat the di dong
8
1.4. Phirong phap Bup nov-Galerkin trong ly thuyet dao dong
9
cac he lien luc
Chuong 2
KHAI NIEM VA TINH CHAT CUA HE SO DONG
12
LUC CUA DAM CAU
2.1. Khai niem
12
2.2. Xac dinh he so dong lire hoc cua dam cau v6i tai trong
13
h4ng so di dong
2.3. Xac djnh he so dong lire hoc ciia dam cau khi co vat the di
22
dong
Chuong 3
XAC DINH HE SO DONG LUC HOC TlTSO LIEU
38
DO DAO DONG
3.1. Thilr nghiem dong voi cau
38
3. L1 Cac phuong phap kich dong
38
3.1.2 Cac duns cu. ihicl hi do "hi
41
3.1.3 Cac yeu cdu doi v6i thu nghiem dong
3.2. Cac dai lirong do dac diroc
41
42
3.2.1 Tdnsorieng
43
3.2.2 He so can
43
3.3. Tinh he so dong cua dam cau tir so lieu do
KET LUAN
TAI LIEU THAM KHAO
44
45
L d l CAM ON
Tac gia xin bay to long biet cm ch^n thanh va sau s4c doi voi thay giao
hudng dSn Pho giao su Tien sy khoa hoc Nguyin Tien Khiem, nguai da het sue tan
tinh huong dSn, giiip da, tao dieu kien thu£ln Icri va c6 cac chi dSn khoa hoc cho noi
dung nghien cuxi cua lu|in van.
Tac gia chan thanh cam on su giiip do dong vien cua cac thay c6 giao cua
Trung t^m Hop tac dao tao va b6i duong co hoc Vien Co hoc. Nhan day, tac gia
cung xin ch^n thanh cam on su quan tarn, dong vien va tao dieu kien thuan Id cua
Ban giam doc Trung t^m cong nghe thong tin Dien luc Viet nam, noi tac gia dang
c6ng tac.
MO DAU
Cong tac ki^m dinh, chan doan va danh gia cau giao thong 6 Viet nam dang
la mot va'n de duoc nhi^u ngudi quan tarn do he thong cau giao thong la he thong co
so ha tang quan trong co gia tri 16n, dam bao hoat dong cua ca nen kinh te quoc dan
nhung lai da dang phu'c tap vi k^'t c^'u va co nhieu nguyen nhan khac nhau dSn den
su xudng c^'p va hu hong. Doi voi cau giao thong, truoc khi dua vao khai thac dat ra
cong tac kiim tra, danh gia, xem xet viec thi cong co dam bao diing thiet ke va cac
thong so yeu cSu hay khong. Doi v6i cau cu, viec chuan doan danh gia cau phai dua
ra duoc cac danh gia v6 kha nang lam viec tiep theo v^ do bin, do cung, do on dinh,
d6 an toan va tu6i tho con lai ciia cau. Ket luan dua ra la can cu vao mot loat cac so
lieu ihu Ihap duoc tir Ho so cong trinh, lir kiem tra thuc nghiem va thu tai w...
M6t trong nhung thong so quan trong can phai xac dinh khi ki^m tra, danh
gia c^u la he so dong luc hoc. He so dong luc hoc la mot dac trung mo ta anh huong
cua cac tai trong dong len cau so v6i tai trong tinh duoc tinh b^ng ty so giua chuyen
vi dong tren chuyen vi tinh. Trong thiet ke, nguoi ta chu yeu dua vao so lieu phan
tich tinh, phan ung dong dudi tac dung cua cac loai tai trong khac nhau chua duoc
nghien cuu mot each day du. Khi kiem dinh, danh gia cau he so dong duoc xac dinh
b4ng phuong phap thuc nghiem dua tren so lieu do. Tuy nhien, 6 nuoc ta viec xac
dinh he so dong b^ng thuc nghiem lai chua co mot phuong phap thong nhat va hieu
qua. Truoc thuc te do, tinh toan, nghien cuu tinh chat va d6 xuat phuong phap xac
djnh he so dong dua vao dong thoi ca ly thuyet va thuc nghiem, chii trong den
phuong phap xac dinh b^ng thuc nghiem thong qua phan tich dao dong ciia dam
dudi tac dung cua vat the di dong la mot viec lam can thiet.
Tir yeu cau dat ra va muc di'ch nghien cuu, luan van co nhiem vu giai quyet
nhung van de sau:
•
De nghien cuu tac dung cua vat the' di dong so voi tai trong tinh va tai
trong h^ng so di dong len dam cau. Ian luoi viet cac phuong trinh dao
dong cua dam cau dudi dang tong qual, phuong trinh dao dong cua dam
Ciiu dudi lac dung ciia lai irong hang so di dong va phucmg irinh dao dong
cua dam Citu dudi tac dung cua vai the di dong.
Tinh toan xac dinh he so dong cua dam cau ducri tac dong ciia tai di dong
•
Tinh toan va nghien cuu tinh chat cua he so dong luc hoc trong trudng
hop tai trong h^ng so di dong va vat the di dong de co ket luan \6 anh
huong cua v$t th^ di dong den he so dong luc hoc.
•
Tim hieu v6 cong tac thu nghiem dong vdi cau, cac dai lugng do dac duoc
tir viec thu nghiem dong va phuong phap xac dinh he so dong luc hoc
dim ciu dang lam viec tir so lieu do dac thuc nghiem.
