RU2487222C2 - Method to increase durability of steel truss - Google Patents

Abstract

FIELD: construction.

SUBSTANCE: truss comprises ascending compressed support and intermediate braces, stands and descending stretched braces. Upper and lower belts of the truss are made of equally stable double-tee sections. In the most stressed zone in the middle of the span the upper belt is converted into a tubular one by means of welding of closing sheets at sides to the double-tee section. All compressed elements of the truss lattice are made of oval pipes with the ratio of larger dimension to smaller one equal to three. Oval pipes are aligned with a larger dimension perpendicularly to the truss plane. Compressed braces are detached by two struts in the truss plane. After installation of the truss into the design position, hoses of concrete pipelines are connected to nozzles of closed tubular elements of the truss. Fine-grained expanding concrete is injected into cavities of elements with formation of guncrete elements after concrete setting.

EFFECT: improved durability and reliability of a truss.

5 dwg

Description

The proposed invention relates to the survivability of steel farms of civil and industrial structures. The need to improve the survivability of farms stems from the current “negative” situation in Russia.

A significant number of structures are operated in a state close to the limit or even in emergency. In Russia, buildings are unsuitable for operation with a high probability of collapse. Collapse more often occurs in the metallurgical complex of structures, and over the past two decades, the rate of collapse has been increasing. This trend confirms the urgent need to solve the problem of increasing the survivability and reliability of structures in extreme operating conditions.

Steel farms are more prone to accidents by violating facility operating rules and increasing the risk of terrorist attacks.

Often, during the design and construction, even spectacular and other structures, where a large number of people accumulate, preference is given to the decisions of an architect with low and even “zero” survivability, rather than an engineer! Such decisions lead to tragic consequences with the death of people, for example, the well-known cases of the collapse of the “Aquapark” and “Covered Market” in Moscow [1, p.26, fig. 28 ... 31], [1, p. 75, fig. 108 ... 111].

Another example of the passion for architectural expressiveness to the detriment of vitality is the project of the “Indoor Skating Center” in Moscow [2]. In the spring of 2007, an emergency occurred at this center, which could lead to the collapse of the entire structure. After the restoration of the structure, the survivability remained “zero”. Obviously, structures with low survivability can lead to death and can not be recommended for spectacular and sports facilities.

Consequently, the relevance of increasing the survivability of structures is undeniable and continues to grow. Undoubtedly, the survivability and reliability of steel coating trusses should be enhanced.

Calculation of steel structures is carried out according to the first and second ultimate state [3, p. 60].

The first group of limiting states of structures is achieved with the following:

- loss of bearing capacity and (or) the complete unsuitability of the structure for use due to: loss of stability by structural elements;

- transformation of the structure into a geometrically variable system of elements (mechanism), which in turn leads to a qualitative change in the configuration of the structure [4, p. 37];

- brittle, sudden failure as a result of the occurrence and development of fatigue cracks during cyclic influences;

- with an excessive increase in plastic deformations, which ultimately leads to the destruction of the material of the structure and structure.

The second group of limit states is characterized by difficulties in the normal operation of the structure or a decrease in durability due to the occurrence of unacceptable displacements (deflections, subsidence of supports, rotation angles, vibrations, cracks, etc.).

Obviously, the first group of limit states is more dangerous than the second, since the destruction occurs suddenly, brittle without visible displacements and deformations.

As an analogue, we take a “typical” truss with a triangular lattice with ascending compressed and descending stretched braces [3, p. 370 ... 378, fig. 13.9], [5, p. 13 ... 26, fig. 2.1]. Support reactions from the farm are transmitted in the zone of its lower extended belt. Each truss is connected to the columns by flange connections. Mounting bolts transmit the calculated interaction forces of the connected elements. We take these solutions as analogues.

