RU2623558C2 - Method of circular pipe reshaping - Google Patents

Abstract

FIELD: construction.

SUBSTANCE: change of cross-section shape of circular plane-oval pipe with size ratio of 1/3.064.

EFFECT: increased pipe strength in bending.

7 dwg

Description

The proposed technical solution relates to the field of metal forming, as well as to the field of construction and can be used in bridge and crane structures, in bearing coating systems (floors) of various buildings and structures.

A rail and beam structure was developed containing a crane beam made of a tubular elliptical in cross section oriented vertically by a large axis [Nezhdanov KK, Tumanov VA, Nezhdanov AK, Karev MA Rail and beam construction. - Patent No. 2192381, 11/10/2002, bull. No. 31]. The cross-section of the rail and beam structure is balanced relative to the main horizontal axis. However, the tubular profile of the elliptical section is not optimized according to the criterion of the maximum bending capacity.

Known technical solution, framed as a method of manufacturing an oval welded straight pipe, which is carried out as follows. After molding and welding, the round longitudinal pipe is oriented along the perimeter of the section at a certain angle relative to the minor axis of the oval (approximately 39 ° 32 '). After the orientation of the weld, the round pipe enters the two-roll gauge, where it is ovalized with a small axis of about 440 mm and a large axis - 650 mm [V. Zarankin A method of manufacturing an oval welded straight pipe. - Copyright certificate No. 1466827, 03/23/1989, bull. No. 11]. This method provides the location of welds in the areas of minimum bending stresses of oval pipes, the cross-sectional shape of which, however, is different from profiles with maximum load-bearing capacity.

Another technical solution (adopted as an analogue) is known in the form of a method of increasing the bending capacity of a cylindrical pipe, which consists in compressing it when heated to 600 ... 650 ° C with the formation of an oval profile optimized by the criterion of maximum load capacity, which is the largest ( maximum) moment of resistance [Nezhdanov K.K., Tumanov V.A., Nezhdanov A.K., Rublev S.G. A method of increasing the bending capacity of a cylindrical pipe. - Patent No. 2304479, 08.20.2007, bull. No. 23]. However, the production of a hot-rolled oval profile in this way has not yet been mastered by industry.

The closest technical solution (adopted as a prototype) to the proposed one is a method for producing an oval profile bending pipe with a circular profile optimized by the criterion of maximum resistance moment and flattened by the distribution of the pipe in the cold state by means of a jack system from the inside between the two matrices [K. Nezhdanov K., Nezhdanov A.K., Zhukov A.N. A method of obtaining an oval profile pipe bending from a cylindrical pipe of a round profile. - Patent No. 2460603, 04/20/2011, bull. No. 11]. The essence of the distinguishing features of the prototype and analogue is that, if the tubular profiles have oval sections with an aspect ratio of 1 / 2,99999≈1 / 3, then the moments of resistance of these sections are maximum, and profiles with such sections have the greatest bearing capacity resources on bend.

The main disadvantage of the prototype, as well as other technical solutions, lies in the complex shape of the oval pipe, the surface of which is characterized by the complete absence of flat sections and the same absence of sections of constant curvature. Such a characteristic increases the complexity of both manufacturing the pipe itself and its use in bar or beam structures, which is accompanied by a certain increase in costs.

So, for example, oval pipes are 3.5 ... 4.4% more expensive than flat oval profiles:

- according to the price list dated April 1, 2008 Moscow Tube Works OJSC (6, Barclay St., Moscow) a ton of oval pipe cost 37,090 rubles, and a flat oval pipe - 35,510 rubles (37090/35510 = 1,044);

- according to the price list dated 11/27/2013 KlassiK PKF LLC (21, Marata St., Kiev) a ton of oval pipe cost 8,800 hryvnias, and a flat-oval one - 8500 hryvnias (8800/8500 = 1,035).

The technical result of the proposed solution is to increase the load-bearing capacity of the tubular profile due to the increase in the strength characteristics of the bend, reduce the difficulties of its manufacture and use in rod or beam structures, as well as reduce costs for them.