Phuong phap nghien cuu duoc ap dung trong luan van la; phuong phap giai
tich d^' giai cac phuong trinh vi phan dao dong cua dam cau va phuong phap thu
nghiem dOng d^ xac dinh cac dac trung dong luc hoc cua dam cau phuc vu viec tinh
toan he so dong lire hoc tir so lieu do.
Noi dung luan van bao gom:
Chuong 1: Phuong trinh dao dong ciia dam dan h6i dudi tac dung cua vat th^
di dong. Trong chuong nay trinh bay \6 cac phuong trinh dao dong t6ng quat cua
dim dan hoi, phuong trinh dao dong ciia dam cau dudi tac dung ciia tai trong h^ng
so di dong va phuong trinh dao dong cua dim ciu dudi tac dung cua vat the di dong.
Chuong 2: Khai niem va tinh chat cua he so dong luc hoc dam cau. Chuong
nay dua ra khai niem ciia he so dong luc hoc dam cau, he so dong luc hoc dam cau
duoc giai ra tir cac phuong trinh dao dong. Tinh chat cQa he so dong cung duoc
nghien cuu.
Chuong 3: Xac dinh he so dong lire hoc dam cau tir so lieu do dao dong.
Cong lac thu nghiem dong cau, cac dai luong do duoc can cho viec tinh toan he so
dong luc hoc va phuong phap xac dinh he so dong luc hoc dua vao so lieu do duoc
irinh bay 6 day.
Ket luan chung dua ra nhimg ket qua chinh va phuong hudng nghien cuu phat
tri^n ciia luan van.
Tinh todn xac dinh he so dong ciia dam edit ducn tdc dong ctia lai di dong
CHUONG1
PHUONG TRINH DAO DONG CUA DAM DAN HOI DUOl TAC DUNG
CUA VAT THE DI D O N G
l.l.Phuong trinh tong quat dao dong uon cua dam dan hoi
Noi dung cua phin nay la thiet lap phuong trinh dao dong uon cua dam dan hoi
vdi tai trong phan bo bat ky.
Gia thiet r^ng:
- dam dong chat vdi do dai /
- mat do khoi luong p
- dien tich tiet dien F
- pF: khoi luong don vi dai
- mo dun dan hoi E
- mo dun quan tinh J
- EJ: do cung chong uon
p(xj}
Hucftig dao dong
Hinh I.}.1 Dao dong uon cua dam
Ta la'y true trung hoa ciia dam khi d trang thai ban dau khi chua cd tac dong
cua tai trong la true x\ true r vuong gdc vdi true x. Dam duoc xet vdi dao dong uon
ve phuong :; bo qua dao dong xodn va dao dong doc true. Tai mat cat A ta cd:
Do vong u=uf.v./i
Gdc xoay (p=(p(xj)
Tinh todn xdc dinh he so dong ctia dam can du&i tdc dong ciia tai di dong
Mo men uon M-M(x,t)
Luc cat Q=Q(x,t)
Y>i thiet lap cac phuong trinh vi phan dao dong uon, ta xet mot phan to ciia
dzim CO chi6u dai dx nhu mo hinh sau day.
p(x,t)
K4
^ ^
M +
A
dx
dx
^
R
Hinh I.J .2 He Ivtc tdc dung len mot phan to cua dam
He lire tac dung len phan to nay bao gom: p(x,t)dx; luc cat Q va
Q-\-~dx;
dx
mo menM va M +
dx; luc quan tinh R = pF—rdx;
dx
'
df
dw
phu thuoc vao van tdc R = (3(pF — )dx, (3 = const.
luc can
Theo nguyen ly D'Alembert, tir di^u kien can bang cac luc theo phuong z
tacd:
dO
•Q-^^dx
dx
hay
d' w
+ Q + p(x,t)dx^pF—rdxdf
-f- + pF(—^ ^P'^)
dx
df
dt
dw
p(pp — )dx = 0
dt
(1-1-1)
= P(''>0
Mat khac mo men uon M tai mat cat bat ky khong chi phu thuoc vao do cong x
ma con phu thuoc vao van tdc bien thien cua nd.
M = EJ(x +
d'w
dx
a^)
dt
:> M = EJ( — r + a — — )
GX
exet
Tinh todn xdc dinh he so dong ciia dam can dudi tdc dong ciia tdi di dong
(1.1.2)
Tir di^u kien can bang cac mo men, bo qua cac dai luong bac cao cua dx ta cd:
Q=
dM
dx
Thay M tir (1.1.2) vao bieu thiic tren ta duoc:
^ c.,.5V
d'w
Q = EJ(—--^-a—T— )
^
dx'
dx'dt
Thay Q vao (1.1.1) ta nhan duoc phuong trinh:
(1.1.3)
dx
dx dt
dt
dt
Day la phuong trinh mo ta dao dong uon cua dam dan h6i vdi tai trong phan bd
bat ky.