If the farm is statically determinable, then its survivability is the lowest - "zero"! That is, with the loss of stability of one of the compressed elements, the farm turns into a mechanism, and it suddenly collapses. The rigid connection of the truss with the upper parts of the columns increases the survivability of the entire system, since the frame system of the structure has become statically indefinable. The limiting state of the farm in this case occurs only when the second weak element of the system is turned off.

The disadvantages of the analogue are as follows: the survivability of trellised trusses of the coating [4, p. 37] is low. The exhaustion of the bearing capacity of trellised trusses occurs as a result of the loss of stability of only one of the compressed elements of a statically determinate truss, which turns it into a mechanism and the truss collapses.

Compressed rather than stretched rods always lose their bearing capacity, since the design resistance of steel BCt3sp5 (C255), assigned by the yield strength, is R _y = 240 MPa [8, p. 64], and assigned by the temporary resistance R _u = 360 MPa . Therefore, the extended rods have more than one and a half margin in bearing capacity with respect to the compressed elements!

This is also confirmed by the studies of Beleni E.I. [4], Belyaeva B.I., Kornienko B.C. [6, p. 98] - the cause of construction accidents in 44% of cases is the loss of stability of one of the compressed elements of the farm [6, p. 17]. Therefore, it is necessary to increase the survivability and reliability of farms of the systems of coatings for buildings and structures by increasing the bearing capacity (stability) of the compressed elements of the farm.

An example of low survivability of steel trusses from the corners is an avalanche-like collapse of the coating of the foundry and reinforcement workshop in Penza [7, p.27, 13, p.16, table 1].

The reason for the avalanche-like collapse was the low survivability of the coating farms and a number of related phenomena - a no-run coating system, flaws in the bond system, the use of boiling steel, excessive load on the roof, etc.

A farm with low survivability was used in the foundry and reinforcement workshop in Penza. Two spans 18 + 24 = 42 m of the temperature block of the building 96 m long completely collapsed in the workshop. People died in the collapse. Collapse of the coating over an area of 42 · 96 = 4032 m ² occurred in an avalanche-like manner.

The farm is made of corner profiles symmetrical in section [5], which form a T-section in the assembly, that is, a “typical” farm developed in the middle of the last century. In Russia and in the former USSR, a large number of structures have been built with such farms with low survivability.

As a result of the low survivability of "typical" farms and the natural aging of structures during long-term operation, the likelihood of emergency situations is constantly increasing. The likelihood of a sudden collapse also increases. For example, on August 13, 2010, in Samara, in the old building, converted to trade in furniture, the roof collapsed. The people in the building suffered!

In the proposed construction of the farm, we will use a new equidistant I-beam profile proposed by KK Nezhdanov and developed with graduate students [9, p. 75, 14]. The main advantage of this profile is its equal stability with respect to the X and Y axes. Due to this property, a significant positive effect arises in coating farms.

The technical task of the invention is to increase the survivability and reliability of steel trusses of civil and industrial structures operating in extreme operating conditions, with the exception of a sudden loss of stability of compressed elements by making them from oval pipe-concrete elements with a larger to smaller ratio of three, and transferring the work of the entire truss structure from dangerous stage of work on the first limit state to a more favorable work on the second limit state.

The technical problem of the invention by a method of increasing the survivability and reliability of a steel truss of a civil or industrial structure is solved as follows.

A method of increasing the survivability of a steel truss with ascending compressed support and intermediate braces, downward stretched braces and compressed struts is as follows.

The upper and lower zones of the truss are made of equally stable I-beams [9, 14].

In the most intense zone in the middle of the span, the upper belt is converted into a tubular closed one by means of welding from the sides to the equally stable I-beam profile of the closing sheets [10, 11].

All compressed elements of the farm are made of oval pipes with a larger to smaller ratio of three [11, 15, 16, 17], orienting them with a large dimension perpendicular to the plane of the farm, and compressed braces in two trusses are unfastened in the plane of the farm, thereby reducing their flexibility by three times and significantly increase their bearing capacity (stability) by 1.5 ... 1.6 times.