The specified technical result is achieved by the fact that in the method of reprofiling a round pipe, including changing the cross section of the pipe, receive a pipe with a cross section of a flat oval shape with a ratio of overall dimensions of 1 / 3,064

The proposed technical solution consists in ovalizing tubular profiles of a round shape by means of technological operations from an analogue or prototype with replacing an oval sectional shape that differs by a 1/3 aspect ratio by a flat oval shape with a ratio of 1 / 3.064.

The proposed technical solution is illustrated by graphic materials, where in FIG. 1 shows a design diagram of the cross section of a flat oval pipe; in FIG. 2 is a design diagram of one of two curved sections of a cross section of a flat oval pipe having the shape of round half rings; in FIG. 3 shows a general view of flat oval profiles; in FIG. 4 shows a design diagram of a cross section of a round pipe; in FIG. 5 is a design diagram of a cross section of a pipe completely flattened in a vertical plane; in FIG. 6 is a design diagram of a cross section of a pipe completely flattened in a horizontal plane; in FIG. 7 is a cross-sectional view of a flat oval pipe crimped by rolls on four sides.

To derive the given ratio and quantify the resources of the bearing capacity for bending, it is advisable to calculate the axial moment of inertia I _X and the moment of resistance W _X , as well as the cross-sectional area A of the flat oval tubular profile. The section of this profile can be considered as a figure that includes two rectangles and two half rings. In turn, a sector of a thin-walled ring with an angular parameter α = 90 ° = π / 2 = 1.57 can be taken as a composite half-ring [Material Resistance Handbook / Ed. G.S. Pisarenko. - Kiev: Naukova Dumka, 1988. - S. 68-69]:

y _c = R sinα / α = R × 1 / 1.57 = 0.6369426 R = 0.6369426 × 0.5 U = 0.3185 U ;

I _X = (2α + sin2α-4 (sinα) ² / α) R ³ t / 2 = (3.14 + 0-4 / 1.57) R ³ t / 2 = 0.2961146 (0.5 U ) ³ t = 0.0370143 U ³ t ;

I _Y = ((2α-sin2α) R ³ t / 2 = (3.14-0) R ³ t / 2 = 1.57 (0.5 U ) ³ t = 0.196250 U ³ t ;

A = 2α Rt = 2 × 1.57 × 0.5 U = 1.57 Ut ,

where y _c , I _X , I _Y , A are respectively the ordinate of the center of gravity of the section, the moment of inertia of the section relative to the xx axis, the moment of inertia of the section relative to the yy axis, the cross-sectional area of the half ring;

R is the radius of the semicircle along its midline, R = 0.5 U ;

U is the smaller dimension of the cross section of the flat oval pipe in the midline;

V is the larger dimension of the cross section of a flat oval pipe in the midline;

n is the ratio of the smaller to the larger, n = U / V.

t is the wall thickness of the flat oval pipe.

To calculate the axial moment of inertia of a section of a flat oval pipe, you can apply the rule of parallel transfer of axes:

I _X = 2 (0.0370143 U ³ t +1.57 Ut ( U (0.5 / n -0.5) +0.3185 U ) ² +8 U ³ (0.5 / n -0.5 ) ³ t / 12) = U ³ t (0,0740286 + 0,785 (1 / n -0,363) ² + (1 / n -1) ^3/6),

where the value of the larger dimension V is replaced by its ratio with the smaller dimension U , that is, V = U / n .

The moment of resistance along the midline relative to the x-x axis:

2 W _X = I _X / V = 2 nI _X / U = 2 nU ³ t (0,0740286 + 0,785 (1 / n -0,363) ² + (1 / n -1) ^3/6) / U = U ² t (0.33333333 / n ² + 0.57 / n +0.0216012 n -0.13982).

The cross-sectional area of a thin-walled flat oval pipe can be calculated along the length of the midline:

A = t (π U +2 ( U / n - U ) = t (3.14 U +2 U (1 / n -1) = Ut (2 / n +1.14).

