DAI
HQC QUOC GIA
HA
NOI
TRlTOfNG
DAI
HOC
KHOA HOC
TlT NHIEN
ON
DINH
DAN DEO CUA KET CAU
CHIU TAI
PHirC
TAP PHU THUOC VAC X
VA
Y
Ma so:
QT-
07 - 03
Chu tn
de tai
: PCS. TS Dao Van Dung
Can bo
tham
gia :
GS.
TSKH.
Dao
Huy
Bich

CN. Bui
Tin Tliuyet
HaN6i-2007
I. BAO CAO TOM
TAT
KET QUA
THl/C HIEN
DE TAX NAM 2007
_- '^
A
f
f
1.
Ten de tai: On dinh dan deo cua ket cau chiu tai phuc tap phu thugc
vao
X
va y.
Mas6:QT-07-03.
2.
Chu
tri d§
tai:
PGS.
TS. Dao Van Dung,
3.
Can bo tham gia:
GS.
TSKH.
Dao Huy
Bich,

can bg
Iruang
DHKHTN.
CN. Bui
TIti
Thuyit,
hgc vien cao hgc truang DHKHTN.
4.
Muc tieu va noi dung nghien
cuu:
Trong
thirc
te, nhieu
ITnh
vuc ky thuat ung dung cac phuang phap va
ket qua cua
ly
thuyel on
djnh
va dao dgng cua cac he dan hoi va deo. Nhieu
f f f
cong
Irinh
trong xay dung, giao thong va cong nghiep c6 dang ket cau lam va
vo.Vi vay nghien cuu do ben,
sir
on dinh va dao dgng cua cac he nhu vay
khong chi c6 y nghTa khoa hgc ma con c6 y nghTa
thirc
lien. Khi nghien cuu

on dinh khia canh chinh ma la can quan lam la xac dinh tai trgng tai han, nhat
la khi ket cau do chiu quy
luat
tai phuc tap phu thugc vao toa do. O nuac la
cac nghien cuu van de nay bat dau
tir
nam 2000
Ira lai
day. De tai QT - 07 -
f
^
03
CO
muc lieu di vao van de con dang thai
sir
nay va nam 2007
d§
dat dugc
r f
mot so ket qua sau day:
X.
f
>
*/
On dinh dan deo cua lam mong chu nhat lam bang vat lieu nen dugc,
>
f
chiu tai phuc tap khong
ihuan
nhat.

*• > >
*/ Tinh loan lam dan deo
chU
nhat bang phuang phap phan
lu
huu han.
*/ Nghien cuu hien
lugng
mat on dinh khi dgng bang phuang phap
nghiem giai
lich
gan dung
ciia
phuang Irinh Vander Pol.
r
\
f f
•>
*/
Cac ket qua tinh loan bang so cho mot so vat lieu cu the.
f
5.
Cac
ket
qua dat dugc.
a,
Bai toan ve on dinh dan deo cua
tdm
mong
chiv

nhat lam bang vat
lieu nen duv'c chiu tai phuc tap khong thudn
nhat.
Da xay dung dugc
he
phuang Irinh on djnh din
ddo.
X6l hai \6p bai
\
f f
loan vai bien
tira
ban
le
va bien ngam tren bon canh. Ap dung phuang phap
Bubnov - Galerkin va phuang phap tham so tai dan den he thuc cho phep tim
lire
tai han.
\
f
y
Da trinh bay thuat loan, tinh loan bang so, xay dung do thi mo la anh
f f
huang
ciia
do
manh,
ciia
tinh nen dugc den
lire

tai han cua lam.
b.
Bai toan ve tinh todn tarn dan deo chit nhat bang
phwong
phap
phan tie
hitu
han.
Da nghien cuu lam theo mo hinh tuang thich
dira
tren ly thuyet qua
\
f
y
\
trinh dan deo. Bai loan dugc giai bang phuang phap bien the nghiem dan hoi.
•^ > f
Qua trinh giai dugc chia thanh k giai doan, moi giai doan gom n buac lap. Ket
•
f f
qua cua buac lap truac la ca sa de tinh loan buac lap tiep theo. Tam dugc
chia thanh cac phan
tii
chu nhat. Chuang trinh dugc thuc hien bang phan
mem Matlab 6.5 Hinh anh mien deo dugc mo ta cu the sau cac giai doan, cho
^
y
f \ f
Ihay
dugc mem deo xuat

hien tir
bien dat
lire Ian
dan vao ben trong tam.
c. Nghien civu hien tuvng mat on djnh khi dong bang phuang phap
nghiem giai tich gan dung cua
phuvng
trinh Vander Pol.
/^ f f
Bai loan dan den viec giai phuang trinh Vander Pol vai he so phu thugc
\
f
\
vao tan so
ciia lire
kich dgng. Cac tac gia da tim nghiem giai tich gan dung
ciia
phuang Irinh nay, thong qua nghiem
Ihu
dugc da phan tich
sir
phu thugc
/• f
rat nhay
ciia
dang dieu
ciia
nghiem vao cac he so
ciia
phuang trinh vao

