Научная статья на тему 'Boundary problems of elastic rods and their solution by finite difference method in various approximations'

Boundary problems of elastic rods and their solution by finite difference method in various approximations Текст научной статьи по специальности «Математика»

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Аннотация научной статьи по математике, автор научной работы — Isomiddinov Anvarjon Inomjonovich

Boundary conditions of elastic rods are considered in the paper on the basis of Lagrange variation principle. The computing algorithm of finite difference method in various approximations is developed. On the basis of comparison and analysis of numeric results an effective computing scheme for the solution of the problems of basic and applied researches are offered.

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Текст научной работы на тему «Boundary problems of elastic rods and their solution by finite difference method in various approximations»

Isomiddinov Anvarjon Inomjonovich, Institute of Seismic Stability of Structures of the Academy of Sciences of the Republic of Uzbekistan,

Senior Scientific Researcher E-mail: isomiddinov_a@mail.ru

Boundary problems of elastic rods and their solution by finite difference method in various approximations

Abstract: Boundary conditions of elastic rods are considered in the paper on the basis of Lagrange variation principle. The computing algorithm of finite difference method in various approximations is developed. On the basis of comparison and analysis of numeric results an effective computing scheme for the solution of the problems ofbasic and applied researches are offered.

Keywords: Thin-walled rod, variation principle, equation of balance, boundary problems, finite difference method, approximation, computing algorithm, cell matrixes and vectors.

On the basis of specified theory of rods, offered by V. Z. Vlasov, G. Yu. Djanelidze, V. K. Kabulov [1], the expressions of displacements of the points of a rod under joint action of longitudinal, transverse and torque forces may be presented in the form:

U = U - za1 - ya2 + $v + alPl + a2P2; U2 = y + zQ; U3 = W - yO

where U ,V,W — are the displacements of central line of the rod; a1,a2 — the tilt angles of tangential line in relation to elastic one; Pi, P2 — the angles of transverse shears; 6 — the angle of torsion; v — the linear twist, j — the function of torsion, coefficients a1 = y1 (y) - y and a2 = y/2 (z) - z in S. P. Timoshenko-V. L. Bider-mann representation [1]:

^ (y)=-f y -4y3], (z) = ±-f z -4z!\ * = 6. 2K1/ 3 h J ' 2K t 3 h J 5

Then, according to Cauchy formula with consideration of (1) one could calculate the components of strain:

_dul. £ _duL +duL _ _duL + du± _ = 0 (2)

*11 ; *12 - , + - ; Z13 - , + - ; * 22 * 33 _ *23 0 . (2)

11 dX dx dy dx dz

Stress and strain components are related by the following way:

dui ( du2 dul ) ( du3 dul ')

an = E£n = E—1 ; o12 = GI —- + —1 I ; a13 = GI —3 + —1 I ;

dx 1 dx dy J \ dx dz )

a22 = a33 = 0^ = 0.

(3)

Relationships (1) and Lagrange variation principle [1] are used to derive a fundamental differential equation of equilibrium of the rod: S(-fi + A ) = 0. (4)

Variation of potential energy¿n in this statement has the form:

8n = j± au8lJV

V i-1

Taking into account Cauchy relationships (2) and Hooke's law (3) one has the following:

sn = {

„ du. , du. f du2 du. | , f du2 du. E—1S—L + Gl —1 + —1 151 —1 + —1 dx dx I dx dy I I dx dy

+Gldui+duL wa^+duL dx dz I I dx dz

dV. (5)

Opening the brackets we could fulfill integration operation in parts:

Sn^Kj^Mj +a12Su2 +a13Su3) + ^a12Su1[ +JJa135u1| -

-J

^ SUl + ^ 8Ui + 5M3 + SUi + SUi dx dx dx dy dy

dV.