The obtained calculation formulas must be tested, since the middle line of the cross section of the flat oval pipe was used to derive them. In addition, the moment of resistance of the cross section relative to the x-x axis is calculated at the level of the same midline. It is advisable to perform a test calculation on the basis of the 12 largest large-caliber and thick-walled flat-oval type A pipes according to GOST R 54157-2010, and its main results are more clearly presented in tabular form (see table). As can be seen, the obtained calculation formulas are correct enough to continue numerical calculations with their application.

If, to continue the computational calculations, we take as the initial data the cross-sectional area A = const with the thickness t = const of the strip (sheet blank), then in the formula for the moment of resistance W _X it is advisable to leave only the last of the two variables U and n :

U = ( A / t ) (1 / (2 / n +1.14));

U ² = (( A / t ) (1 / (2 / n +1.14)) ² ; = ( A ² / t ² ) (1 / (4 / n ² + 4.56 / n +1, 2996));

W _X = U ² t (0.33333333 / n ² + 0.57 / n +0.0216012 n -0.13982) = ( A ² / t ² ) (1 / (4 / n ² + 4.56 / n +1.2996)) t (0.33333333 / n ² + 0.57 / n +0.0216012 n -0.13982) = ( A ² / t ) (0.33333333 / n ² + 0.57 / n +0.0216012 n -0.13982) / (4 / n ² + 4.56 / n +1.2996).

To find the extreme value of the moment of resistance W _X, its formula must be differentiated with respect to the variable n :

dW _X / dn = d (( A ² / t ) (0.33333333 / n ² + 0.57 / n +0.0216012 n -0.13982) / (4 / n ² + 4.56 / n +1 , 2996)) / ( dn ) == ( A ² / t ) (0.7599988 / n ⁴ -1.98496 / n ³ -1.1191368 / n ² + 0.1970028 / n +0.0280729) / ( 4 / n ² + 4,56 / n +1,2996) ² .

Equating the derivative to zero ( dW _X / dn = 0), we can obtain an equation of the fourth degree

0.7599988 / n ⁴ -1.98496 / n ³ -1.1191368 / n ² + 0.1970028 / n + 0.0280729 = 0

or

0.0280729 n ⁴ +0.1970028 n ³ -1.1191368 n ² -1.98496 n + 0.7599988 = 0

with roots

n ₁ = -10.216734; n ₂ = -1.7543858; n ₃ = 0.32641796; n ₄ = 4.6271586.

Of these roots, the third is of practical interest, the value of which can be rounded to n ₃ = 0.32642 = 1 / 3.064, thereby obtaining the above ratio.

With n = U / V = 0.32642 = 1 / 3.064, the tubular profile with a flat oval cross-section has the following parameters:

U = ( A / t ) (1 / (2 / n +1.14)) = ( A / t ) (1 / (2 / 0.32642 + 1.14)) = 0.1376069 A / t ;

V = ( A / t ) (1 / (2 + 1.14 n )) = ( A / t ) (1 / (2 + 1.14 × 0.32642)) = 0.4215640 A / t ;

n = 0.1376069 / 0.4215640 = 0.3264199≈0.32642 = 1 / 3.064;

W _X = ( A ² / t ) (0.33333333 / n ² + 0.57 / n +0.0216012 n -0.13982) / (4 / n ² + 4.56 / n +1.2996) = ( A ² / t ) (0.33333333 / 0.32642 ² + 0.57 / 0.32642 + 0.0216012 × 0.32642-0.13982) / (4 / 0.32642 ² + 4.56 / 0 32642 + 1.2996) = 0.0897905 A ² / t ;

I _X = U ³ t (0,0740286 + 0,785 (1 / n -0,363) ² + (1 / n -1) ^3/6) = (0,1376069 A / t) ³ t (0,0740286 + 0,785 ( 1 / 0,32642-0,363) ² + (1 / 0,32642-1) ^3/6) = 0,0189529 A ³ / t ^2;

I _X = W _X V / 2 = (0.0897905 A ² / t ) (0.4215640 A / t ) / 2 = 0.0189262 A ³ / t ² ,

where two values of the same moment of inertia I _X make it possible to compare them as a kind of control check, the error of which is 100 (0.0189529-0.0189262) / (0.0189529 ... 0.0189262) = 0.1409 ... 0 , 1411%.