sir
f f f
luang
tac giua cac yeu to phi tuyen va
lire
kich dgng. Da chi ra nhung hieu
f
•>
• r
ung dac biet lien quan den chuyen dgng xoay, den hien tugng Galloping.
r
-y
-*
Cac ket qua nghien
ciiu
cua de tai
duac
the hien tren cac bai bao
va bao cao khoa hoc.
1.
Dao Van Dung, Bui Thi Thuyet. Elastoplastic stabibity of thin
rectangular plates made of compressible material under nonhomogeneous
complex loading (to appear in VNU. Journal of Science, 2007).
2.
Dao van Dung,
Nguyfin
Cao San. Tinh loan tam dan deo chu nhat
bang phuang phap phan
tii
huu han. Tuyen tap cong trinh hoi nghi ca hgc

loan quoc
Ian
thu 8, Ha Ngi ngay 6-7 thang 12 nam 2007.
3.
Dao Huy Bich, Nguyen dang Bich, Nguyen Anh Tuan. Nghien cuu
f
t
y
^
hien tugng mat on dinh khi dgng bang phuang phap nghiem giai tich gan
f f
y
f
\
f f
diing
phuang trinh Vander Pol. Hoi nghi quoc te
Ian
thu nhat ve thiet ke, xay
f f
dung va bao tri cac ket cau, ngay 10-11 thang 12 nam 2007, Ha Ngi,
Viet Nam.
6.
Tinh hinh
kinh
phi.
+ Cac bai bao, bao cao khoa hgc va thu
lao
chuyen mon:
12.000.000d

H-
Hoi thao va xemina khoa hgc:
4.000.000d
+ Chay chuang trinh va che ban: 1.600.000d
+ Quan ly ca sa
SOO.OOOd
+ Van phong pham va cac chi khac 1.600.000d
Tong cong 20.000.000d
7. Nhan xet va danh gia ket qua
thyc
hien de tai.
*/
De tai vai thai gian thuc hien mot nam da hoan thanh vugt muc ke
hoach so vai chi lieu dat ra ve so lugng bai bao va bao cao khoa hgc. Da c6
1
f
\
01
bai dang trong tuyen tap hoi
nghj
ca hgc loan quoc
Ian
thu 8, nam 2007 va
f
01
bai nhan dang a tap chi khoa hgc Dai hgc quoc gia Ha Ngi va 01 bao cao
f f
khoa hgc a hoi nghi Quoc te nam 2007.
*/
Cac van de nghien cuu c6 y

nghia
khoa hgc va hgc thuat, gop phan
t
f f
djnh huang ung dung trong viec xem
xel sir
on dinh
ciia
cac ket cau.
*/
De tai gop phan nang cao chuyen mon
ciia
can bg, cao hgc va nghien
cuu sinh
ciing
nhu sinh vien nganh ca
hg^.
Thong qua cac xemina va hoi thao
' f
khoa hgc da
ciing
c6 va trang bi them
nhiing
kien thuc chuyen sau cung nhu
huang ung dung
ciia
Bg mon Ca hgc, Khoa Toan - Ca - Tin hgc, Truang dai
hgc Khoa hgc Tu nhien, Dai hgc
Qu6c
gia Ha Ngi.

*/ Da huang dan
1
cao hgc, 1 sinh vien theo huang de tai.
*/ Nhom de tai kien nghi trong thai gian tai se dugc nang cap de tai
theo phuang huang nay .
Xac nhan cua ban chu nhiem khoa
• •
Ha Noi ngay 5 thang 1 nam 2008.
Chu tri de
t^i
CH
TS
Uc.^c
K-^
D^
PGS.TS.
Dao Van Dung
Truang Dai hoc khoa hoc Tu uhien
I^U
tPUONC*
i^^^.^^^:^iv^^^
n. BAO CAO TOM TAT BANG
TIENG
ANH
1.
Title: Stability of elastoplastic structures subjected to complex loading
depending on x and
y.
Project's code:
QT-07-03;

Duration: 2007.
2.
Head of research group: Assoc.
Prof.
Dr. Dao Van Dung.
3.
Paticipants:
Prof Dr. Sc. Dao Huy Bich
B.
Sci. Bui Thi Thuyet
Duration: 2007
4.
Resume on the aim and main contents of project
In practice, there is a lot of technical domains applying the research
methods and results of the theory of stability and oscillation of the elastic
systems and elastoplastic systems. Many structures in
contruetion,
transportation and industry are the form of the plate and shell. Therefore, the
investigation on the durability, stability and oscillation of these systems is not
only scicntillc sens but also practical sens. The main aim of stability problem
is to find critical loads, in particular for structures subjected to the complex
loading defending on x and y. In our country, the studies on this orientation
have been beginning since 2000 up to now. The project QT-07-03 has the
purpose to investigate these hot current problems.
In this project, our staff have investigated the following topics:
a. Elastoplastic stability of thin rectangular plates
-made
of
compressible material under nonhomogeneous complex loading.
b.