1.2.Phirong trinh dao dong cua dam vdi tai h^ng so di dong
Xet tai trong hang so cd gia tri FQ
De mo ta tai hang sd P„di dong tren dam ta sir dung ham Delta-Dirac
S(x)
Theo dinh nghia, ham Delta-Dirac xac dinh bdi he thuc:
S(x-a)
0
ox^a
coc^x = a
cu the:
b(x -a) = lim bjx - a)
i-*0
I
<=> |x-a|
bjx-a) =
0
o
\x-a\ > €
e>0
va cd tinh chat:
CO
l8(x-a)dx=I
—00
\f(x)8(x-a)dx=f(a)
-co
Cd the' xem tai trong tap trung P(, di dong lac dong len dam nhu la mot luc phan
bd/;(.v.;j cd cudng do P„ tren khoang [vf-i:, M+c]inc la tai vi irf lure ihdi cua tai
trong di dong tai thoi diem /. Sir dung ham Delta-Dirac ta c6 p(xj)-
Tinh todn \ac dinh he so dong ciia dam can diccn tdc dong ciia tdi di dong
F„d(x-\t}.
That vay, theo tinh ch^^t ham Delta-Dirac:
P = p(x,t)-2€ =
= P.5(x-vt)-2£
= limP,SJx-vt)-2e^PJim^-2E
c-*0
£-*0
1
= P^
2£
Tir do, phuong trinh dao dong cua dam vdi tai trong hang sd di dong nhu sau:
d^w
d^w
EJ(^^<^^)^f>^(^^P^)
dx
dx dt
d'w
dw
df
dt
= P(^'0-PoS(x-vt)
(1.2.1)
1.3.Phirong trinh dao dong cua dam vdi vat the di dong
U.l.Phuang trinh dao dong ctia vat vol nen dao dong
m
Y
'^ X
/ / / / / / / / / / / / / /
Hinh 1.3.} Vat vcri nen di dong
Xet vat the cd khdi lugfng m dat tren mot 16 xo do cutig k va bo giam
cha'n he sd c. Ca he dat tren n^n cd dich chuyen thing dung la x„.
Lay gdc toa do la nen dung yen ban dau va dua vao cac ky hieu sau day:
Xi - do dai 16 xo khong bi nen (hang sd)
X() - khoang each ban dau tir vat den nen (hang sd)
A XQ= A'/ - Xo - do nen tinh cua 16 xo sau khi vat da dat vao vi tri ban dau
(hang sd)
A- khoang each tu'c thdi (tai thdi diem /) tir vat den nen ban dau (ham cua
thoi gian)
x„ - la djch chuyen tuyel ddi ciia nen (khoang each cua nen tuc thdi den
nen ban dau)
Nhu vay ta cd the' tinh cac dai luong sau day lai thdi diem /
1 .Chuyen dich luyel ddi cua vai, ky hieu la x = xft) - x„ ~x(t)
Tinh todn xdc dinh he so dong ciia dam can didri tdc dong ciia tdi di dong
2.Chuy^n dich tuong ddi ciia vat so vdi n^n, ky hieu la
z = x-x^ ^x(t)~xjt)
=
x(t)-x^-xjt)
3.Tai trang thai ban dau can bang tinh cua vat ta cd:
kAx,=mg
(1.3.1)
4.Luc dan hdi sinh ra trong 16 xo bang he sd k nhan vdi chenh lech do dai
ciia 16 xo so vdi trang thai tu do, tu'c la:
P^=k-(x-x^-x,) = k-(x-x^-x,^-AxJ = k(x-xJ-kAx,
^
P^=kz-mg
(1.3.2)
5.Luc can nhdt sinh ra trong bo giam chan:
P^=c-(x-xJ^c-(i-xJ
=c-z
(1.3.3)
Dua vao cac dai luong da duoc tinh toan, ta c6 the thiet lap phuong trinh
chuyen dong ciia vat. That vay, tir cac luc tac dung len vat sau day:
quan tinh ciia vat: - mx(t)
trong luc cua vat: -mg
luc dan hoi trong 16 xo: -kz-\- mg
lire can nhdt trong bp giam chan: -cz
Theo nguyen ly D'Alembert ta duoc phuong trinh:
- mx(t) - mg -kz -\- mg -cz -0
hay
mx(t)-\-kz-hcz = 0
nhung vi: x = z-\-x^ nen phuong trinh dao dong cudi ciing cd dang:
m'z + cz-¥kz = -mx^
(1.3.4)
L32.Phi(ang trinh dao dong ciia dam ditai tdc dung ciia vat di dong
Trong muc 1.1. ta da thiet lap phuong trinh dao dong udn ciia dam vdi
lire tac dung phan bd bat ky va d muc 1.2. la phuong trinh dao dong ciia dam vdi
tai di dong. Nhu vay, de thiet lap phuong trinh dao dong ciia dam dudi tac dung
cua vat the di dong, ta chi can tinh luc tac dung vao dam khi vat the dao dong tai
vi tri CLia vat la A=*^=V/.