It is interesting to optimize the cross-sectional oval pipe section (proposed technical solution) in this way with a similar oval pipe section (prototype). For the correct comparison, it is advisable to repeat the numerical calculations in relation to the oval section, taking the section area A = const with the thickness t = const and using the calculation formulas of its geometric characteristics [Reference book of the designer of industrial, residential and public buildings and structures. Settlement-theoretical. - In 2 kn. Prince 1 / Ed. A.A. Umansky. - M .: Stroyizdat, 1972. - S. 364]:

A = (π / 2) / ( U + V );

I _X = (π / 32) V ² t (3 U + V );

W _X = (π / 16) Vt (3 U + V ),

where from

U = 2 A / (πt (1 + 1 / n ));

V = 2 A / (π t (1+ n )),

where U is the smaller dimension of the cross section of the oval pipe in the midline;

V is the larger dimension of the cross section of the oval pipe in the midline;

n is the ratio of the smaller to the larger, n = U / V.

The moment of resistance along the midline relative to the x-x axis:

W _X = (π / 16) Vt (3 U + V ) = (π / 16) 2 A / (π t (1+ n )) t (3 × 2 A / (π t (1 + 1 / n ) ) +2 A / (π t (1+ n ))) = ( A ² / (4π t )) (3 n + 1 / n +4) / ( n ² +3 n + 1 / n +3).

To find the extreme value of the moment of resistance W _X, its formula must be differentiated with respect to the variable n :

dW _X / dn = d (( A ² / (4π t )) (3 n + 1 / n +4) / ( n ² +3 n + 1 / n +3)) / ( dn ) = ( A ² / (4π t )) (- 3 n ² -8 n + 1 / n ² -6) / ( n ² +3 n + 1 / n ² +3) ² .

Equating the derivative to zero ( dW _X / dn = 0), we can obtain an equation of the fourth degree

-3 n ² -8 n + 1 / n ² -6 = 0

or

3 n ⁴ +8 n ³ +6 n ² -1 = 0

with roots

n ₁ = -1-0.000001561 (-1) ^1/2 ;

n ₂ = -1 + 0.000001561 (-1) ^1/2 ;

n ₃ = -1;

n ₄ = 0.333333333333333333.

Of these roots, the fourth is of practical interest, the value of which can be rounded to n ₄ = 0.33333 = 1/3, thereby obtaining the ratio given in the prototype.

With n = U / V = 0.33333 = 1/3, the tubular profile with an oval section has the following parameters:

U = 2 A / (π t (1 + 1 / n )) = 2 A / (3.14 t (1 + 1 / 0.33333)) = 0.1592356 A / t ;

V = 2 A / (π t (1+ n )) = 2 A / (3.14 t (1 + 0.33333)) = 0.4777081 A / t ;

n = U / V = 0.1592356 / 0.4777081 = 0.3333324≈0.33333 = 1/3;

W _X = ( A ² / (4π t )) (3 n + 1 / n +4) / ( n ² +3 n + 1 / n +3) = ( A ² / (4 × 3,14 t )) (3 × 0.33333 + 1 / 0.33333 + 4) / (0.33333 ² + 3 × 0.33333 + 1 / 0.33333 + 3) = 0.089570 A ² / t ;

W _X = (π / 16) Vt (3 U + V ) = (3.14 / 16) (0.4777081 ( A / t )) t (3 × 0.1592356 A / t +0.477708 A / t ) = 0.0895703 A ² / t ;

I _X = (π / 32) V ² t (3 U + V ) = (3.14 / 32) (0.4777081 A / t ) ² t (3 × 0.1592356 A / t +0.4777081 A / t ) = 0.0213942 A ³ / t ² ;

I _X = W _X V / 2 = (0.0895703 A ² / t ) (0.4777081 A / t ) / 2 = 0.0213942 A ³ / t ² ,

where the two values of the same moment of inertia I _X coincided, and the two values of the same moment of resistance allow you to compare them as a kind of control test, the error of which is 100 (0.0895703-0.089570) / (0.0895703 ... 0.089570) = 0.00033%.