After the farm is installed in the design position, the hoses of the concrete pipelines are connected to the nozzles of the closed tubular cavities of the farm, fine-grained expanding concrete is pumped into the cavity of the elements with the formation of concrete elements after concrete setting.

They increase the survivability of the entire steel truss by 1.5 ... 1.6 times, exclude the possibility of a sudden loss of stability of each of the reinforced compressed braces, and transfer the work of the entire truss structure from the dangerous stage of work in the first limit state to more favorable work in the second limit state.

Figure 1 shows a diagram of a statically determinate farm [12, p.209] for covering a span of 30 m. The farm has 17 nodes. On the left, the farm rests on a hinged-motionless support (node 1), on the right - on a hinged-motion support (node 13).

The upper belt is made of an equally stable I-beam profile (from node 2 to node 12), that is, there are ten panels. The extreme panels (2-3 and 11-12) are zero. The farm perceives the vertical concentrated forces applied in nodes 3 through 11. The central section of the upper belt from node 5 to node 9 (four panels) is the most compressed.

A truss without a lantern is shown, so the snow load is evenly distributed along the span. The constant load is also distributed evenly, i.e. the load is symmetrical.

Support braces (rods 1-3 and 11-13) are the most compressed. The pairs of ascending braces are equally compressed: 5-17 and 9-14, as well as 7-16 and 7-15.

The pairs of descending stretched braces 3-17 and 11-14 are equally stretched. Similarly, a pair of braces 5-16 and 9-15 are equally stretched.

All racks (rods: 4-17, 6-16, 8-15 and 10-14) are compressed. All compressed braces are made of oval pipes in the section with a ratio of 3: 1, with the larger dimension oriented perpendicular to the plane of the truss.

To equalize the flexibility of the compressed braces, they are unfastened by double trusses coming from even nodes (2, 4, 6, 8, 10, and 12). Sprengels divide each squeezed brace into three equal parts. The Central section of the upper zone from the node 5 to the node 9 is made of concrete.

An equally stable I-beam is turned into a concrete element as follows.

Figure 2 shows the most compressed section of the upper belt from node 5 to node 9 (four panels: 5-6, 6-7, 7-8, 8-9). Figure 3 is a section of this pipe-concrete section (5-6-7-8-9).

Equal-stable 1 I-profile is closed in tubular by two elements from a steel strip of 6 steel. Two strips 6 are welded by automatic welding with continuous seams. The result is two closed cavities 7. The flanges of the I-beams 1 are connected to the strips 6 by continuous welds 8.

Figure 4 shows the node And the abutment of the supporting compressed brace (rod 1-3) and the extended brace (rod 3-17) to the upper one at the belt 1 (solid rod 2-3-4-5-6-7-8-9- 10-11-12). Subsequently, the central section from node 5 to node 9 is turned into a concrete pipe.

Figure 5 shows the node B adjoining the extended brace (rod 3-17), the compressed strut 5 (rod 4-17) and the compressed brace 3 (rod 5-17) to the stretched lower belt 1, consisting of five rods: 13-14 , 14-15, 15-16, 16-17, 17-1.

In addition, it should be noted that the supporting brace 2 (element 1-3) is unfastened in the plane of the truss with spreels (elements 2-18 and 2-19), and the supporting brace 2 on the right side is unfastened with spreels (elements 12-28, 12-29 )

Other compressed braces 5-17 and 9-14 and around the middle of 7-16 and 7-15 are also loosened with sprengels.

Concrete implementation example

Collecting loads on the farm (no lantern construction)

We find the nodal forces from a constant uniformly distributed load on the coating q = · 123.84 gN / m (Fig. 1, a):

P _q = qd = 123.84 · 3 = 371.5 gN;

from evenly distributed snow load (Penza) s = 192 gN / m;

S = s · d = 216 · 3 = 648 gN;

Definition of efforts in truss rods

Calculation of the farm is carried out according to the SNFERMA program from the action of independent loads: constant and snow loads. The force in the elements of the farm is determined by the concentrated forces applied in the nodes of the farm. The calculated values of the effort, in each rod of the farm, we find by performing an unfavorable combination of influences.