Calculating elastoplastic rectangular plates by finite element
method.
c. Investigation of Aerodynamical instability phenomena by using a
method for the approximated analytical solution of the Vander Pol equation.
5.
Results
a. Scientific activities:
-01
research
paper
accepted
to publication in VNU. Journal of science, 2007.
-01 research paper have been pulished in the proceedings of 8-th National
Conference
on Mechanics, 6-7, December, 2007.
-01 scientific report in the 1-th Intemational Conference on Modem Design,
construction and Maintenance of structures,
10-11,
December, 2007, Hanoi,
Vietnam.
b.
Training activities: 01 M.Sci and 01 B.Sci.
c. Scientific papers and reports
- Dao Van Dung, Bui Thi Thuyet. Elastoplastic stability of thin
rectangular plates made of compressible material under nonhomogeneous
complex loading (to appear in VNU. Journal of Science, 2007)
- Dao Van Dung, Nguyen Cao Son. Calculating elastoplastic
rectangular plates by finite element method. Proceedings of the Eigth
National Conference on Mechanics, Ha Noi 6-7, December, 2007.
- Dao huy Bich, Nguyen Dang Bich, Nguyen Anh Tuan. Investigation

of Aerodynamical instability phenomena by using a method for the
approximated analytical solution of the Vander Pol equation. The
1^*
International Conference on Modern
design.
Construction and Maintenance of
structures,
10 -
11,
December, 2007, HaNoi, VietNam.
III.
NQI
DUNG CHINK CUA BAO CAO
7
1.
Ma
dau
f
\ •'
•>
f f ^
Van de on dinh khi dgng va on dinh
ciia
cac ket cau dan deo chiu
tai,
dac biet la chiu tai phuc tap dugc quan tam nghien cuu vi khong
nhGng
c6 y
nghTa khoa hgc ma con c6 y nghTa thuc
tiln.

»
^
•»
^
De giai quyet bai toan on dinh dan deo can phai xay dung mo hinh phu
f
t
s
y
f
hgp,
thiet lap cac phuang trinh on djnh va dieu kien bien, de xuat phuang
phap giai, tim bieu thuc de xac djnh
lire
tai han . Lap chuang trinh may tinh
' f f
va tinh loan cho mot so dang ket cau
tir
do phan tich ca sa khoa hgc va y
f
nghTa
co*
hgc
ciia
cac ket qua do.
•»
f
\
•>
ist

f
De nghien cuu van de on djnh khi dgng ta dan den phuang trinh phi
' f '
tuyen, cho nen tim nghiem giai tich rat kho. Vi vay ngoai phuang huang giai
^ > f
gan dung bang phuang phap giai tich so, nguai ta con di tim nghiem dang giai
tich
gan
diing.
•»
Ngoai ra hien nay do
sir
phat trien manh me
ciia
tin hgc, nen trong ca
hgc ung dung nhieu phuang
.phap
phan
tii hiiu
han de giai cac bai toan xac
f f
\
f f
djnh trang thai ung suat va bien dang va mien deo trong ket cau.
> y f
Da
CO
nhieu cong trinh nghien cuu on djnh dan deo
ciia
tam va v6

chju
tai thuan nhat. Tuy nhien khi quy luat tai phu thugc vao toa do thi bai toan
rv
r f
dan den nghien cuu he cac phuang trinh dao ham rieng vai he so la ham
ciia
x
va y. Vi vay lop bai toan nay hien nay con it dugc nghien cuu.
* \
f f f y y
De tai nham giai quyet mot so khia canh trong nhung van de on dinh
dan deo, on djnh khi dgng nhu vay.
2.
Noi dung chinh
a. Van de on dinh dan deo cua
tdm
mong
cliir
nhat lam bang vat lieu
nen
dwac
chiu
taiphivc
tap khong thudn nhat
Bai toan nay vai vat lieu khong nen dugc da dugc tac gia VG Cong
Ham nghien cuu nam
2003.
O day cac tac gia trong de tai nay nghien cuu vai
vat lieu la nen dugc. Da thiet
lap

phuang trinh on dinh cho tam. Khao sat hai
16'p
bai toan voi bien
tira
ban le bon canh va ngam bon canh, Ap dung
8
phuang phap Bubnov - Galerkin va phuang phap tham so tai. Xay dung he
thuc tim
lire
tai han
Det(Hik)
=0
Da trinh bay thuat toan va tinh bang s6 cho 4
k6t ciu
cu
thi:
- Xet tai trgng phu thugc vao tham s6 tai t
oQi
ni.
(283 +
0.1//
p,
=
283
+
0.1/,
p^=
^—^
.
'