A comparison of the results allows us to conclude that the calculated parameters of the prototype in absolute value exceed the calculated parameters of the proposed solution with the exception of the strength characteristics W _X :

Δ U = 0.1592356 / 0.1376069 = 1.1571774;

Δ V = 0.4777081 / 0.421564 = 1.1331804;

Δ I _X = 0.0213942 / 0.0189262 = 1.1304012;

Δ W _X = 0.0895703 / 0.0897905 = 0.9975476,

that is, under given conditions, when A = const and t = const, and W _X = W _max , cross-sectional pipes are 13.3 ... 15.7% more compact for oval pipes than oval pipes, and axial moments of inertia for oval pipes the cross sections are 13.04% larger, and the moments of resistance are 0.245% less than those of flat oval.

The findings can be illustrated by the example of tubular profiles with a circular cross section, the geometric characteristics of which are [Handbook of resistance of materials / Res. ed. Pisarenko G.S. - Kiev: Naukova Dumka, 1988. - S. 60-61]:

d = U = V , i.e. n = 1;

A = π dt = 3.14 dt ;

I _X = I _Y = π d ³ t / 8 = 0.3925 d ³ t ;

W _X = W _Y = π d ² t / 4 = 0.7850 d ² t ,

where d is the diameter of the section along its midline.

If a prototype is used to strengthen the tubular profiles, the geometric characteristics of their sections will change as follows:

U = 0.1592356 A / t = 0.1592356 × 3.14 dt / t = 0.4999997 d ≈0.5 d ;

V = 0.4777081 A / t = 0.4777081 × 3.14 dt / t = 1.5000034 d ≈1.5 d ;

I _X = 0.0213942 A ³ / t ² = 0.0213942 (3.14 dt ) ³ / t ² = 0.6623461 d ³ t ;

W _X = 0.089570 A ² / t = 0.089570 (3.14 dt ) ² / t = 0.8831273 d ² t ;

Δ U = 0.5 / 1 = 0.5;

Δ V = 1.5 / 1 = 1.5;

Δ I _X = 0.6623461 / 0.3925 = 1.6875059;

Δ W _X = 0.8831273 / 0.7850 = 1.1250029,

that is, the width dimension decreased by 2 times, and increased 1.5 times in height, while the stiffness characteristic I _X increased by 68.8%, and the strength W _X - 12.5%.

If we use the proposed solution to strengthen the tubular profiles, then the geometric characteristics of their sections will change somewhat differently than in the case of the prototype:

U = 0.1376069 A / t = 0.1376069 × 3.14 dt / t = 0.4320856 d ;

V = 0.4215640 A / t = 0.4215640 × 3.14 dt / t = 1.33237109 d ;

I _X = 0.0189529 A ³ / t ² = 0.0189529 (3.14 dt ) ³ / t ² = 0.5867652 d ³ t ;

W _X = 0.0897905 A ² / t = 0.0897905 (3.14 dt ) ² / t = 0.8852981 d ² t ;

Δ U = 0.4320856 / 1 = 0.4320856;

Δ V = 1.3237109 / 1 = 1.3237109;

Δ I _X = 0.5867652 / 0.3925 = 1.4949431;

Δ W _X = 0.8852981 / 0.7850 = 1.1277682,

that is, the width dimension decreased 2.31 times, and height increased 1.32 times, while the stiffness characteristic I _X increased by 49.5%, and the strength W _X - 12.8%.

Such results once again confirmed that under the given conditions, when A = const and t = const, and W _X = W _max , the cross-sectional pipes have cross sections of 13.6 ... 15.5% (1.5 / 1.32 ... 2.31 / 2 = 1.1363636 ... 1.1550) is more compact than that of oval pipes, while in oval pipes the axial moments of inertia of the cross sections are 12.88% (1.6875059 / 1.4949431 = 1.1288094) more, and resistance moments are 0.246% (1.1277682 / 1.1250029 = 1.002458) less than for flat oval.