Trong 1.3.1 coi dam la nen dao dong tuc la:
x^=w(vtj)==xjt)
Tinh todn xdc dinh he so dong cua dam can di(&i tdc dong ciia tai Ji dong
(1.3.5)
Khi do, luc tac dung len n^n bao gdm:
iuc dan hdi trong 16 xo:
P^ = -kz -\- mg
luc can nhdt:
P^ = -cz
vay tong luc tac dung len dam tai vi tri x=vt la:
P = P-\-P= -kz -\-mg-cz
Ne'u tinh da'n phuong trinh (1.3.4) ta cd:
P = mg-kz-cz
= mg - m'x^ - m'z
= m(g-x„-z)
(1.3.6)
vdi A„ cddang (1.3.5)
Nhu vay, phuong trinh dao dong cua dam dudi tac dung cua vat the di dong cudi
cijng cd dang:
d^w
d w
EJ(^
+ a^)+pF(-—
dx
dx dt
d'w
df
dw
+ p—)
dt
= m(g~x„-z).5(x-vt)
(1.3.7)
1.3.3. He phirang trinh dao dong ciia dam vai vat thedi dong
Ghep hai phuong trinh (1.3.4) va (1.3.7) vao thanh mot he vdi ky hieu
.. ^ , d'w(vt,t)
..
X (t) =
= wta se duoc
dfm
d w
d'w
dw
+a-—)
+ pF(^
+ l3—) = m(g-w-z).5(x-vt)
dx
dx dt
df
dt
mz + cz + kz = -mw
EJ(
d w
(1.3.8)
(1.3.9)
TrKong hap rieng: Khi vat khong tach ra khoi dam, tu'c la vat Ian tren dam thi
z = 0 ,\c phai phuong trinh (1.3.8) chi edn la: m( g - w).S(x - vt)
luc do, ta chi can giai mot phuong trinh:
d w
d w
EJ(~^a^)^pF(^
dx
dx dt
d~ w
dw
+ p^)
= m(g-w).5(x-vt)
df
dt
(1.3.10)
Trudng hop tdng quat, he phuong trinh dao dong ciia dam vdi vat the di dong cd
dang (1.3.8) - (1.3.9) vdi di^u kien bien khdp tai hai dau:
w{0}=w"{0)=0
yX'(l) = H"(l)=0
Tinh todn xdc dinh he so dong cua dam can dudi tdc dong ciia tdi di dong
m
I
Y
C
K
^
Hinh 1.3.3 Mo hinh vat thedi dong tren dam
1.4 Phuong phap Bup nov- Galerkin trong ly thuyet dao dong cac he lien tuc
Xet phuong trinh I6ng quat ciia he lien tuc cd dang:
,a^w(A,/) „5W(A,/)
+
+B
dt^
dt
Cu{xj)^q(xj)
(1.4.1)
trong do A, B,C Xh. cac toan tir vi phan tuyen tinh theo toa dp A; wfA./j-chuye'n vi
ciia he tai A va thdi diem /, q(x,t) la mat do phan bd luc ngoai; cac phuong trinh
tren xac dinh trong doan (0, L).
Dieu kien bien cua bai loan cd dang:
BO{U(XJ)1__^=BM^^OI^,=0
(1.4.2)
BQ va Bi ciing la cac loan tir tuong ung.
*
Ddi vdi cac ham sd (p(x), i//(x) xac dinh trong doan (0, L) ta xac dinh tich v6 hudng
nhu sau:
{g>,f^)=\(p(x)M'X)dx
(1.4.3)
va
\d = {(p,(p)- =j
Vdi khai niem nay, toan lir A goi la ddi xung neu
(A(p, y/) = {(p,Ay/)
va A la xac dinh duong neu {A
Phuong phap Bup nov- Galerkin ddi vdi phuong irinh (1.4.1) d6 nghi vice
lim nghiem cua (1.4.1) d dang:
Tinh todn xdc dinh he so dong ciia dam can du&i tdc dong cua tat di dong
10
u(xj) = f^^^(t)(pj(x),
(1.4.4)
trong do |^^ (A), j = 1,2,...} la mot he ham thoa man di6u kien bien (1.4.2) va true
giao ddi vdi cac toan tLri4,5, C neu tren. Dieu do cd nghIa la;
m. neu j = k
0
{B(PJ,
neu j "^ k
m-\\ = n . neu j ~k
0
{C(Pj,(p,) =
\(p\
\\~ f Wc
(1.4.5)
neuj ^ k
= /,. neu j = k
^
0
neuj ^ k
khi do thay (1.4.4) vao (1.4.1) tadupc:
J
nhan hai ve phuong trinh tren vdi (Pi^ (A) ta duoc:
X[(^^,,^J-^'(o+K,,^J-^,(/)+(c^,.^J-^,(o]=^,(0
J
L
^k(0= j(p^(x).qixj)dx
0
Theo tinh chat cua (1.4.5) cua cac ham
rrijl it) + n^4j (0 + ij^, (0 - ^. (0
(1 -4.6)
j=I,2,3...
Nhu vay bang phuong phap Bupnov-Galerkin ta dua phuong trinh (1.4.1) v^
he cac phuong trinh vi phan thudng (1.4.6). Giai cac phuong trinh (1.4.6) ta duoc
Ndi chung cac ham nay thudng duoc chon la cac dang dao dong rieng ciia he, tu'c
thoa man phuong trinh:
- (o'Aipix) + C(p{x) = 0
hay
{C-(o\4)p{x) = ^
(1.4.7)
va cac dieu kien bien. Do tinh chat ciia cac loan lir A va C (ddi xung va xac dinh
duong) nen cac dang rieng deu true giao \i\ do do ihoa man dieu kien ciia phucrng
Tinh todn xdc dinh he so dong ciia dam can diari tdc dong ciia tdi di dong
11
phap Bupn6v- Galerkin. Trong trudng hop nay, cac dang rieng phai gia thiet la
ciing true giao ddi vdi ma tran B.
