*/ Nghien cuu hien lugng mat on dinh khi dgng bang phuang phap nghiem giai lich gan dung ciia phuang Irinh Vander Pol.. \ f y Da trinh bay thuat loan, tinh loan bang so, xay dung do th
Trang 1DAI 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 t n de tai : PCS TS Dao Van Dung
Can bo tham gia :
GS TSKH Dao Huy Bich
CN Bui Tin Tliuyet
H a N 6 i - 2 0 0 7
Trang 2I BAO CAO TOM T A T 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
M a s 6 : Q T - 0 7 - 0 3
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
*/ 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
Trang 3Da xay dung dugc he phuang Irinh on djnh din ddo X6l hai \6p bai
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
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
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)
Trang 42 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
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
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
Trang 5*/ 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
T r u a n g Dai hoc khoa hoc Tu uhien
I^U tPUONC*
- i ^ ^ ^ ^ ^ ^ : ^ i v ^ ^ ^
Trang 6n 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: Q T - 0 7 - 0 3 ; 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
Trang 7-01 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
Trang 8III NQI DUNG CHINK CUA BAO CAO
7
Trang 91 M a 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
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
Trang 10phuang 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
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
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
Trang 11Bai 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
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
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
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
Trang 12+ Tim nghiem giai tich gan diing thu nhat
+ Tim nghiem giai tich gan diing tiep theo
+ Bieu dien nghiem ciia phuang trinh can nghien cuu ban dau
+ 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
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
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
Trang 13*/ 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
Trang 14IV PHU LUC (Cac bai bao va bao cao khoa hoc)
13
Trang 15ELASTOPLASTIC 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 p n ^ A i V ' ^ ) ' P '^^- P22V ^^\)^
Trang 17It 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
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 t h a t ^ ^ =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
in where
Trang 18Se',, - the small increment of strain of the middle surface,
5x,i - the small increment of curvature and torsion They have the form
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
Trang 19The 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
Trang 20+ 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
/=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
(-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
Trang 21III
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
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 + / ; , 3 J - l : 3 J = l : 4
f 511-^ + ' M : ! ' " ^ ^513' ~ T ' ^ ' ^ : i - ^ ^ ' 5 2 : J " + '523'
Trang 22e) 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:
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
Trang 23The 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
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
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
Trang 24p,=(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
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
Trang 25p, =(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
<^ u
528.4 521.2 468.2 435.5 413.4 392.3 Table 1
V
0.2 0.3 0.4 0.5
^u
508.1 552.3 589.9 634.4 Table 2
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
Trang 26P\
52.50 49.25 40.75 38.00 32.51 28.00
^ u
465.5 455.7 443.2 437.7 418.2 415.6
491 481.3
470
P\
610.4 598.2 548.9 502.4 498.5 420.5
o-„
509.3 497.4 481.9 471.5
462 430.5
P\
(plasticity) 603.5 578.8 508.6 453.2 413.6 337.3 312.4 302.4 298.5 278.3 256.8
^ u
(plasticity) 565.9 549.2 502.9 468.8 445.6 412.6 398.4 282.3 275.5 243.7 223.5
(elasticity) 670.4 623.5 598.7 532.3 480.7 420.5 398.4 282.3 275,5 243.7 223.5
Table 6
12
Trang 27The graphics of the above problems are presented as follow/^j