All the above calculations of flat oval, oval and round sections of tubular profiles were performed under the assumption that they have the same initial sheet blank (strip) with a constant cross-sectional area ( A = const) and constant thickness ( t = const). This premise limits the change in dimensions to frames, which are determined by the complete flattening of the tubular profiles:

- in general

U _min = t ≤ U at V ≤ V _max = 0.5 A / t or V _min = t ≤ V at U ≤ U _max = 0.5 A / t ;

- in case of round section

U _min = t ≤ U at V ≤ V _max = 1,57 d or V _min = t ≤ V at U ≤ U _max = 1,57 d .

In the analogue (patent No. 2304479) and prototype (patent No. 2460603) after the conversion of round pipes into oval profiles, the overall dimensions exceeded the indicated frames:

1. round pipe 1020 × 8 mm ( A = 254 cm ² ; t = 0.8 cm)

d = 254 / (0.8 × 3.14) = 101.1464 cm;

V _max = 1.57 d = 1.57 × 101.11464 = 158.74998 cm < V = 178.594 cm;

Δ I _X = 2.08>1.6875059;

Δ W _X = 1.213>1.1250029;

2. round pipe 1220 × 9 mm ( A = 342 cm ² ; t = 0.9 cm)

d = 342 / (0.9 × 3.14) = 121.0191 cm;

V _max = 1.57 d = 1.57 × 121.0191 = 189.99998 cm < V = 213.75 cm;

Δ I _X = 2.087>1.6875059;

Δ W _X = 1.209>1.1250029;

3. round pipe 1320 × 11 mm ( A = 452.3579 cm ² ; t = 1.1 cm)

d = 452.3579 / (1.1 × 3.14) = 130.96638 cm;

V _max = 1.57 d = 1.57 × 130.96638 = 205.61721 cm < V = 231.136 cm;

Δ I _X = 2.125>1.6875059;

Δ W _X = 1.197>1.1250029;

4. round pipe 1420 × 10 mm ( A = 443 cm ² ; t = 1.0 cm)

d = 443 / (1.0 × 3.14) = 141.0828 cm;

V _max = 1.57 d = 1.57 × 141.0828 = 221.49999 cm < V = 249.184 cm;

Δ I _X = 2.125>1.6875059;

Δ W _X = 1.197> 1.1250029.

As can be seen, such an excess of dimensions is accompanied by an increased increase in the axial moments of inertia and moments of resistance of the cross section, which can be explained by a certain elongation of the midlines of the round profiles during their ovalization.

If it is permissible to calculate the length of the midline before ovulation using the formula of round section

L = 3.14 d ,

then, after reprofiling, the oval section formula is applicable

L _eff = 1.57 ( U + V ),

then, it is obvious that with a constant cross-sectional area ( A = const), the marked elongation is accompanied by a corresponding narrowing, which in this case represents a decrease in thickness ( t ≠ const), that is, thinning the profile wall to a certain value that is acceptable as a calculated premise:

t _eff = t (1-μ),

where μ is the Poisson's ratio, for steel μ = 0.24 ... 0.30 [Guide to the resistance of materials / Res. ed. Pisarenko G.S. - Kiev: Naukova Dumka, 1988. - S. 165].

Here, of particular practical interest is the calculation of tubular profiles, taking into account their elongated middle lines and the corresponding thinning of the walls:

d _eff = L _eff / 3.14,

where d _eff is the diameter of the conditional section along its elongated midline;

- for oval sections

U = 0.5 d _eff ;

V = 1.5 d _eff ;

I _X = 0.6623461 d _eff ³ t _eff ;

W _X = 0.8831273 d _eff ² t _eff ;

- for flat oval sections

U = 0.4320856 d _eff ;

V = 1.32371096 d _eff ;

I _X = 0.5867652 d _eff ³ t _eff ;

W _X = 0.8852981 d _eff ² t _eff .

Calculated calculations according to the given formulas and comparison of their results are more clearly presented in expanded form.