Trong thuc te, chudi v6 han trong (1.4.4) la khong thi tim duoc nen ngudi ta
han ch6' 6 mdt sd hChi han cac thanh phan, tiie la ehi tim
u{x,t) = f^^^(t)
(1.4.8)
D'\6u nay khong lam giam y nghia ciia Idi giai vi: Mot la ta chi quan tarn den
mot dang dao dong nao dd thi chi can
Phuong phap nay duoc ap dung trong phan tiep theo cua luan van.
Tinh todn xdc dinh he so dong ciia dam can du&i tdc dong ciia tdi di dong
12
CHl/ONG 2
KHAI NI$M VA TINH CHAT CUA H$ SO DONG LlTC CUA DAM CAU
2.1.Khai niem
Dudi tac dung ciia tai trong tinh, dam cau chiu cac luc tap trung va do vong
do tai trong tinh gay ra duoc goi la do vong tinh. Tai trong tinh phu thuoc vao vi tri
dat tai va do vay do vong tinh cung phu thuoc vao vi tri dat tai. Do vong tinh dat
ciic dai khi tai trong tinh dat d giua nhip (ddi vdi dam cau gian don khau do / ma
chung ta dang xet thi dd la vi tri 112). Gia six xet do vong tinh ciia dam tai mat cat A
dudi tac dung cua tai trong tinh P,, tac dung d diem XQ la WQ(X). DO vong tinh tai
mat cat X phu thuoc ca vao x la vi tri dat tai.
Ddi vdi tai trong dong, do vong dong con phu thuoc vao thdi gian /. Gia sir
xet tai trong dong Pacoscot. Dudi tac dong ciia tai trong dong, do vong tdng the ciia
ddm tai mat cat A la w(x,t). Do vong nay ngoai do vong tinh con cd phan gia tri
lh6m vao do tac dong cua hieu ung dong luc hoe cua tai trong. Khi dam chiu lac
dung cua tai trong di dong, do vong cue dai dat duoc khong phai khi tai trong d vi
tri giiJa nhjp ma sau khi tai da di qua vi tri giira dam.
Su anh hudng cua tai trong dong den do vong da dua den viec xac dinh mot thong
sd rat quan trong can duoc xac dinh dd la He sd dong lire hoc. He sd dong luc hoc
la dac trung mo ta anh hudng cua lai trong dong len dam cau so vdi lai trong tinh
va duoc tinh bang ty sd giira chuyen vi dong tren chuyen vi tinh. Khai niem He sd
dong lire hoc duoc dinh nghIa nhu sau:
Dinh nghia: He so dong lire tong quat ciia dam duac xdc dinh bang ty so:
w(x,t)
. ^
>V/A)
Neu hieu diin do vong dong w(x,t)=WQ(x)+ w(x,t) tuc la w(x,t) la phan do vong
them vao do tdc dong cua hieu i(ng dong life hoc cua tdi trong thi:
fi(x,t) = } + fi(x,t)
vai fi(x,t) = ——
Id he so dong lire thong thudng cua dam cau vd noi chung la
VV/A>
ham cua cd x va t.
Trong thuc te, ngudi ta thudng xet: // - ma x fi( x. t) la hang sd.
Tinh todn xdc dinh he so dong ctia dam can du&i tdc dong cua tdi di dong
13
2.2.Xac djnh he sd dong lire hoc cua d a m cau vdi tai trong hang sd di dong
Di xac dinh he sd dong luc hoc cua dam cau vdi tai trong hang sd di
dOng, ta phai giai phuong tfmh:
„,,5'w
d^w ^
r-yd'w
^dw
^ ^
tinh fifxj) hoactinh Ji(x,t) va ju^^^
Mudn giai duoc bai toan nay dau tien giai bai toan tinh
d'w
EJ-—- = P,.S(x-xJ
dx
tinh
wjx)
(2.2.2)
roi giai tiep:
dw
dw
d'w
^•^^^^^"hw,^^P^(^^PT^
OX
dx dt
df
dw
dt
= P''-5(x-x„-vt)
(2.2.3)
de tinh VV^A,V,/^
Chiing minh w/x) = lim max w(x,v,t)
I
\-*0
Va cudi cung
w(x,v,t)-w(x,0,t)
w(x.Oj)
*Trudc tien, ta ap dung phuong phap Bupnov-Galerkin dua phuong trinh dao
ham rieng:
d'w
dw
d'w
EJ(^^a^^)+pF(^+^—)
dx'
dx'dt'
' df
dw
= P,Mx-vt)
dt^ '
^
vi he phuong trinh vi phan thudng.