450-
- Xet tai trgng phu thugc vao
/
va toa do
, (283.0
10-^^^3.
b ^'
450'
a
p,
=(283 +
0.1/)(l
+
0.35f),
P2^
ZV (1-0.25-)
Xet lai trong vai quy luat
ji;,
=(246 +
0(1+^),
p,=20t\
b a
Nghiem
cua 3 bai toan tren tim
duai
dang chuoi
lugng
giac,
- Xet tai trong vai quy luat
p,

=(283
+
0.10(1
+
0.35^),
(283 +
0-'0
(i_0
25J^).
'
b ~
450'
a
Nghiem a day tim duai dang chuoi
luy
thira.
Tinh toan va xay dung do thi mo ta anh huang
ciia
tinh nen dugc,
ciia
do
f f
manh den
lire
tai han
ciia
tam.
f f f
r
Ket qua cho thay tam chiu tai phuc tap thi

lire
tai han nho han khi tam
chju tai dan gian. Da chi ra tinh nen dugc c6 anh huang dang ke den
lire
tai
f
han
ciia
tam.
^
>
s
r
•y
b. Van de ve tinh toan tdm dan deo
elm
nhat bang
phuang
phap
phdn
tir hivu
han
> y f
Phuang phap phan tu huu han, hien nay dugc
sii*
dung nhieu va rat eo
hieu qua, nhat la khi giai cac bai toan c6 hinh dang phuc tap hoac cac bai toan
dan den cac phuang trinh khong the tim dugc nghiem duai dang giai tich.
^
f

1^
f f
Bai
toan tam dan deo chiu tai phuc tap dan den phuang trinh phi tuyen
VI
vay cac tac gia da chgn phuang phap phan tu
hiJu
han de tinh toan va xay
, ^
f
f
r
dung
mien
deo
ciia
ket cau. Da nghien cuu tam theo mo hinh tuang thich, giai
^
ft
y
bang phuang phap bien the nghiem dan hoi, qua trinh giai dugc chia thanh k
giai doan, moi giai doan gom n buac lap. Ket qua
ciia
buac truac la ca sa de
f
tinh buac lap tiep theo:
f f
Buac k>o, tinh ung suat va bien dang do dai cung theo buac lap
.v"''*'=i'"'-"+c/.v"''*-"
.

y
f f y y f
Da tinh toan bang so cho bai toan tam lam bang vat lieu tai ben tuyen
tinh
CO 4)'(s)/3G =
0,2;
cr,
=
400 (MPa), v = 0,3; 3G = 2,6.
lO"
"(Pa),
kich
thuac a
=
1,6 (m), b
=
1,2 (m); h
=
0,003^
(m).
f y f f
Quy luat tai dang Parabol va phan bo deu. Ket qua cho thay hinh anh
deo xuat hien sau giai doan 6 va sau giai doan
15
thi
Ian
ra hau nhu toan bg
f
t t
f f

tam. Phuang phap c6 uu diem dung de giuai cac ket cau c6 hinh dang phuc
tap,
cac ca he c6 lien he phi tuyen.
c. Vdn de mat on dinh
khi
dong
i,v
f
y
f
Hien tugng nay dan den dugc mo ta bang phuang trinh phi tuyen dang
f
y
f
Vander Pol voi he so phu thugc vao tan so cua
lire
kich dgng:
—J +(2u +
3o v^)i-i-a^r^
~\~aqx^
-\-kx~
qcoscoi.
y y
Da nghien cuu phuang trinh nay bang phuang phap, gom cac buac
f
+
Tach thanh hai phuang trinh vi phan thuang cap
1.
f
»vr

'
>
'\-
Xay dung phiem ham dan xuat va dieu kien rang bugc.
10
+
Tim nghiem giai
tich
gan diing thu nhat.
\
f
+ Tim nghiem giai tich gan diing
tiep
theo.
+
Bieu
dien nghiem
ciia
phuang trinh can nghien cuu ban dau.
>
f
+ Chi ra cac dieu kien cho phep xap xi
Da giai mot so vi du cu the vai quy trinh:
+ Xay dung cac buac giai
+
L§y
cac tham
s6
k=l,
V 0.025,

a 0.0033,a=0.005,q=0.01,
ty=0.4,
f y
+ Da xet 4 bg tham so, xay dung cac do
thj
tuang ung.
+ Phan tich va thao
luan
cac ket qua, cho thay phuang phap cho ta tim
nghiem giai tich gan diing
ciia
phuang trinh vai cac tham
s6
khong
cSn
phai
f
la be. Cac ket qua thu dugc khong
nhiing
c6 y nghTa hgc thuat khoa hgc ma
con chi ra dugc nhung hieu
ling
dac biet nhu la hien tugng xoay, hien tugng
Galloping.
3. KET LUAN
>
De tai QT - 07 -03 vai thai gian mot nam da thuc hien diing vai ban
f f
dang ky va da hoan thanh tot Cac ket qua dat dugc c6 y nghTa khoa hgc gop
phan vao viec nghien