1. Round pipe 1020 × 8 mm ( A = 254 cm ² ; t = 0.8 cm):

L = A / t = 254 / 0.8 = 317.5 cm;

L _eff = 1.57 ( U + V ) = 1.57 (59.531125 + 178.594) = 373.85643 cm;

t _eff = t (1-μ) = 0.8 (1-0.24) = 0.608 cm;

d _eff = L _eff / 3.14 = 373.85643 / 3.14 = 119.06255 cm;

- for oval section

U = 0.5 d _eff = 0.5 × 119.06255 = 59.531275 cm;

V = 1.5 d _eff = 1.5 × 119.06255 = 178.59382 cm;

I _X = 0.6623461 d _eff ³ t _eff = 0.6623461 × 119.06255 ³ × 0.606 = 679694.99 cm ⁴ (97.9% of the prototype);

W _X = 0.8831273 d _eff ² t _eff = 0.8831273 × 119.06255 ² × 0.608 = 7611.6221 cm ³ (98.3% of the prototype);

- for flat oval section

U = 0.4320856 d _eff = 0.4320856 × 119.06255 = 51.445213 cm;

V = 1.3237109 d _eff = 1.3323109 × 119.06255 = 157.60439 cm;

I _X = 0.5867652 d _eff ³ t _eff = 0.5867652 × 119.06255 ³ × 0.608 = 602134.39 cm ⁴ ;

W _X = 0.8852981 d _eff ² t _eff = 0.8852981 × 119.06255 ² × 0.608 = 7630.3321 cm ³ .

2. Round pipe 1220 × 9 mm ( A = 342 cm ² ; t = 0.9 cm):

L = A / t = 342 / 0.9 = 380.0 cm;

L _eff = 1.57 ( U + V ) = 1.57 (71.25 + 213.75) = 447.45 cm;

t _eff = t (1-μ) = 0.9 (1-0.24) = 0.684 cm;

d _eff = L _eff / 3.14 = 447.45 / 3.14 = 142.50 cm;

- for oval section

U = 0.5 d _eff = 0.5 × 142.5 = 71.25 cm;

V = 1.5 d _eff = 1.5 × 142.5 = 213.75 cm;

I _X = 0.6623461 d _eff ³ t _eff = 0.6623461 × 142.5 ³ × 0.684 = 1310948.5 cm ⁴ (97.9% of the prototype);

W _X = 0.8831273 d _eff ² t _eff = 0.8831273 × 142.5 ² × 0.684 = 12266.174 cm ³ (98.3% of the prototype);

- for flat oval section

U = 0.4320856 d _eff = 0.4320856 × 142.5 = 61.572193 cm;

V = 1.3237109 d _eff = 1.3237109 × 142.5 = 188.62879 cm;

I _X = 0.5867652 d _eff ³ t _eff = 0.5867652 × 142.5 ³ × 0.684 = 1161355.0 cm ⁴ ;

W _X = 0.8852981 d _eff ² t _eff = 0.8852981 × 142.5 ² × 0.684 = 12296.325 cm ³ .

3. Round pipe 1320 × 11 mm ( A = 452.3579 cm ² ; t = 1.1 cm):

L = A / t = 452.3579 / 1.1 = 411.23445 cm;

L _eff = 1.57 ( U + V ) = 1.57 (77.045453 + 231.136) = 483.84487 cm;

t _eff = t (1-μ) = 1.1 (1-0.24) = 0.836 cm;

d _eff = L _eff / 3.14 = 483.84487 / 3.14 = 154.09072 cm;

- for oval section

U = 0.5 d _eff = 0.5 × 154.09072 = 77.04536 cm;

V = 1.5 d _eff = 1.5 × 154.09072 = 231.13608 cm;

I _X = 0.6623461 d _eff ³ t _eff = 0.6623461 × 154.09072 ³ × 0.836 = 2025912.4 cm ⁴ (95.9% of the prototype);

W _X = 0.8831273 d _eff ² t _eff = 0.8831273 × 154.09072 ² × 0.836 = 17530.024 cm ³ (97.3% of the prototype);

- for flat oval section

U = 0.4320856 d _eff = 0.4320856 × 154.09072 = 61.572193 cm;

V = 1,3237109 d _eff = 1,3237109 × 154,09072 = 188,62879 cm;

I _X = 0.5867652 d _eff ³ t _eff = 0.5867652 × 154.09072 ³ × 0.836 = 1794733.7 cm ⁴ ;

W _X = 0.8852981 d _eff ² t _eff = 0.8852981 × 154.09072 ² × 0.836 = 17552.094 cm ³ .