Assignment of truss rod sections

The purpose of the sections of the rods of the upper compressed zone

The section of each compressed rod is assigned so that the first limit state, loss of stability, is not achieved. If the farm is statically determinable, then switching off only one rod from the work turns the farm into a mechanism and it collapses. Assign a tubular section of the upper truss belt (rod 6-7) and compare it with the section from the corners.

- Estimated compressive force in the upper belt N = -11244.5 gN

The actual bearing capacity of the centrally compressed upper truss belt depends on the coefficient of working conditions γ, the stability coefficient φ _min , the calculated steel resistance R _y and the cross-sectional area A

F = γ · φ _min · R _y · A.

- We select the necessary cross-sectional area of the upper compressed zone from the new I-beam profile of equal stability with respect to the x and y axes.

$$A=\frac{{\rm N}}{\gamma \cdot \varphi \cdot R_y}=\frac{\mathrm{one}\mathrm{one}2\mathrm{four}\mathrm{four},5}{0,95\cdot 0,8\cdot 230}=6\mathrm{four},33\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

We assign an equally stable I-beam profile PI23K1 with a cross-sectional area of A = 66.69 cm ² , inertia radii i _x = i _y = 7.17 cm and a unit mass of length m = 52.35 kg / m.

We find the flexibility of the upper compressed belt from the equally stable I-beam profile PI23K1 with respect to the x and y axes when the ends are articulated:

$${\lambda }_x={\lambda }_y=\frac{{\ell }_{xef}}{i_x}=\frac{300}{7.17}=41.84⊲60\Rightarrow \gamma =0.95$$

Then reduced flexibility

$${\overline{\lambda }}_{\text{x}}={\lambda }_x\sqrt{\frac{{\text{R}}_y}{\text{E}}}=41.84\sqrt{\frac{230}{206000}}=\mathrm{1,398}⊲\mathrm{2,5}$$

Then the minimum coefficient of stability is more convenient to determine by building codes [8, p. 9] depending on the reduced flexibility of the rod according to one of three formulas:

$$PR\mathrm{and}0⊲\overline{\lambda }\le \mathrm{2,5}\Rightarrow \phi =\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }};$$

$$PR\mathrm{and}\mathrm{2,5}⊲\overline{\lambda }\le \mathrm{4,5}\Rightarrow \phi =1.46-0.34\cdot \overline{\lambda }+\mathrm{0,021}\cdot {\overline{\lambda }}^2;$$

$$PR\mathrm{and}\overline{\lambda }⊳\mathrm{4,5}\Rightarrow \phi =332/\left[{\overline{\lambda }}^2\left(51-\overline{\lambda }\right)\right].$$

We find the minimum stability coefficient by the formula

$${\varphi }_{x\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot \mathrm{1,398}\cdot \sqrt{\mathrm{1,398}}=0.891⊳0.8$$

Correct the calculated resistance R _y when the thickness of the elements is less than 10 mm, R _y = 240 MPa.

Checking stability [8, p.9]

N <F = γ · φ _min · R _y · A⇒N = 11244.5 <F = 0.95 · 0.891 · 240 · 66.69 = 13547.9 gN

Conclusion: the effective compressive force N is 83% of the bearing capacity of the upper belt F = γ · φ _min · R _у · A. There is an excess margin of sustainability of 17%. Reduce the cross section.

Assign an equally stable I-beam [9], [14]

PI20K2, A = 59.86 cm ² , i _x = i _y = 6.3 cm, m = 47 kg / m.