Sought for functions U, a, a2, v, Pi and P2 are the functions of x, therefore, boundary conditions in y and z drop off. Here with

consideration of Hooke's law (3) similar items are given and an integral in rod section is singled out:

5n=JJ

„du. ~ J du2 du Ï f du3 du. ,

E—1 Su. + Gl —1 + —1 \Su2 + GI—1 + —1 \Su, dx I dx dy I I dx dz

S

ff

d2u

d\ d2u, d2u, d\

- + -

E—2" + Gl 2 2

dx i, dx dy dy dxdz dz

2- + -Ï + -

\ ^ Su,

dF + (6)

"J

G

d u3 d u, . —f + —- |Su2 dx dx dy ^

dF + J[ G[ ^

d2u3 d 2u,

+

dx dxdz

Su3 dF

dx.

Variation of the work of external forces SA in (4) is taken in the form:

8 A = {]T PSUdV + {£ f SUtdS1 +{£ qt SU\dF\x, (7)

V i=l S i=1 F i=1

where Pi, fi, qi — are volume, superficial and end components of the force.

Single out an integral in rod section:

8A = j j P.Su.dF + j P28u2dF +

x L F F

+jP38u3dF + jf18u1dl + jf.\8u2dl + jf38u3dl dx +

F lll _

+ J qi5u1dF + J q25u2dF + j q3Su3dF I

-F F F J*

Variations of potential energy (6) and of the work of external forces (8) are substituted into Lagrange variation principle (4):

«5(-n + A ) = j

e dU+G

dx2

d u2 d u.

_2__,__1 _

dxdy dy2 d2u, d\

_i__3__i__L

dx dz dz2

\ \

+Pi SuL dF +

J

( if

t-J fSu.di-

G

W ff

d2u, + d2u,

dx dxdy

+ P

52u, 52u. 1 GI —^ +-1 + P3 Su3

V- vv dxdzJ J j

Su2

J J

\ A

dF + J f2Su2dl-

dF + J f3Su3dl

dx +

^JJ

-E+ q, \8u, + dx 1 \ 1

-Gl^ + dUl.| + q2 1 dx 5y 1 2

Su2

+i -g f+f \+q.

dzdy = 0

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equation (9).

F

y

Expressions u from relationship (1) are substituted into Lagrange variation principle (9). We will introduce the following di-mensionless values: x = lx, y = b0y, z = h0Z, U = h0U, V = h0V and W = h0W, and perform the integration operation in rod section. As a result in variation equation of equilibrium (9) there are the following sought for functions: W, al, ¡3l, V, a2, P2, U, 0 and v . Formed equation ofvariation exists in any volume of discussed object.

Besides, variations of sought for functions do not equal to zero. Therefore, from formed variation equation of equilibrium (9) one obtains the system of equations with nine unknowns and corresponding boundary conditions.

Introducing the displacement vector V and the vectors of external forces F, Q in (9), we would obtain the system of differential equations with natural boundary conditions in a vector form: d_ dx

and boundary conditions:

dV -A— + BV dx

dV - -

+C-+DV=F

dx

(10)

A— + BV -Q ¡>SV dx

= 0,

(11)

where V — is a sought for vector ofthe function; A, B, C and D — quadratic matrices of the ninth order with corresponding elements ai,, b, c and d (i = j = 1,...9 ), presented in details in [2]. To il-

lustrate it we would present the elements a11 and a18

l2

3n = — F ; 11 3I

,o , here I. — is an inertia of cross section, the area of

lo o 1 T z 0

3V 0

cross section of the rod - F and static moments - Sz are determined from the following integral relationships:

F = jj dF ; Sz = \\ ydF

F F •

Usually there are geometric, static and mixed boundary conditions.

Boundary conditions (11) are written at x = 0 and x = l :

dV - -Ko A — + W = Qo, at x = 0; dx

dV - -

-KNAN-+ TNBNV = QN , at x = N,

N N i N N ^-N ' '

dx

(12)

here K0, KN, T0 and TN — are forming coefficients of boundary conditions.