ciiu
hgc thuat va phuang phap, dong thai da chi ra
bang nhu'ng vi du, nhung bai toan cu the ve y nghTa ea hgc, ve djnh huang
f f
ung dung. Day la tai lieu cho cac nha thiet ke, xay dung va ky su tham khao.
De tai da gop phan dao tao sinh vien, hgc vien cao hgc va NCS. Theo
huang nay da huang dan 1 hgc vien cao hgc, 1 sinh vien he
eii
nhan tai nang
CO
hgc. De tai ciing gop phan
thiic
day su phat trien chuyen nganh ca hgc vat
f r f f
ran bien dang, dac biet la bg mon ca hgc. Vai 1 bao cao a hoi nghi Quoc te, 1
bai dang a tap chi khoa hgc DHQG, 1 bai dang a
tuj^en
tap Hoi nghi ca hgc
r
y
•»
f
y
toan quoc
Ian
thu 8. De tai da hoan thanh tot muc tieu de ra.
4.
CAC CONG TRINH CONG BO
*/ Dao Van Dung, Bui Thi Thuyet. Elastoplastic stability of thin
rectangular plates made of compressible material under nonhomogeneous

complex
loading (to appear in VNU. Journal of Science, 2007)
*/
Dao Van Dung, Nguyen Cao Son. Calculating elastoplastic
rectangular plates by finite element method. Proceedings of the Eigth
Nafional
Conference on Mechanics, Ha Noi 6-7, December, 2007
11
*/
Dao huy Bich, Nguyen Dang Bich, Nguyen Anh Tuan. Investigation
of Aerodynamical instability phenomena by using a method for the
approximated analytical solution of the Vander Pol equation. The
1^*
International Conference on Modem
design.
Construction and maintenance of
structures, 10 -
11,
December, 2007, Hanoi, Vietnam.
12
IV.
PHU LUC (Cac bai bao va bao cao khoa hoc)
13
ELASTOPLASTIC STABILITY OF THIN RECTANGULAR PLATES
MADE OF COMPRESSIBLE MATERIAL
UNDER NONHOMOGENEOUS COMPLEX LOADING
Dao Van Dung, Bui Thi Thuyet
Department of Mathematics, Mechanics, and Informatics
College of Science, VNU
Stability problem of the elastoplastic plates subjected to the nonhomogeneous complex

loading with incompressible materials is investigated in
[4],
In this paper studied a
above problem with compressible materials, established the stability equation and solved
one by the Bubnov-Galerkin
method.
Have been calculated and compared the critical
forces between the compressible materials and incompressible materials.
1.
Stability problem
We will consider a thin rectangular plate which has the biaxial dimensions a, b and the
thickness h. An orthogonal coordinate system
Qx^x-^z'xs
attached to the plate so that the
plane Ox^Xj
coincides with its middle surface.
Wc assume that the plate is acted by biaxial compressive forces
Py^=
p^\t,Xjj^
P22 ~PiiV^^])
depending arbitrarily on t,
x^^x^
and shear force
p^^
=
Pnid-
The
problem
is proposed that have to establish elastoplastic stability system of equations of the plate
and to define critical value

/*
and critical forces
pn^AiV'^)'
P
'^^- P22V
^^\)^
P
'2 "=
P\2\)^
respectively.
First wc use the stress- strain relationship according to the theory of elastoplastic process
[K3J
^ll==(Al^Il-A2^22)^^+A3(2^n+^22)-A4(^lI+2^22)
G
^l2=A|f^,2-^^-A2(2^M+^22)-Alfe,+2^22)+A4^l2
cr'
in where
A. =
C
,
B
^
aa.j
1 + — + A ^^
3K 3Ka
"
J
;A,=^
+
A

"
3K
3>Kal J
C
V
+ ^A ^^
3/:
ZKa
" /
;A.=|
+ ^ "
-iK
ZKGI
J
"
c
-^
A
^^
3K 3Ka
"
J
•D
-^
'
"
C
1+ +
A~ "
3^

3/:c7,!,
A3 =
5^
c
+ /1 "
3^
3KaJ,
••^"-~c
+
+
A-
"
3>K 3Kal J
->
I'_2
C
3A'CT
C
3A'cr.
C
2>K(7
(1-2)
C
= 1+ -\-A ^ + /1 ^
.
3K
l>Kal
3Ka^
A
= ^'-N -B^-N ;N = ^

3
X
2.
Stability
equations
2.1.
Prc-buckling stage
At any moment t in the
pre-buckling
stage, there exists a membrane stress state
^22
=-P22{^-^'i)'
t^l2=-Pl2
(0,
^23
=^13 =^33
=0'
(2-1)
a
-
^11
+<^22
P\\
+P
II
'
/'22
(2-2)
a„
=

V^M
-^11^^22+^'2+3a|'2
=
^|p'^
-
/J,
1/^22
+
;^22
+
3/;
(2-3)
9
It is easy that the chosen stress values satisfy equation of equilibrium,
boundar>'
condition of the problem.
The arc -
lengh
of the strain trajectory is given respectively by the formula
^ =
^[(^1.
-^22)'
+(^22 -
^33)'
+
{^33
-
^n)'
+6ef,f =
F{s,t)

