THE STABILITY OF THIN-WALLED OPEN- PROFILE BARS WITHIN THE NONLINEAR ELASTIC DEFORMATION Текст научной статьи по специальности «Физика»

THE STABILITY OF THIN-WALLED OPEN- PROFILE BARS WITHIN THE NONLINEAR

ELASTIC DEFORMATION

Sadigov I.R.

Department of Mechanics, Azerbaijan University of Architecture and Construction, Ayna Sultanova 11, Baku, Azerbaijan Phone: +994125390609

ABSTRACT

The paper considers researches dealing with the stability of thin-walled open-profile bars. The widespread use of thin-walled bars in engineering constructions is resulted in a significant reduction in the weight of these systems. Considering the relevance of the given problem, the stability of nonlinear deformation to the central axis direction of the thin-walled bars has been investigated. The physical nonlinearity of the bar's material, dependence of the normal tension in its cross-section on the relative linear deformation has been taken as the form of the dual cubic polynomial. An appropriate nonlinear differential complex equation for a single torsion angle has been composed for the determination of the normal and touching tensions at bar's cuts in the non-free torsion of the longitudinal compression of the bar subjected to nonlinear deformations, and free touch tensions in free torsion towards the direction of the thickness of the bar. In order to use the small parameter method for the solution of this differential equation, the small parameter expression is composed of the elastic characteristics of the bar material. The solution line of the form of the nonlinear differential equation due to the small number of parameters is divided into differential equations, so that their solution is easily carried out. As a result, the expression of thin-walled bar's tension is obtained in the third approximation.

Keywords: Thin-walled bar, nonlinear deformation, open -profile, deplanation, non-free torsion, bending, curling moment, sectorial field, sustainability.

INTRODUCTION

The tap of the thin-walled bars in different constructions, especially in shipbuilding, aviation industry, and construction of high-mile buildings, etc., caused a creation of the new computation theory. The famous scientist, Vlasov's fundamental works had an irreplaceable role in the sphere of the creation and development of this theory [1]. Taking into account that the thin-walled bars squeezed in the longitudinal direction are problematic ones, the significant investigations of Peres N., Goncalves R., Camotim D. and others along with Vlasov's survey had a great impact on their work on calculations for sustainability [2-4, 9].

Unlike the closed contoured or the whole cut thin-walled bars, the open-profile bars are slightly resistant to torsion. According to the general theory of open

profile thin-walled bars, in the torsion of such bars their cuts are bent, thus various points take different movements in the direction of the central longitudinal axis of the bar. Such longitudinal displacements are called deplanation.

PROBLEM STATEMENT If the deplanation of the cuts of the bar doesn't occur freely, it implies that normal tensions arise in non-free torsion. In this case touch tensions also arise in the points of the cut of the bar. These touching tensions are indicated as t?s-, they are accepted like regularly disseminated in wall thickness of the shaft [1]. In the free torsion the tensile stresses varying by linear law in the direction of bar thickness are called free touching tensions, and are indicated as ts (see Fig. 1).

a) Non-free torsion

Figure 1. The touching tensions.

b) Free torsion

When we indicate the momentum that is born of internal touch forces in the free torsion with Mb, and the momentum that is born of touch forces in the nonfree torsion with Mb, the full torque momentum is taken as follows:

Mb=Mb + Mb

(1)

The shift (deplanation) u of any point of the cut of the bar to the longitudinal axis x can be taken as follows [2]:

u = -a(x) • w(s),

(2)

here a(x) — is the relative torsional angle of bar , which is the function of x variable, w(s) —is the sectorial area of S function. Sectorial area as rotation of radius-vector that takes its beginning from any polar point k is assumed as double area resulting from the movement of the last (the second) point on the middle line of the bar wall (Fig. 2).

Figure 2. The sectorial area.