Flexibility of the upper compressed belt from an equally stable I-beam profile with respect to the x and y axes:

$${\lambda }_x={\lambda }_y=\frac{{\ell }_{xef}}{i_x}=\frac{300}{6.8}=44.12⊲60\Rightarrow \gamma =0.95$$

Then reduced flexibility

$${\overline{\lambda }}_x={\lambda }_x\sqrt{\frac{{\text{R}}_y}{\text{E}}}=44.12\sqrt{\frac{240}{206000}}=\mathrm{1,506}⊲\mathrm{2,5}$$

Then $${\varphi }_{x\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot \mathrm{1,506}\cdot \sqrt{\mathrm{1,506}}=0.878⊳0.8$$

Checking stability [8, p.9].

Correct the calculated resistance R _y when the thickness of the elements is less than 10 mm, R _y = 240 MPa.

N <F = γ · φ _min · R _y · A⇒N = 11244.5 <F = 0.95 · 0.878 · 240 · 59.86 = 11983.3 gN

The effective compressive force N = 11244.5 gN is 93.8% of the bearing capacity of the upper belt F = γ · φ _min · R _y · A. That is, a margin of stability of 6.2% is provided.

Top belt from a pipe

We set the coefficient of working conditions γ with flexibility

и коэффициент устойчивости φ=0,9 [8, табл.72]. $${\lambda }_x=\frac{{\ell }_{xef}}{i_x}⊲60\Rightarrow \gamma =0,95$$

and the stability coefficient φ = 0.9 [8, Table 72].

The required area of the cylindrical section of the upper belt.

$$A=\frac{{\rm N}}{\gamma \cdot \varphi \cdot R_y}=\frac{\mathrm{one}\mathrm{one}2\mathrm{four}\mathrm{four},5}{0,95\cdot 0,9\cdot 230}=57,\mathrm{one}8\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

We check its stability relative to any axis. Effective compressive force should be less than the limit value, limited by the norms.

- Assign a pipe Ø245 · 8, A = 59.54 cm ² , i _x = 8.39 cm, m = 46.76 kg / m.

Flexibility of a cylindrical rod with respect to any axis:

$$\lambda =\frac{{\ell }_{ef}}{i}=\frac{300}{8,39}=35,76⊲60,7\Rightarrow \gamma =0,95$$

Reduced flexibility

$$\overline{\lambda }=\lambda \sqrt{\frac{R_y}{E}}=35,76\sqrt{\frac{230}{206000}}=\mathrm{one},\mathrm{one}9\mathrm{four}8⊲2,5$$

по формуле [8, с.9]:The stability coefficient φ _{min is} found by the reduced flexibility $$\overline{\lambda }=1.1948⊲\mathrm{2,5}$$

according to the formula [8, p. 9]:

$${\varphi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot 1.1978\cdot \sqrt{1.1978}=0.9138$$

Stability test

N <F = γ · φ _min · R _y · A⇒N = 11244.5 <F = 0.95 · 0.9138 · 230 · 59.54 = 11888.0 gN.

Stability is ensured, since the calculated compressive force N = 11244.5 gN in the upper zone of the cylindrical section ⌀245 · 8 is less than the actual bearing capacity.

Stability testing can also be performed at compressive stresses.

$$\sigma =\frac{N}{A}=\frac{11244.5}{59.54}=188.9<\gamma \cdot {\varphi }_{\min }\cdot R_{\mathrm{at}}=0.95\cdot 0.9272\cdot 230=202.6MP\mathrm{but}$$

stability provided.

In the future, we will compare the effective compressive force N with the actual bearing capacity of the rod N <F = γ · φ _min · R _y · A.

For comparison, we analyze how the stability of the upper belt changes [8, p. 9] with changing sections.

- With a pipe ⌀273 · 8 (cross-sectional area A = 66.62 cm ² , radius of inertia i _x = 9.39 cm and mass m = 52.28 kg / m).