Ta tim nghiem cua phuong trinh dudi dang:
w(x,t) = Y,qJt)sin
rTDC
I
vdi nghiem n h u tren ta cd:
dj^_f ^/-Tr^
dx'
dw
Yt
r=\ I
rTDC
q^ sin
I'/. sin rirc
•
-
/
Tinh todn xdc dinh he so dong ciia ddm can dudi tdc dong ciia tai di dong
(2.2.4)
14
d'w
^..
rTDC
— - =y a sin
dt'
t',
d^w
_ ^
dx'dt
~
I
rwc
q.sin^
^
Thay vao (2.2.1) ta cd:
z
EJ
. rjix
q^sin—+
^
a^
v/y
q^ sin
V' y
+ pF 2.q,sm—--\-(32^q^sin~—
r=l
-^
. rTDC
r=l
/
EJ
r=/
q^ +aEJ
I
« z
I)
£7 'ri:^
A-;D:
^.+ a.—
sin —
/
rjDc
p,.(^rx-vr;
/
' rir^
-q,+ pF.q^ + &pF.q^
V' y
+0
P,rS(X-Vt)
EJfrir^
qr +
V ' J
b(x - vt)
^ r
pF V ' J
pF
STDC
Nhan hai ve vdi sin
rdi lay tich phan tir 0 den / va de y rang:
0 neu r ^ s
rTDC
. STDC
.sin — dx
I
I
\sin
i
I
neu r = s
Tacd:
Sin
0 r=l
rTDC . STDC
sin — Qr^
/
0_
O
q^ + a
/
\sin
r\
EJ(rn^
a
pF
+ y9 9.+
I/
pF V ' /
df A - v ^ ^ A
I
Ej(s7t^
pF V / J
+P
EJ
(], + —
pF
2^ r . STDC ^ .
ds
ipFi
\sin
n
d(x-vt
*
vdi s = I,n
Xet tich phan: jsin
S(x-vt)d.
r
>/i.v ^,,,
,
,.
f I
S7VC
STD:
sin
(> (X - vt kix = urn
— sin
dx
I ,
I
STDC
— lim — cos
Tinh todn xdc dinh he so dong ciia ddm can dim tdc dong ciia tdi dt dong
^ ,
)ax
15
1
=
,.
} {
STt(vt->r£)
lim — {cos
STT '-^ 2e[
STt(vt-~£)\
- cos
>=
/ J
/
] J. I
. STn^t . STtE ^ ^ ,
7 ,
/
STTVt
STt£
lim — sin
. sin
(-2) = — lim — sin
. sin
STT ^--^ 2s
I
I
sTt ' ^ 2e
I
I
. STtE
sin
. STtVt ,
/
= sm
lim
— =
=
I
'^^^
SK£
. STtVt
I
= sm
Dan ra he phuong trmh vi phan
^ >
kyhitu:2h
a
EjfsTT^
pF I /
+ >^ ^s +
EJfsir^'
=a—
pF V I' y
'
EjfsTT^
2/1
pF
IpF
V ' y
. STTVt
Sin
2P.
EJ(STT^'
pF K l )
:Po
IpF
STDC
-•^
'
I
(2.2.5)
Tacd he: q^ +2h^q^ +^5^5 =PfjSinvJ
vdi s=I ,n
Nhu vay ta da dua phuong trinh dao ham rieng \i n phuong trinh vi phan
thudng. Ta se sir dung ket qua nay cho viec giai phuong trinh dao ham
rieng \i sau.
*Bay gid ta giai bai loan tinh tim w„(x)
EJ^
dx
= P,-S{x-xJ
Ta se tim nghiem gdn dung dua vao phuong phap bien doi 0 tren, ap dung
cho w khong phu thuoc /.
Nghiem can tim cd dang w(x) =
^q/r)sin
rTDC
r=I
/
vdi nghiem nhu tren ta cd:
^rx^'
d'w
rTDC
q^ sin
7
=
Z
dx
v/ y
Thay vao (2.2.2) ta duoc:
EJT
^rx^
q^ '^'" — ~ K' ^{-^ ~ -^.)
\ ' y
Tinh todn xdc dinh he so dong ciia ddm cau du&i tdc dong ciia tdi di dong
16
Nhan hai ve vdi sin
EJ
^ sir^
STtX
va tich phan tir 0 den / ta nhan duoc:
2'c
=-\PrMx~xJsin—-dx
^s
i 0
V ' y
I
Nhu da biet, tich phan d ve phai cd gia tri bang:
fo-.
STDC
jPo6(x-xJsin—~dx
. STDC
=
0
^
P,sin—^
I
I
EJ - J q^=P,sin—^
2P^ sin -—^
I
(2.2.6)
lEJ
v/y
EJ
— 2P
va /I = — - va nhan ca tir va miu cua
/pF
I J
'STT^
Neu ky hieu co^ =
pF
(2.2.6) vdi pF thi (2.2.6) se la:
^•sm'-^
I
CD'
(2.2.7)
Do vay W(,(x) = /^,2.~T'^''^~7^'^'"^~
s=l 0)^
I
I
Ket qua nay se diing de xac dinh he sd dong luc hoc sau khi tinh duoc
w(x,v,t)
*Tinh wfx,v,/^tir:
^,.d'w
d''w
^A
,
r-/^'^
av or
df
o^^' i
r>
c/
dt
Tim nghiem gan diing dua vao phuong phap bien ddi d phan tren.
Nghiem cd dang: w(x,v.t)= ^q/t)sin
vdi nghiem nhu tren ta da cd:
5St
^VT^^
7 = r=l1'v
dx
q. ^"»'"
rTDC
/
Tinh todn xdc dinh he so dong cua dam cau du&i tdc dong cua tdi di dong
17
dw
—
^
,
= /
. rTDC
Qr Sin
dt tt^'
d^w
^..