The negative symbol in Eq. (2) indicates the counterclockwise rotation of the radius-vector. Considering that the bar material is non-linear elastic, we find normal tension in its most extreme non-free torsion in the cut of the bar as follows [5]:

(v < 1), p = 1/e1mh , em± — is the relative deformity of the material due to the range of the tolerance of the material [6].

Using Koshi dependences and considering Eq. (2), we can write the following:

ox = e0£x eIex,

(3)

here E0,E1 — are elastic constants of the bar material, £x is the relative longitudinal linear deformation.

Choosing the Method of Solution Let's make the last expression as follows:

ox = eo£x(1 - VPs'2),

(4)

here v = — e;n h — is the small parameter drawn

Eo

from the elasticity of the bar material

da(x) , ^

£r =---Wis),

x dx v '

(5)

here the single torsion angle a(x) equals to derivative of 9 — through x variable:

a =

dQ dx

Taking into account the last equation, we can substitute Eq. (5) with Eq. (4) and have:

aT = -E,

£»*>- vp № •(.•*»

(6)

Considering the following equilibrium Eq. (6) we determine the touching tensions:

—^ +— =0, from here

dx ds

rs drrx , _

T = -fo -£ds = Eo

(7)

We take the last equation and multiply it with the thickness of the bar wall t and get the intensity of the flood of the forces touching along its wall:

Tt = -fcS d-ttds = Eo

d39 rs

wtds

(8)

In Eq. (8) we mark tds = dF andr ■ t = q, but —]0> = f w2dF — sectorial inertial momentum

mtegrak are mdicated as follows: (unit of measurement sm6).

= f udF — sectorial static momentum Considering these signs, we make Eq. (8) in the

(unit of measurement sm4), following form [7]:

3

q = E0

d (d2e\3

dx3 to f rfv V J JF

dx \dx2/

(9)

non-free torsion. As it is seen from Fig. 3, sm6 is polar momentum of elemental force qpds = q ■ dw (here dw = pds - is the growth of the sectorial area).

We define the Mb momentum due to arrow passing through the k pole of the tensile forces in the

Figure 3. Determination of momentum of the touched force. Momentum alternative Mb is written as follows:

— r \d39 r r d fd28\3 r rs i

Mb = Jr qdw = E0—3)F dwJF wdF-vp ■ — [—2) ■}„ dwL w'dF

here integration is carried out on all F areas. We get this equation through partial integration:

Mb=Eo

'gCH adF-jJ-vfi-ifë)3-^!, *>3dF-JF œ4dF)

d x d x2

(10)

In the definition of the sectorial area the starting position of the radius-vector is determined by the fact that the exact sectorial static momentum of the field is zero, that is:

S0>.F = IF œdF = 0

(11)

Realization of the Method

Taking into consideration the above-mentioned symbols, we put Eq. (10) in this form:

Mb = -Eo

d (d20\3 d x d x2

-3dF-l,

œ4dF)

(12)

We can write the momentum of the tensile forces of the profile that are created by the free torsion as follows:

de

Mb = GJk-~

d x

(13)

here GJk is rigidity of profile in torsion, Jk is inertia momentum of torsion. We can write the equation in the following way (if profile consists of rectangle):

Jk=1vl?=i Sft?,

(14)

here Si is the length of the i small wall, ti is the thickness, and t] - is the ratio that is the basis of the shape of the cut. The unit of Jk measurement is sm4.

According to Eq. (1) the general torsional momentum equals to the sum of Eq. (12) and Eq. (13):

Mh = -En

d3e d (d2e\3 , r 3 r

—z-vB—(—-) - (œ I œ3dF - I

dx3 dr\rir2J v Jf jF

d x d x2

œ4dF)

+ GkTx (15)

to

This equation (Eq. 15) is the nonlinearial differential equation of the non-free torsion of the open we get: profile thin-walled bar.