Where
[1,3]
(2-4)
^M
=
N
Pu+-P
22
-n{s,i
Pu p
22
Pu+Pn
9K
^22
= —
N
P22+-P,
n(v,/) P22-TA
P^^+Pn
9K
^i2=-
3^2
3
2N 2
n(^,0A2>
(2-5)
_
(7
.
^33

~
"7^
~^H
~ ^-~
'
ii(.u) =
1
11
A 1 r
f
-vj^
1.1.
,
.
PuPu
+
PiiPn PuPn PuPii
+^PnPi
'{'•
=
<t>\s)-
N
a.
Hereafter, we will use the criterion of bifurcation of equilibrium state to study the stability
of plate according to physical relationship (1-1).
2.2.
Stabihty
equations
Wc use the assumption of
A.A.

Iliusin
[2] said
that^^
=0
and
don't
consider the
unloading domain, then the stabiltily equation of elastoplastic rectangular thin plate
'L
4.
jsj^^Sy.,
-
0
dx,dXj
(2-6)
When the
plate is
cur\cd,
we can get the increments of
deformation&*,^.
Using the
Kirchhoffs
a
ssumplion .
wc
liavc
^^n
=^K
^>./
(2-7)

in where
Se',, - the small increment of strain of the middle surface,
5x,i
- the small increment of curvature and torsion. They have the form
Ss„
=-
'
2
dSu.
ddu
+
V
^^j ^^i
SXi,
=
dx,dXj
(2-8)
The increments
Su^,
S\v
are the funtions of
x^
and
x^.
The increment of stress, according to the theory of elastoplastic process, we have
Sa,,
=
{D,,a,,
-D,,a,,)^^^
+

D,,{2S£,,
+
SE,,)-D,,{S£,,
+2Se,,)
UUSE
Sa,,
={-D„a,,+
D,,a„
)^^
-
D,,
{2Se,,
+
.^^22) +
A^
(^^,,
+ 2 ^^22) (2-9)
cju.Se
Sa,,=D,,<j,,-^^-^-D,,{25e,,^5e,,yD,,{5s,,
^2Ss,,)+D,,5s
12
a.
he increment compoments of moments, don't depend on z, are defined
<5/V/||
=
\zSa^,dz
hi
12
(A
1^11

^2^22)^^
+^3(2^^'.,,+^^.22)
-A4
(^'',11+2*^22)
]
SM,, =
\zSa,,dz
= -
12
a„Sv
(-
D„a,,
+
D22C722) " ,"
-
A3
(2<5iv,,
+
^v 22)
a
+ A4(^*',n+2Av22)
]
(2-10)
/;'
SM,,
= jzSa„dz =
a„Sw
A>^,2-V^-A2(2<5iv„ +Av,J-A3(5>v„ +2^V,J
+
A,,(5ii;|j

]
Writing
(2-10)
in
the
dc\clopment, leads to
h'
SM„ =
12
A,
r-
+
A.
+
A
d'Sv
^
ex:
^ dxl '" dx.dx
(2-11)
l^-^2
J
dM22 = -T^i
A^
—;-
+
A^
r-
+ A
\2\^

dx '
dx\
dx,dx
2
J
SM„
=
h'
( .
d^av
.
d'Sv
.
d'av
^
12
A,
V ^-^i
2
+A-Z-l-
+ ^9
ax:
5^15^2
y
where,
we
denote
^,
=(D„^„-D,2CT,J^
+

2D,3-D„,
^^=(D„a„-D,2C722)^
+ A3-2A4>
^3
=2^(D„a,,
-A2f^22),
^4
=(-A.^ +A2^22)^-2A3+A4.
'^.S=(-A,^M+A2^22)^-A3+2D2,,
A-2^(-D2,C7„+Z)22a22),
^7 -
A
0'I20'M
'^12^22
2D
-D A =D
—^ ~-D
-2D
^^n
^33
'
^h
^31
2
32
•^'^33
'
a.
a.
(2-12)

The coefficients
A^
(/
=
1 —>
9) arc the
functions
of
.v,
and
x^.
Putting
the
expression
of
5M,^
in
(2-11) into equation
(2-6)
after series
of
calculation,
we
get
dx\
dx^dx.
+
D
d'd\v
^

d'd\v
^
d'a
'
dxldx]
"" ^'
+
A
V ~
d 3w
5^19x2
^-'^2 ^A
+
A
d'Sv
dx^dx.
+
A
d'Sw
^
d\v
-^
5'Av
^
d'^Sw
^
d^Sv
dx,
dx
+

D„ —r
+
£>
2
5^2
5X,
2+A,
5.r,9x2
+ D
12
:^
2
-:—(cJ|.
^+2cr
+ 0-,.
r-J
= 0.
/;^A^^
"
ax
dv,5A*2
22
^
2
(2-13)
wiicre
'
4iV ^ 47V
^
3