To compare and analyze the results we would solve the formed boundary problem (10)-(11) by finite difference method in approximations [3; 4]:

I. Central differential scheme (CDS);

II. A. A. Samarsky - I. V. Fryazinov modification (SFM). Consider computing algorithms of solution of a boundary

problem with geometric boundary conditions.

I. CDS. Re-write the formulated problem (10)-(11) with geometric boundary conditions, when P = (B + C) :

AdV +pdv+Dv = p ; (13)

dx dx

V = 0 ; VN = 0. (14)

Using approximation of central difference scheme, the boundary problem (13)-(14) is written in the form [3]:

AVi+1 - BV + CV 1 = F ; (15)

"v0 = 0, V^n = 0, ' (16)

where A. =1 A. +1 P. ; B = - A. - hD. ; C . =1 A. -1 P. and F = hF. ' h ' 2 ' 'h ' ' 'h ' 2 ' ' Boundary problem (15)-(16) is solved by run method, with

the following recurrent formula:

V = aV+i + A, i = N-1,...1 ; _ (17)

here: a={B - Cfli1)1 A, - Ctat ) (Cfiil - F). (18)

II. SFM. In A. A. Samarsky - I. V. Fryazinov modification the vector equation (10) is approximated by the difference scheme devided in two groups. Introducing analytical grid mh = { = ih } ( i = 1,2,...n -1) inthe domain G, we approximate the equation (10) by the following difference schemes up to the order of O (h2 ) [4]: 1 f A+1 + A yff-V(1) _A, + A 1 yOj-V®* h 1 2 h 2 h

B,,V (2_ - b.v[I_>_+cv}1! - c.V

?(2)

(2).

r( 2)

- + DV. (i)= F. (i);

h h

1 f A+t + A V-+2>- V-(2) _ A + A l V-(2)- V-h 1 2 h 2 h

+

2 h 2 BVV.(I)- b V(i_ c V(I)- c V(1_

(19)

.+D. V(2) = F.(2).

h h In the system of equations (19) we will introduce the following

denotations:

A., = ; A., =

/2 2 ' >2 2

(20)

then, the equation (19) acquires the following form:

--A.

11 V1 - V(1) V(1) - V

h [ i+>2 h f>2 h

B.,V.(2>- BV.(2) c.V(2>- CV(2)

h

h

+ DV. (I)= Ü(I);

^+9-V(2> - ^(2)-^ h l f+>2 h fX h

+ BV( (1)- BilVi + cv(1) - CV('i) + DV0) = pu

hh

(a) = 0, W = 0 ; 0 N

(21)

(22)

where a = 1,2.

Introduction of the vector = (1) V{(2) J , into the system of equations (21) is presented in the form:

IiUM -BiÜi +CuÜi-1 = Fi; i = 1,2,...,N -1;; (23) U. = 0 at i = 0 and i = N; (24)

here A i, B i ,C i - are cell matrices, F i — cell vectors in the form:

A +1-

B i =

A.

A f =

A

Ü 1 i>2 ! h

h Q

- hD.

A

Ï2

B + C.

A

A

C i =

-Bi - C

Ax

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h

-B , - C.

A

12

h

; F i =

h

hFi (I)

hF(2)

-- hD.

To solve the formulated Cauchy problems (23)-(24) the method of matrix run is used.

The solution is sought in the form

Ut =a iUM + at i = N -1,...1 ; (25)

where ai and — are yet undetermined cell matrices and vectors. For the calculation of these parameters one may get the following recurrent formulae:

ai -CiOd-1) ' Ai, 0. =(%i-Cia) 1 (Ci0,i-1 -JF^). (26)

Boundary problem is solved with geometric boundary conditions under the following initial data: geometric and mechanical

r

h

h

0

characteristics of the rod: l = 200 cm , h0 = 20 cm , b0 = 15 cm , E = 2 • 106 kg I cm1 , the values of external load: = 12 ; f- = 6; f:=10 ; f-= 2 ; f+= 8 ; f; = 4 kg /cm2 .