Let's express touching forces with the following new B(x) function of the momentum of the torsional forces in non-free torsion:

dB 77

— = Mh

dx h

In the process of comparing Eq. (6) and Eq. (12)

(16)

here B is called bending - torsional bimoment (bumper), or simply bimoment, its unit of measurement

is kN-sm2.

Mh =

dx »

(17)

While comparing Eq. (16) and Eq. (17) we get:

°x = — (18)

J u>

We can see from here that, the normal tensions in the non-free torsion are proportional to the bimoment, and while it is or = 0, B = 0 is obtained.

Placing Eq. (16) in Eq. (12) we integrate according to x and get the following:

B = -En

M4dF)

(19)

We differentiate both sides of Eq. (15) according to x and get:

E

i d4!- ß d2 (d29\3 » ■ dx4 VP ■ dx2 \dx2)

("5 "3dF-jp "4dF)

-GJk

d29 dx2

dMh d x

(20)

Here mb is the intensity of the external bending forces and we accept it as a positive quantity, because Mb decreases while the value of x increases.

First of all, let's look at the existence form of the two symmetry arrows of the bar cut (double-headed

form) (Fig. 4, a). Such bar with length of l is influenced by the squeezing P force in the direction of the centre axis x [7].

D-the centre of bending Figure 4. a) Double-headed cut; b) About computing the torque of the squeezing force.

Let's assume that all the longitudinal fibers except the central fibers are bending from the given force (to the direction of x arrow), i.e. the form of the loss of the tolerance in the torsion of the bar. When looking through the free edge of the bar at the x arrow we accept that the positive direction of the 9 rotation angle of any cut of the bar is turning counterclockwise [8].

Before deformation, accepting the fact that dF elemental pitch fits to any fiber in the cut of the bar parallel to x axis, after the torsion the bending radius of the very fiber will have the curve shape on the surface of the pcircular cylinder (Fig. 4, b). Let's mark the vertical fiber and angle of the touch to this curve with^. The o ■ dF elemental force that effects the fiber is spinning like $ angle, creating the momentum around

the x arrow, will also be expressed as aippdF, and the intensity of the full torque momentum will be expressed as follows:

mh = -fr <Jdc P dF, (21) or considering that pdd = tydx it will be like:

P2dF(22)

here Jp is the polar inertia momentum due to the centre of the cut. Writing Eq. (22) for Eq. (20), we get the following nonlinear differential equation [10,11]:

E

d4e 0 d2 (d2e\3 ( r

œ3dF - I

F

œ4dF)

+ (°Jr-GJk)dx2=0,

(23)

We solve this complex differential equation by using the small parameters method. For this purpose we put Eq. (23) in the following form:

d4e p d2 d2ey ( c 3 c 4 a-jP-Gjk d2e —7 v---t(—-) (w I w3dF - I w4dF)+--p---r=0

dx4 lu dx2\dx2) v JF JF ' e0]u dx2

We take the solution of the last equation in the following order for a small parameter:

e = 6o + ve1+..=Yz=0 vnen (n>0)

(23')

(a)

We write (a) in the same equation and obtain the following linear differential equation system (the first two equations of the system were shown):

d4e0 ab(oyJp-GJk d2e0 _ 0

dx4 E0]u dx2

d^ei + ab(oylp-Glk - d^e^ = £—d2 idle°) (œl œ3dF - | œ4dF)

dx4 E0]u dx2 lu dx2\dx2)y JF JF '

The following substitution was accepted in Eq. (24):

ab(o)'Jp-GJk _ ,2

kn

EoJu

(24)

(25)

(26)

We obtain its solution through the following way:

d2e,

dx2

0 = C1 sin k0x + C2 cos k0 x (27)

Since the boundary conditions are

C2 = 0, C2^0

we get s in k0l = 0; k0l =nn or k0 = —

Accepting n=1, we write k0 =n/l in Eq. (26) and find the initial cost of the crisis tension:

_ ^2eoju . GJk nm

°b(0) = — + !;, (28)

Similarly to the strongest fasteners of the sharpest ends of the bars, we can write Eq. (28) in the following

way:

ab(0)=ZIvTp + l-k, (28)

Here the length coefficient of the bar may be equal ton = 0,5. If one of the cutting edges of the bar is tightly fastened and the other one is rolling n = 0,7 is accepted.