I
rTDC
dt' t!^'
I
d'w ^yn A^^>
q^sin
dx'dt ~ hV /' y
rTDC
/
Thay vao phuong trinh tren rdi bien doi ta cd:
2,
sin--
r=l
/
Nhan hai ve vdi sin
^,+
a
EjfrTT^
^.+
a
pF V ' J
STDC
Ejfsn^
Ejfrrt^
+P
=
pF V ' J
-^S(x-x„-vt)
pF
roi la'y tich phan hai ve tir 0 den / ta nhan duoc:
+P
EJ_ '
^s +
V / J
2P^ V . STDC „^
^,
=——\sin-—S(x-x^-vt)d}
IpF i
I
STT ^
pF \ ' y
q^
vdi 5=/,n
Tich phan d ve phai cd gia tri bang:
\5(x -Xg-vt)
sin
n
dx = P^ sin
SK(
Xfj
t
+vt)
I
Dan ra he n phuong trinh vi phan thudng:
^.+
Ejfsir^'
a
pF
2P, .
q^ =—-sin
pF \ ' y
IpF
EJ ' sir^
+^
K t y
sTr(x,+vt)
(2.2.8)
I
vais=Ln
ky hieu: 2h^ = a
EJ_ 'sic^'
pF
V =
STDC
,
\
It
," 0 =
+ ft- cj;
pF V i y
/
SK(X^
T a c d he: cj^ -^^h^q^ '^^l^s
Ej(sTc\
^
p
2P
=zi±IpF
^-Vt)
=PoSin
(2.2.9)
vdi s=I ,n
vi vai Iro ciia s trong timg phuong trinh trong he la nhu nhau, ta chi can xet phuong
trinh thu s. Ket qua suy ra cho ca he.
CA: i-*^.-'- vjo'" of i v/ . H
'•--.-.i
Tinh todn xdc dinh he so dong cua ddm can dudi tdc dong cua tdi di dong
\'-UlJ04
j
18
Xet q,+2h^q^^(j)]q^ =PoSin<f>^
(2.2.10)
Day la phuong trinh vi phan tuyen tinh cap 2. Ta phai tim nghiem q^ cua (2.2.10)
dudi dang:
q^ (t)=A sin (t>^ + B cos^^
vdi nghiem tren la cd
q,(t)=Av^ cos(p^-Bv^ sin(t>^
qJt)=-Av] sin^^ - Bv'^ cos<j>^
Thay vao (2.2.10) thuc hien mot sd bien ddi so cap nhdm cac sd hang theo sin^^ va
cos^^ia rut ra:
l(a?:-v-:)A-2v^Bh^=P^
\(co;-v-)B-h2vJh^
(2.2.11)
=0
Giai he (2.2.11) de tinh he sdy\, B
2vh
.'
B
,
\A
CO' - v ;
A = P,
A=
B
CI.
(co; -v-y
+4h;vi
2hvA
(co] -v] ) ' +4h-v]
(oy]-v])P,
sin (j>^
((o]-v\y
^4hlv]
''-'"
(CO] -vj-
-.cos^,
+4h;v;
B
Viet 7^dudi dang q^=asin(<f>^ -^cp^)'^ ci=yJA' -\-B' ;
P„sin(<t>^ -\-(pJ
<=>
Qs
voi
^ = -arctg
^J(co; -V- ) ' -^4h;v~
2h V
CO' - v ;
Tinh todn xdc dinh he sddpng cua ddm can du&i tdc dong ciia tdi di dong
19
'^x,v,t) = Y,qJt)sin
STDC
S=I
_^
—/
P,sin((l>^+cpJ
. STDC
I
sin
*Tim lim w(x,v,t)
lim w(x,vJ) =
lim±-,^^Mi,l£j=sin^
'^' ^->^(co:wy+4h:v: i
v-*0
Ta cd: lim v, = lim
v-frO
'-^0
V->0
V-¥0
lim cp = lim
"
=0
I
2hv^
arctg —,
co: -V
/
S TDC
lim w(x,v,t)= P^Y^^sin
v-*0
s=i co^
I
0
V TTX
-sin
I
= vv/A^da xac dinh trong (2.2.7)
Nhu vay, bieu thire do vong tinh cd the' xac dinh duoc tir bieu thirc do vong
dong bang each cho van tdc cua tai trong v -> 0.
. .
j ^ ^
I
. STDC,.
. STDC
,.
w.ix) ~ r.y —- sin
sin
= lim w(x, v, /)
'tto)'
I
I
-^
*Tiep theo, ta phai xac dinh h^ sd dong lire hoc cua dam cau vdi tai trong hang
sd di dong.
M(xJ) =
w(x,vj)-w(x,0,t)
Mx.Oj)
^
P^sin(6-\-(p)
. STDC — ^
2_ .
\
' = sin
s=i ^(col -v] y ^4h;v]
Poll
sin(
\
1
1 CO
.
i
STDCn
-sin
.
STDCQ
.
STDC
sin
I
I
. STDC
sin —
'
/
/
STDC^
^ ^ ^ ^ ^ = ^ ^ =
s= l
J
Po2^~T sm
s=i CO]
yj(co] -v] y
^
Sin
+4lvv]
I
CO'
. STDC,.