Estimated Length $${\ell }_{ef}=\mu \cdot \ell =\mathrm{one}\cdot 300\mathrm{from}m$$

$${\lambda }_x=\frac{300}{i_x}=\frac{300}{9.39}=31.95={\lambda }_{\mathrm{at}}⊲60.7\Rightarrow \gamma =0.95$$

$$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=31.95\sqrt{\frac{230}{206000}}=\mathrm{1,0675}⊲\mathrm{2,5}$$

$${\varphi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot \mathrm{1,0675}\cdot \sqrt{\mathrm{1,0675}}=0.9272$$

N <F = γ · φ _min · R _y · A⇒N = 11341.5 <F = 0.95 · 0.9272 · 230 · 66.62 = 13497.2 gN.

- With a pipe ⌀299 · 8 (A = 73.12 cm ² , i _x = 10.3 cm, m = 57.41 kg / m),

$${\lambda }_x=\frac{300}{i_x}=\frac{300}{10.3}=29.13={\lambda }_{\mathrm{at}}⊲60.7\Rightarrow \gamma =0.95$$

$$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=29.13\sqrt{\frac{230}{206000}}=0.9732⊲\mathrm{2,5}$$

$${\varphi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot 0.9732\cdot \sqrt{0.9732}=0.937$$

N <F = γ · φ _min · R _y · A⇒

N = 11244.5 <F = 0.95 · 0.937 · 230 · 73.12 = 14970 gN.

The stability of the upper belt in the middle of the span in the truss plane with respect to the x axis and from the truss plane with respect to the y axis is ensured.

- With a pipe ⌀325 · 8 (A = 79.64 cm ² , i _x = 11.22 cm, m = 62.54 kg / m)

$${\lambda }_x=\frac{300}{i_x}=\frac{300}{11.22}=26.74={\lambda }_{\mathrm{at}}⊲60.7\Rightarrow \gamma =0.95$$

$$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=26.74\sqrt{\frac{230}{206000}}=0.8934⊲\mathrm{2,5}$$

$${\varphi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot 0.8934\cdot \sqrt{0.8934}=0.944$$

Checking the stability of the rod [8, p.9] relative to the x axis

N <11341.5 <γ · φ _min · R _y · A = 0.95 · 0.944 · 230 · 79.64 = 16431 gN.

The stability of the upper belt in the middle of the span in the truss plane with respect to the x axis and from the truss plane with respect to the y axis is ensured.

- With a pipe ⌀351 · 8 (A = 86.19 cm ² , i _x = 12.14 cm, m = 67.67 kg / m)

$${\lambda }_x=\frac{{\ell }_{xef}}{i_x}=\frac{300}{12.14}=24.7={\lambda }_y⊲60.7\Rightarrow \gamma =0.95$$

Reduced flexibility $$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=24.7\sqrt{\frac{230}{206000}}=0.8257⊲\mathrm{2,5}$$

Then $${\varphi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot 0.8257\cdot \sqrt{0.8257}=0.95$$

Checking the stability of the rod of the upper belt relative to any axis [8, p.9]

N <F = γ · φ _min · R _y · A⇒N = 11341.5 <F = 0.95 · 0.866 · 230 · 86.19 = 16309 gN

N = 11244.5 (69.5%) F = 16309 gN (100%) margin of stability 30.5%.

For comparison, we take a section of two symmetric isosceles corners. In this case, we set φ = 0.8 [8, Table 72] and the coefficient of working conditions γ with flexibility $${\lambda }_x=\frac{{\ell }_{xef}}{i_x}⊳60\Rightarrow \gamma =0.8$$

The required area of symmetrical section from two isosceles corners

$$A=\frac{{\rm N}}{\gamma \cdot \varphi \cdot R_y}=\frac{11244.5}{0.8\cdot 0.8\cdot 230}=\mathrm{76,4}\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

2∠160 · 14, m = 64 kg / m, A = 2 · 43.3 · = 86.6 cm ² , i _x = 4.92, i _y = 7.05 cm