4. Round pipe 1420 × 10 mm ( A = 443 cm ² ; t = 1.0 cm):

L = A / t = 443 / 1.0 = 443.0 cm;

L _eff = 1.57 ( U + V ) = 1.57 (83.06 + 249.184) = 521.62308 cm;

t _eff = t (1-μ) = 1.0 (1-0.24) = 0.760 cm;

d _eff = L _eff / 3.14 = 521.62308 / 3.14 = 166.122 cm;

- for oval section

U = 0.5 d _eff = 0.5 × 166.122 = 83.061 cm;

V = 1.5 d _eff = 1.5 × 166.122 = 249.183 cm;

I _X = 0.6623461 d _eff ³ t _eff = 0.6623461 × 166.122 ³ × 0.76 = 2307703.4 cm ⁴ (97.9% of the prototype);

W _X = 0.8831273 d _eff ² t _eff = 0.8831273 × 166.122 ² × 0.76 = 18522.14 cm ³ (98.3% of the prototype);

- for flat oval section

U = 0.4320856 d _eff = 0.4320856 × 166.122 = 71.778924 cm;

V = 1.33237109 d _eff = 1.3323109 × 166.122 = 249.183 cm;

I _X = 0.5867652 d _eff ³ t _eff = 0.5867652 × 166.122 ³ × 0.76 = 2044369.4 cm ⁴ ;

W _X = 0.8852981 d _eff ² t _eff = 0.8852981 × 166.122 ² × 0.76 = 18567.669 cm ³ .

As can be seen, the calculation of tubular profiles, taking into account the elongation of the midline and minimal thinning (Poisson's ratio μ = 0.24) during ovalization of their cross sections, is quite correct, since with an acceptable error not exceeding a 4.1 percent threshold, it coincided with the calculated calculations, given in the analogue and prototype. Moreover, the increased growth of axial moments of inertia and moments of resistance of the cross-section takes place not only in the analogue and prototype (oval profiles), but also in the proposed technical solution (plane-oval profiles). The reprofiling of round pipes into oval and plane-oval profiles with a similar increase in the geometric characteristics of the cross sections leads, however, to inelastic (residual) deformations, the magnitude of which negatively affects the resources of strength, stiffness and stability, which is extremely undesirable for such supporting structures as beams and beam system. Therefore, reprofiling of round pipes seems to be more preferable, leaving the length of the middle lines of oval and plane oval sections unchanged, that is, L = const at A = const and t = const. Under such conditions, there is also a significant increase in geometric characteristics, and inelastic (residual) deformations are local (local) in nature and practically do not affect the bearing capacity reserves.

Thus, the obtained results of computational calculations and their comparison allow us to conclude that, to increase the strength characteristics, the proposed solution is more efficient than the prototype, since the increase in the moment of resistance of the cross section, ceteris paribus, is higher for the oval profile than for the oval profile, and in terms of overall dimensions, the flat oval section shape is obtained more compact. Moreover, as far as the oval section in large size exceeds the plane-oval, the growth of the stiffness characteristic, that is, the axial moment of inertia of the section, is also approximately higher. If the large size is located vertically, then it can have very substantial practical value, which determines the dimensions of the building heights of the beam structures, which makes the choice of a flat oval cross-sectional shape for tubular profiles more preferable and justified. If we add to this that this form is simpler, the complexity of manufacturing and use in structures is less, and the price is lower, then the positive effect of the proposed technical solution may be more visual and significant.

Claims (1)

A method of reprofiling a round pipe, including technological operations for changing the cross section of the pipe, characterized in that the cross section of the pipe is given a flat oval shape with a ratio of overall dimensions of 1 / 3,064.

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