Taking into account C2 = 0, Eq. (27) takes the following form:

2

= C1 sin k0 x (29)

Considering Eq. (29), the following complex differential in Eq. (25) is defined as:

72 (dx2) = Ci(—^k2sink0x + 9k0>sin3k0x) (30)

Subsequenty, placing Eq. (30) in Eq. (25) we get:

+ =■ (—3k2sink0x+ 9k2sin3k0x) (œ fp w3dF — Jp œ4dF) (31)

here

_ °b(iyJy-GJk EoJ(à

(32)

we accept the solution of the differential in Eq. (31) in the following way:

^r = D1 sin k1x + D2 cos k1x + C3k0,(a sin k0x + b sin 3 k0x) f w3dF — f w4dF) (33) by substituting Eq. (33) in Eq. (25), we get equations a and b:

a =---

b = 9-.

(34)

here

k1

ak = V k0

Let's assume that the cutting edges of the bar do not rotate in the flat shape. In this case, the boundary conditions of the equation will be as follows:

x = 0, x = I olduqda 9 = 0;

ae

x = 0, x = I olduqda — = 0

^ dx

(35)

We write Eq. (29) and Eq. (33) equations to their places in expression a and get :

d29 d29n d29i

+ v ——— = C sin knx +

^ur M3,

Ju

dx2 dx2 ' " dx2 ,x + b sin 3

+v[d1s ink1x + C3k0(asinkox + b sin3k0x)-^(w Jp œ3dF — Jp œ4dF)] (36)

We get the last equation by integrating it:

d9 d90 d91 Cr — = — + v— = — — cos k0 x d x d x d x k

— I

1 _ 7/a b \ß f f

— D1 cos k1x + C3k2 I—cos k0 x + ——cos 3k0x) — ((o\ œ3dF — I k1 \k0 3k0 ' Jw\ jf 'F

9 = 90 + v91 = — — sin k0x — v

k 2

œ4dF

Kx) ■

œ3dF — $f M4dF)]

(37)

Substituting Eq. (37) in the boundary conditions of Eq. (35), we get:

x = 0

C1

= 0; ——^ — v

= 0; —c os k0l + v x = k

d9

d x d9

d x

9\x=o = 0;

C1

9\x=i = 0; ——sink0l — v k 2

œ4dF

t+C13k0(a+b)T('°l m'dF~Ip

D1 i b \

— c os k11 + C?k0 la cos k0l +— cos 3k0l)

f(af œ3dF — f Jw \ Jf 'a

œ4dF

= 0;

D1 b

-j s in k1l + C3 (a sin k0 I + —sin3k0l )

k1 \ 9 J

= 0;

31

1

4 ar,-1

4 ar,-1

C

1

1

1

0

1

0

Eepa3uucKuu COK>3 YneHbix (ECY) #12 (69), 2019 œ3dF-}F œ4dF)] = 0 (38)

From the first of the conditions of Eq. (38) we get:

C13 =--R/ h\ ko kl--(39)

1 Vk^JL(a+ly(o LP o3dF-Ip o4dF) v '

Having written the last expression in the place of other conditions of Eq. (38), we obtain the following algebric equations for Ci and Di constants:

C / a - cosk0l +Îcos3k0l\ £ ( a - cos k0l + Îcos3k0l' — ( c os k0 I--;-) + v — ( cos k1 I--3-I = 0

U a+3^ ) k\ a + 3

. a-sinkol+—sin3kol\ n / a-sinkol+—cos3kol\

^[sink0l--^--j + v^isin^l--o-3g- ) = 0 (40)