Flexibility of a compressed rod relative to the x and y axes:

$${\lambda }_x=\frac{{\ell }_{xef}}{i_x}=\frac{300}{4.92}=60.98⊳\mathrm{60,}$$

$${\lambda }_y=\frac{300}{7.05}=42.55⊲60.7$$

The stability coefficient φ is taken for maximum flexibility,

Приведенная гибкостьi.e $${\lambda }_{x\max }=60.98⊳60\Rightarrow \gamma =\mathrm{0.8.}$$

Reduced flexibility

$${\overline{\lambda }}_x={\lambda }_x\sqrt{\frac{{\text{R}}_y}{\text{E}}}=60.98\sqrt{\frac{230}{206000}}=\mathrm{2,037}⊲\mathrm{2,5}$$

Then $${\varphi }_{x\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot \mathrm{2,037}\cdot \sqrt{\mathrm{2,037}}=0.808⊳0.8$$

Checking stability [8, p.9]

N <F = γ · φ _min · R _y · A⇒N = 11244.5 <F = 0.8 · 0.808 · 230 · 86.6 = 12876.9 gN.

The effective compressive force N is 88% of the bearing capacity of the rod F = γ · φ _min · R _y · A. That is, a margin of stability of 12% is provided.

The stability of the upper belt in the middle of the span in the truss plane relative to the x and y axis is ensured.

Thus, the material consumption of the compressed upper belt from the corners 2L160 · 14 is 1.45 times higher than the material consumption of the belt from the pipe ⌀245 · 8. The efficiency of replacing the corners of the compressed upper belt with a cylindrical pipe is high!

Assignment of cross sections of the rods of the upper compressed belt from PI20K2 equally stable I-beams to their further transformation into pipe-concrete elements

Concrete grade 250 300 350 400 450 500 550 k _bya 1.92 1.83 1.73 1.66 1,59 1.55 1,50 R _PR , MPa 10.79 13.24 15,2 17.17 19.13 21.09 R _PR , kgf / cm ² 110 135 155 175 195 215

We fill two cavities of an equidistant I-beam PI20K2 (A = 59.86 cm ² , i _x = 6.8 cm, m = 47 kg / m) of the support brace with fine-grained expanding concrete of the M250 brand with prismatic strength R _Pr = 10.79 MPa.

Concrete core area in I-beam

A _ba = 2 (a × b) = 2 (12.725 × 13.75) = 349.94 cm ² . The moment of inertia of the concrete core

$$J_{b\mathrm{I\; am}}=2\cdot \left[\frac{b\cdot {\mathrm{<figure-callout\; id="3"\; label="h"\; filenames="00000122.png,00000125.png"\; state="\{\{state\}\}">h</figure-callout>}}^3}{\mathrm{one}2}+A_{b\mathrm{I\; am}}{\left(\frac{t_{c\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="00000122.png"\; state="\{\{state\}\}">t</figure-callout>}}}{2}+\frac{a}{2}\right)}^2\right]=2\cdot \left[\frac{\mathrm{one}2,725\cdot \mathrm{one}3,75^3}{\mathrm{one}2}+3\mathrm{four}9,93\cdot {\left(\frac{0,87}{2}+\frac{\mathrm{one}2,725}{2}\right)}^2\right]=3785\mathrm{one},77\mathrm{from}{m}^{\mathrm{four}}$$

Radius of inertia of concrete core $$i_{b\mathrm{I\; am}}=\sqrt{\frac{J_{b\mathrm{I\; am}}}{{\mathrm{BUT}}_{b\mathrm{I\; am}}}}=\sqrt{\frac{\mathrm{37,851.77}}{349.93}}=\mathrm{10,4}\mathrm{from}m$$

$$\mu =\frac{A_{Тр}}{A_{бя}}=\frac{\mathrm{59,86}}{\mathrm{349,93}}=\mathrm{0,171}$$