> —-sin
ttco
I
STDC
sin
. STDC
sin —
I
Tinh todn xdc dinh he sddpng ciia dam cdu dudi tdc dong ciia tdi di dong
(2.2.12
20
vdi:
2h.=a
EjfsTT^'
^P;
pF \ I' y
\4
2 _EJ_f STt
"
IpF'
STtV
V
=
STt(Xf^ -k-Vt)
1
*Tinh chat cua he sd dong trong trudng hop tai di dong
Tir (2.2.12), de don gian ta xet he sd:
Y,^(o),,h^,v^
^(x.t)
=
)sin(vj
+ cp^ )sin
.
.
STDC
l-h^(xj)
^
I
STDC,.
> —- Sin
1^1 0)-
STDC
sin ——
I
I
STDC,
voi cp^=(p^ +
I
H(co^,h^,vJ
=
4{oy:-v:y -4hivi
va dimg lai d dang dao dong thu" nhat, tu'c s=L Khi dd:
1
l + fu(x,t) =
\i-f.y
sin(Vjt -^-cp,)
4h]7]
TO,
sin
co\
I
hay
(2.2.13)
i^K
V-r,)
4h]y]
sin
Ta.
0)\
vdi Yt' = —T - d<*i lugng nay thuc te ludn nhd hon 1
CO,
Theo bieu thirc (2.2.13) ta thay, he sddong phu thuoc vao vi tri tren dam.
tu"c la d nhtrng vi tri khac nhau, he sd dong khac nhau. He sd dong cang ktn d
nhimg vi tri cang gan vdi gdi va he sd dong nhd nhat d giira dam va cd gia tri lii:
Tinh todn xdc dinh he sddpng ciia ddm cdu du&i tdc dong ciia tdi di dong
21
^^ = ^ + >",
'y
(2.2.14)
4^j]
\^-f^
(o\
Trudc khi xem xet su phu thuoc cua he sd dong vao cac dac trung khac ciia
dam, ta ki^'m tra dieu kien:
7+27
I
7
=
max
Vr])
>1
4h\Y]
CO.
That vay, di6u kien cudi tuong duong vdi:
[l-f,y
+^^<}h^yy:-2y]+^^<0
CO,
1-2
yj<2(I-2—^):
co:
v6\ /r nhd diau kien nay luon thoa man vi y;<7. Nhu
r 7\
vay /J^>I hay //,
u
>o
Cong thu'c (2.2.14) cho thay: He sd dong phu thuoc vao tdc do di dong
cua tai, tan sd rieng va he sd can. He sd dong cang tang neu tdc do cua tai di
dong tang va cang gan vdi tan sd rieng. Nhu vay tai di dong cang nhanh thi he
sd dong cang Idn. Dieu nay phii hop vdi viec han che tdc do phuong tien giao
thong chay tren cau.
Mat khac cong thu:c (2.2.14) ciing chi ra rang, ta cd the xac dinh duoc he
sd dong neu biet tdc do tai di dong, tan sd rieng va he sd can cua dam cau. Day
la ket luan rat bd ich de xac dinh he sd dong ciia cau bang thuc nghiem ma
khong can do chuyen vi dong va chuyen vi tinh.
He sd dong cung phu thuoc tuan hoan vao thdi gian vdi chu ky:
2Tr
2Td
21
V
TTSV
SV
vdi s=l thi T^ =211 v. Nhu vay he sd dong d giira dam cau Idn nhat vao luc tai
trong di het chieu dai ciia dam cau.
Tinh todn xdc dinh he so dong ctia ddm cdu dudi tdc dong ciia tai di dong
22
2.3. Xac dinh he sd dong luc hoc cua dam cau khi co vat the di dong
Trong luan van nay ta chi gidi han xet trudng hop rieng trong viec xac
dinh he sd dong luc hoc cua dam cau khi cd vat the di dong: Trudng hop vat
khong tach ra khoi dam, tu'c la vat Ian tren dam 'z = 0. Luc nay, d6 xac dinh he
sd dong luc hoc ciia dam cau khi cd vat the di dong, ta chi can giai mot phuong
trinh (1.3.10):
d'^w
d^w
d'w
EJ{—T + cc~j-)+QF{--^^P—)
ox
ox dt
Of
dw
=
dt
m(g-w).5(x-vt)
Sau khi giai (1.3.10) thu duoc ket qua w(x,v,t), chov ->• 0 ta cd:
WQ(x,t) = lim max w(x,v,t)
va //("A,/^ duoc tinh ra tir w(x,v,t)vkWf)(x.t)
*Trudc tien, ap dung phuong phap Bupnov-Galerkin de dua phuong trinh dao
ham rieng
EJ(—^ + a—^-) + pF(—^ + /3—) =
ox
dx dt
dt'
dt
m(g-w).S(x-vt)
ve he phuong trinh vi phan thudng.
Ta tim nghiem cua phuong trinh nay dudi dang:
rTDC
M^J) = Y.^r(Osin
r=l
7
De don gian cho cong viec tinh toan ta lay gidi han n=l va bi^u thu'c gan dung
cd dang:
w(x,t) = q(t
)sin
TDC
I
vdi nshiem nhu tren ta c6
d'w
'TT^
dx'
TDC
qsin
V I' /
dw
TDC
— = qsin
—
d'w
TDC
dt
^
..
I
— - = q sin —
df
^
I
a- vv
dx'dt
TDC
1/
qsin
Tinh todn xdc dinh he so dong cua ddm cdu dudi tdc dong ciia tdi di dong
(2.3.3)