Making the Eq. (40) system's determinant equal to zero for getting the smallest value of the ki, we obtain the following complex algebraic equations system:

( cos ko I

k k

a cos k0 I +^cos 3 k0l '

,b a + 3

1 I asink0l +bsin3knV

+ Vk\Sinkol---

a sin k0 I + bsin 3 k0l ' s in k11--^- I +

a + 3

a cos k0 I + bcos 3 k0l ' c os k11--^- I = 0

a + 3

Defining the minimum equation for the coefficient k1 through numerical methods from the last equation and writing it in Eq. (32) we determine the crisis tension - ab(1) in the first approach:

°b(1)

_ KEoJu + GJk h

(41)

Analogically, as described above, by keeping the first two boundaries of the expression (a) and having written in the differential Eq. (23') we get appropriate k2 = 2 n/l coefficient, and the crisis tension ab(2) according to the n = 2 condition of the small parameter, i.e. due tov2 - a. Thus, we determine the crisis tension in the second approximation of thin-walled bar:

Qi) i i 2

= ab(o) + vab(1) + v2ab(2)

(42)

Numerous calculations have shown that, the difference between the sum of the first two limits of Eq. (42) and (a(S> - the first approximation) the second approximation is 1,64%. Therefore we can be satisfied with that the equation can be solved by the solution in the second approach.

CONCLUSION

The problem of clamping resistance in the centre of the thin-walled open profile bars has been extensively studied. For the first time, the nonlinear elastic property of the material of the bars is taken into

account, in addition, the nonlinear differential equilibrium equation for the determination of crisis tension has been compiled. The smallest parameters method, which is most optimal for determining the crisis tension in the differential equation, has been used. As a result, the complex nonlinear differential equation is divided into several simple linear differential equations and their solution provides the satisfactory results specially in the second approximation.

References

[1]Vlasov V. Thin-walled elastic beams.2nd ed. Springfield, Va.: National Technical Information Service; 1984.

[2]Peres N, Goncalves R, Camotim D. First-order generalized beam theory for curved thin-walled members with circular axis. Thin-Walled Structures 2016; vol.107: 345-361. DOI: 10.1016/j.tws.2016.06.016

[3]Fouzi MSM. Jelani KM. Nazri NA. Sani MSM. Finite Element Modelling and Updating of Welded Thin-Walled Beam. International Journal of Automotive and Mechanical Engineering 2018; 15(4): 5874-5889. DOI: 10.1007/s40430-018-1475-z

[4]Bebiano R, Eisenberger M, Camotim D, Goncalves R. GBT-Based Buckling Analysis Using the Exact Element Method. International Journal of Structural Stability and Dynamics 2017; 17(10): 17501255. DOI: 10.1142/s0219455417501255

[5]Ronaldo I, Borja. Plasticity. Springer, Berlin; 2013.

[6]Rousselier G, Quilici S. Combining porous plasticity with Coulomb and Portevin-Le Chatelier models for ductile fracture analyses. International Journal of Plasticity 2015; vol.69: 118-133. DOI: 10.1016/j.ijplas.2015.02.008

[7]Walter L. Nonlinear Structural Mechanics: Theory, Dynamical Phenomena and Modeling. Springer US; 2013.

[8]Leipholz U. Theory of elasticity, Springer Netherlands; 2014.

[9]Pastor MM, Bonada J, Roure F, Casafont M. Residual stresses and initial imperfections in non-linear

analysis. Engineering Structures 2013; vol.46:493-507. DOI: 10.1016/j.engstruct.2012.08.013

[10]Sadigov IR Phisical nonlinear elastic deformations of smooth ring. Transactions of NAS of Azerbaijan 2016; vol. 36:74-80.

[11]Zhang RJ, Wang C, Zhang Q. Response analysis of the composite random vibration of a highspeed elevator considering the nonlinearity of guide shoe. Journal of the Brazilian Society of Mechanical Sciences and Engineering 2018; vol.40: 4.