(.4,
+
2/0,
A =777(4+^^4+2^)
;D,=—(^,+2/(J,
4iv
4/V
A= A
;
-
4A'
•
A
=
4.V
5/i,
-
dA,
—- +
2—-
^
c'.v,
dx.
D.
'
4.V
2^
+
2^
+

2^
+
2^
V
5-^'i
5.V2
ar,
av
2 /
4N
Ao =
dA, ^dA,
^aA aA
V
^-^'i
+ 2
dx.
+
2^^
+ 2
r
r.2
4N
^dxf
Dx, Dx.
2
A
\
,
A

47V
^dx^
av,
dx^dxj
8x2
J
D.,-'
f
':^1
4N
I
y
2
d'A,
^,
a%
^
a^^,
^
v^
5xf aX|5x2
5.V2
y
(2-14)
The expression (2-13) is the stability equation of the elatstoplastic plate made of
compressible material under nonhomogeneous complex
loads.
Remarks
+
when

p^j
only depends on t, the coefficients
A^i
= 1 ->
9)don't depend on
x,
and
x^,
then the experssion (2-10) gives us
D,=D,=D,^D,=D,,=D,,=D,,^0,
equation (2-13) returns to the stability equation of thin rectangular elastoplastic plate
under complex homogeneous loading [3].
+ If the material is incompressible, then (2-13) gives us the results in [4].
3.
Method flnding the critical forces
It is difficult to
solve
directly stability equation, so we shall use
Bubnov-Galenkin's
method
to find critical forces. The common diagram of the method is:
a) The deflection
Sw
is expressed in the form
M
S\v
~
y^R^dw^
(3-1)
/=l

where M - term of a series,
R^-
the nontrivial terms of
series,
<5ir-
linearly independent
functions and satisfying the given boundary conditions.
b)
Denoting
by
F{SW)
the left side of the equation (2-13) and putting
d\v
of (3-1) into
F{(5\\'),
the result is
F{S^v)^f^R,F{S^v,)
(3-2)
(-1
c) Multiplying both sides of (3-2) equation by
S'A\{k
= 1 -^
M)
and integrating both sides
of the received equation all over the domain of the plate, we obtain
III
0
0
'='
\\Y,R,F{5w,)av,dxdy = Q

(3-3)
d) Taking
k
from
1 to M, we get a
system
of M
linear algebraic equations with
the
unknowns
R„R„
,R^^
.This
system
has the
form
M
Y,H„R,=0,k
= \,2 M
(3-4)
i
= \
where
a
b
H„ = IJF{av,)S,,dxdy
, (3-5)
0
0
Because

/^^/e^,
,7?,^
must
be non-
trivial solutions
of the
system (3-4),
the
condition
for
this
is
det(//,J-0
(3-6)
This
is an
expression
to
define critical forces.
4.
Method
for
determining coefficients
H^^
For
the
plate made
of
incompressible materials
had the

results
in [4]. By the
same
way,
wc calculate
the
coefficients //^^
of
plates with compressible material subjected
to the
complex
loads depending
on
coordinates.
This
method
consists
of
following
steps
a) Dividing
the
plane
of
the plate into
N
rectangular pieces
by
nodal lines parallel with
its edges, respectively

and
denoting
j-th
piece
is
Q^.
b) Because this nodal lines
are
before fixed
so the
coordinates
are
known. Therefore,
we
can calculate
the
value
of
quantities
Z),^,
a^,
o-,
cr^
at
those nodes.
c)
At the
internal points,
by
reason

of
the
continuit>*
of
loading function
and
dividing
on
enough
small
of
each node
of
piece
Q^
may be
approximate
the
belows quantities
as
linear function
of
,v,
= .v
and
.v.
^
v:
A;
=^;i-v^^':V

+
/;,3J-l:3J=l:4.
f
511-^
+'M:.!'"^
^513'
~T'^'^:i-^^'52:J"
+
'523'
a.
cr.
^n^22
2
~ '53
1-^
"^
'532>'
"^ ^533 '
^nQ'i2
=
r^Ax^
+
r,,,y
+
K,.,
^12^X2
'"55I^
+ ''5527 +
''553.
12

=
''561^
+
^562>^
+
^563-
(4-2)
The coefficients
^^^ ,
r^^^^
in (4-2) are found by the minimum square method using the
known values
ofZ),^,
a^^
,CF,CT^
at the nodes of the piece
Q^
d) After determining the quantities in (4-2) we shall find
A^
from the
(2-12),
D^
(/
= 1 -^ 12)
from
(2-14),//,,
from (3-5).
e) Integrate (3-5) by the Gaussian quadric method [6].
5. Finding the expression of the increment of deflection
Using the boundary condition, we find the increment of deflection. Although there are

many choices of the increment of deflection function but always the expresion of the
increment
deflection
function are choosen two forms:
a) Trigonometrical series
mxn . nyn
^^*^-ES^'""S'"
sin^
or
dsv^YT.^^
m=\
n=\
a
b) Power series
mn
m=\
n=\ \
1-cos
2m
JVC
a
1-cos
2n7ty
S\v
= yBXY
m=l
where
^„.
=
T^,