Vector components F in differential equation (10) are formed as follows:

f = ^ +)( +; /2=^ [b. f-)] ;

2EJ„

/3 =

/ 3b-

EhJ 0

h2 )+/-( hi

14

; *=mja [(+}(/;+;

/S J

b (+- f-) 74

=

• f - ^ ; f6 EJ0

f+i-12 > 4 k

EhJ 0

[(h0 + b0 )(( + /;)]

/9 = [(h0+b0 )(-/;)].

SEKJ 0

Table 1. - Numeric results of design of boundary problems in approximations: CDS and SFM

4

max (( (x )) I. CDS II. SFM Approximate solutions under different values of nodes number

N=10 N =20 N =40 N =80 N = 160

w (0.5) I. 0.264553 0.404896 0.471792 0.492521 0.498021

II. 0.482354 0.495291 0.498693 0.499582 0.499809

(0.2 ) I. 0.671438 1.076926 1.269323 1.328727 1.344461

II. 1.349723 1.349757 1.349778 1.349787 1.349789

A(0-1) I. 0.085271 0.148715 0.178511 0.187567 0.189950

II. 0.186052 0.189709 0.190524 0.190701 0.190743

y (0.5) I. 0.322373 0.590028 0.758660 0.818209 0.834672

II. 0.809080 0.832241 0.838231 0.839785 0.840183

ai (0.2 ) I. 0.868041 1.672228 2.177690 2.355813 2.405001

II. 2.421581 2.421735 2.421813 2.421844 2.421854

P2 (0.1) I. 0.076457 0.197457 0.272994 0.299344 0.306593

II. 0.304416 0.308341 0.308956 0.309051 0.309069

u (0.5) I. 0.026250 0.026250 0.026250 0.026250 0.026250

II. 0.026250 0.026250 0.026250 0.026250 0.026250

0(0.5) I. 0.091964 0.092554 0.092754 0.092811 0.092826

II. 0.092153 0.092608 0.092769 0.092815 0.092827

v(0.1) I. 0.060032 0.063666 0.064754 0.065045 0.065120

II. 0.061358 0.063994 0.064836 0.065066 0.065125

Table 1 gives the approximate numeric results of design of the problem under the following boundary conditions V| = 0 and = 0. Maximum values of sought for functions V. -102 are given by two values. The first of them corresponds to CDS solution, and the second — to SFM solution with different values of the grid step-h at N = 10, 20, 40, 80 and 160.

On the basis of the Table data one may conclude that at N = 10 and N = 20 the deflection W and turning angle ar coincide up to two or three signs, at N = 40 and N = 80 coincide up to three or more signs in SFM. In CDS at N = 40 and N = 80 near the ends ( x = 0.1) the values coincide up to two signs, and in the middle of the rod up to one sign. From the table and the graphs it is seen that to obtain more accurate solution of the boundary problem in

CDS it is necessary to choose the number of nodes N = 80 and more. In SFM it is sufficient to take N = 10 or N = 20.

Based on results obtained (Table 1) one may conclude that numeric results in approximation by A. A. Samarsky-I. V. Fryazinov Modification approach faster a stable (asymptotic) solution compared with central difference scheme of the second order of accuracy.

Figure 1 shows the parameters of the vector of displacement V¡ along the length of the rod. The graphs are built on the basis of numeric results in SFM at N = 20.

So, the use of approximation by A. A. Samarsky-I. V. Fryazinov modification shows its merits in solving the problems of fundamental and applied nature, described by differential equations of linear and non-linear types.

Fig. 1. State of design values W, V , U, a, a2, ft, ft along the length of the rod

References:

1. Kabulov V. K. Algorithmization in the Theory of Elasticity and Deformation Theory of Plasticity. - Tashkent: FAN, 1966. - 394 p.

2. Abdusattarov A., Yuldashev N., Isomiddinov A. et al. Numeric Methods of Design of Bearing Elements of Thin-Walled Structures under Spatial Loading. - Tashkent: Uzbekistan Publ., 2015. - 160 p.