A-=0
X
a
i;
=
Z^.
*=0
Kby
6. Some results of
numerical
calculation by
soft^vare
Matlab[5]
a).
In the bellow problem, we consider the boundary condition of plates having simply
supported along the four edges, that means
S\v
=
0,
dx'
= 0 at
x=0,
x=a,
S\v
= 0 ,
d'S\v
^7~
=
0
at y=0,

y=a
8
The plane of plate is divided into 16 pieces by the lines parallel with the edges of plate.
We take the increment of
deflection Sw
for parts in 6.1,6.2 and 6.3 in the trigonometrical
series
r,
y^j^
„
.
mTDc
. nny
dw =
2.Z.^-"Sin
sin—^
m=I
n=\
a
b
6.1.
Problem 1
Given the loading function depending on t loading parameter
oc7^ni.
(283 +
0.10'
A
=283 +
0.1/
,

y;^
=
V_-_^
.
An yield point is equal to
a^
=
400{MPa).
n h
The ratio — varies from 20 to 45 with the arithmetical ratio equal to 5,
— = 35
and the
h h
Poission coefficient
K
=
0.4.
The function
^'(.y)
is given in [1], 3G
=
2.6x
10^A//^a.
The
results are presented in table 1
I1ie
ratio
—
= 40;
—

^^35,
Poission coefficient v varies from 0.2 to 0.5 the arithmetical
h h
ratio equal to 0.1. The function
<f}'{s)
is given in [1],
3G =
2.6x
lO^M/^a.
The results are
presented in the table 2
6.2. Problem 2
Let's consider the loading function depending on t loading parameter and coordinates in
form
(283 +
0.1/)'^
-^
y
450'
0.25-
a)
/;, =(283
+
0.1/
n +
0.35^
An
yield
point is equal to
a^

=
400{MPa).
The ratio - varies from 20 to 45 with the arithmetical ratio equal to
5,-=
35 and the
h
f^
Poission
coefficient
v^
=
0.4.
The function
^^0 is
given in [1],
3G =
2.GxWMPa.
The
results are presented in table 3.
6.3.
Problem 3
Let's
consider the loading function
depending
on t parameter and coordinates
p,=(246
+
//l+|'
p,=20t
The ratio

-
varies from 20 to 45 with the arithmetical ratio equal to 5,- =
35
and the
h
Poission coefficient 0.3. The function
f{s)
is given in [1], 3G =
2.6x10'MPa.
The results
are showed in table 4.
b).
Problem 4
The problem is consided with the four clamped edges
S\v
=
0,-^
= 0
atx=0,x=a,
S^v
=
0,^-^
= 0
aty=0.y=b.
ox Qy
Let's consider the loading function depending on t loading parameter and coordinates
p,
=(283 + 0.1/
n
+

0.35^
' ri
Pi
=
^'''*0-"y'1-025^'
450-
aj
The function of the increment of deflection has the form [4]:
4
S\v
= yB
X
Y
/N-l
where
x' x'
v' v'
/
v' v'
CI
a a a a
cr
ci
x'
.x'
^
••
A-
^x'
A^'

x^
^
y^
^y^
y'
r=^-3^+3i^->^
r->L_4i^^6^-4iL^z:
' y
b' b' b"'
'
b' b' b' b' b'-
n
h
The ratio
-
varies from 20 to 45 with the arithmetical ratio equal to
5,-=
35 and the
/;
h
Poission coefficient
v
= 0.4
.
The function
(f)'{s)
is
given in [1],
3G =
2.6x\Q^MPa

,
The
results
presented in table 5.
7.
The influence of the slcndcrncss on the stability of plate
Let's consider
the problem 6.2 with the loading function depending on
t
loading parameter
and coordinates
10
p,
=(283 +
0.1/
)|
1 +
0.35^
^,=(Hl±^Y,_o.25i^
450
a)
The ratio - varies from 20 to 45 with the arithmetical ratio equal to 5,-=
35
and the
n
1^
Poission coefficient
K
=
0.4.

We solve this one in two cases theory of
elasticit
and theory
of elastoplatic processes. The result is presented in table 6.
a
h
20
25
30
35
40
45
P\
All
412
457
444
434
423
Pi
568
544.7
478.7
426.5
389.3
351.4
<^
u
528.4
521.2

468.2
435.5
413.4
392.3
Table 1
V
0.2
0.3
0.4
0.5
P\
531
562
577
591
P\
468.3
541.6
601.9
668.5
^u
508.1
552.3
589.9
634.4
Table 2
a
h
20
25

30
35
40
45
P\
518.9
513.3
497.5
482.9
471.9
460.6
P*2
603.5
578.8
508.6
453.2
413.6
337.3
^u
565.9
549.2
502.9
468.8
445.6
412.6
Table 3
11