3. Samarsky A. A. Introduction into the Theory of Differential Schemes. - Moscow: Nauka, 1971. - 532 p.

4. Samarsky A. A., Fryazinov I. V. On Differential Methods of Approximation of the Problems of Mathematical Physics//Journal. Success in Mathematics Science. - Moscow: Nauka, 1978. - Volume 31, Issue 6(192). - P. 167-197.

Kambarova Nigora, Tashkent State agrarian University, assistant, the faculty of agroengineering, Uzbekistan E-mail: komil_uzmei@mail.ru

Determine the resource elements that affect confidence level of two lined cotton harvesting machines "Case-2022"

Abstract: In the article the results of the research work are brought about determine the resource elements that affect confidence level of two lined cotton harvesting machines "Case-2022" in Uzbekistan.

Keywords: cotton collecting machine, resource elements, spindle, extractor drum, humidifying unit, a ball lifter, air system,

spindle drum.

In the desire to determine the resource elements (item, node) that affect confidence indicator line two cotton collecting "Case-2022" machine's testing results were sorted. The testing was done between 1996-2000 by the Certification and testing center of the Agriculture technologies and technics of the Republic of Uzbekistan [1; 2].

It is seen that by the data, below are the conclusions which were made according to results (table-1) of the experiments (1996, 1999, 2000) of two and four lined HSM "Case-2022" that were taken at the "Testing and certification center of irrigation technologies and technics of Uzbekistan".

The main part of the total 299 breakings happened with the spindle (66.2 %), the extractor drum (17.4 %) and the humidifying unit. Share of the prop shaft, jack pneumatics and spindle

drum is 3.25 %. And the share of problems in the system, receiving chamber of the machine, basket and wheels are 0.33-0.66 %. Malfunctions in the spindle: Conical gear rod was broken to 27 mm. in length (in 166 cases), teeth of conical gear which turns the rod (in 13 cases) and the cartridge which is stem of the rod was mutilated (in 19 cases).

All 52 malfunctions in the extractor drum are involved with corrosion of the trapezoid teeth of the extractor disks.

Plastic support of a humidifier pad was broken 22 times. This figure belongs to all breakings that happened in the humidifier unit.

Unhealthy spindles will be changed only to new ones, because their design is not meant to be repaired, its basement conic gear and the surface cogs are part of the same body and they are made in a very complex way.

Table 1. - The number and percentage of problems in the components that have arisen during the experiment of two and four lined horizontal spindle machines «CASE-2022»

№ Component parts 1996 year 1999 year 2000 year Sum

1 Spindle 80 47 71 198 66.20 %

2 Extractor drum 7 0 45 52 17.40 %

3 Humidifying unit 20 0 2 22 7.30 %

4 Cardan shaft of cotton collecting machine 0 1 9 3.25 %

5 A ball Lifter 3 0 2 5 1.60 %

6 Air system 3 0 2 5 1.6 %

7 Spindle drum 0 3 0 3 0.66 %

8 Hydraulics 2 0 0 2 0.33 %

9 Cable unit of a device 1 0 0 1 0.33 %

10 Basket 1 0 0 1 0.33 %

11 Wheel 0 1 0 1 0.33 %

12 Sum 125 51 123 299 100 %

It is very difficult job to change pump spindle with the new one; in order to change one spindle it must be removed from the spindle cassette (rotor) drum, washed, unhealthy spindle will be removed, and replaced with a new one and the cassette is will be installed back to its place.

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If the extractor drum's teeth are mutilated at all, they will be changed to only new ones, because the disk and lower trapezium

teeth are made of a rubber material. In order to change the unhealthy disc, the extractor drum must be removed from the frame unit, disks are separated from the shaft, instead of unhealthy disc will be set new ones and the drum will be set back into its place.

Machine operator or for replacement of one spindle, extractor drum with mutilated teeth and for a broken plastic support of the humidifier pad spends an average of20, 62 and 11 minutes respectively.

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