Odds $$k=\frac{k_{b\mathrm{I\; am}}\cdot R_{b\mathrm{I\; am}}}{R_y}=\frac{1.92\cdot 10.79}{240}=\mathrm{0,08632};$$

$$\mu =\frac{A_{TR}}{A_{b\mathrm{I\; am}}}=\frac{59.86}{349.93}=0.171$$

Reduced flexibility

$${\lambda }_{PR}=\frac{{\ell }_{ef}}{i_{b\mathrm{I\; am}}}\sqrt{\frac{k+\mu }{0.25k+0.5\mu }}=\frac{300}{\mathrm{10,4}}\sqrt{\frac{\mathrm{0,08632}+0.171}{0.25\cdot \mathrm{0,08632}+0.5\cdot 0.71}}=\mathrm{26,625}\Rightarrow \varphi =0.942$$

Bearing capacity of a concrete bar

(349,93·10,79·1,92+59,86·240)0,942=20362, 11 гНN≤ (A R _{os os os} + A k _Tr R _y) φ⇒11244,5 $$⊲$$

(349.93 · 10.79 · 1.92 + 59.86 · 240) 0.942 = 20362, 11 gN

The stability of the concrete pipe rod increased in comparison with a steel rod in 20362.11 / 11244.5 = 1.81 times.

Let us compare the material consumption of the compressed upper belt from the equally stable I-beam profile PI20K2 and from the tubular profile ⌀351 · 8, that is, the second and third lines of the table. It is easy to see that with almost the same material consumption, the stability of the belt from the equidistant I-beam profile EI20K2 is only slightly inferior to the tubular profile ⌀351 · 8. The results of comparing the variants of the upper belt with a compressive force N = 11244.5 gN in it are given in Table 10.

Table 10 Top Belt Options Comparison A cm ² i _{x min} m kg λ _x γ

$${\overline{\lambda }}_x$$

φ _xmin F, gN N / F EI20K2 59.86 6.8 47 44.12 0.95 1,506 0.878 11,983.3 0.938 ⌀2458 59.54 8.39 46.76 35.76 0.95 1,195 0.914 11888.0 0.954 2∠160 · 12 86.6 4.94 68 61 0.8 2,037 0.808 12,876.9 0.88 Pipe EI20K2 349.93 10,4 92.18 26.62 0.94 - 0.942 20362.11 0.552

Compare the cross section of the pipe and two corners. The upper belt from the pipe ⌀245 · 8 is more stable than the belt from the corners 2 ∠ 160 · 14, 1.267 times.

The purpose of the cross section of the compressed reference brace

Estimated compressive force in the support brace (rod 1-3)

N _1-3 = -6582.4 gN (Table 1).

We set φ = 0.9 [8, Table 72] and the coefficient of working conditions γ with flexibility $${\lambda }_x=\frac{{\ell }_{xef}}{i_x}⊲60\Rightarrow \gamma =0.95$$

The actual bearing capacity (stability) F = γ · φ _min · R _y · А depends on the accepted section of the upper belt.

We find the required cross-sectional area (Fig. 4) from a pipe round in cross section: $$A_{\mathrm{one}-3}=\frac{{{\rm N}}_{\mathrm{one}-3}}{\gamma \cdot \varphi \cdot R_y}=\frac{6582,\mathrm{four}}{0,8\cdot 0,8\cdot 2\mathrm{four}0}=\mathrm{four}2,85\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

We select a section from an equidistant [9] [14] I-profile [8, p.9]

- PI20K1, A = 52.69 cm ² , i _x = i _y = 6.39 cm, m = 41.36 kg / m.

The flexibility of the rod relative to any axis with a length of 4.02 m, µ = 1:

$${\lambda }_x=\frac{{\ell }_{xef}}{i_x}=\frac{402}{6.39}=62.94⊳60\Rightarrow \gamma =0.8$$

The stability coefficient is found by the given flexibility [8, p. 9]: $$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=62.94\sqrt{\frac{240}{206000}}=\mathrm{2,1483}⊲\mathrm{2,5}$$

then $${\varphi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot \mathrm{2,3427}\cdot \sqrt{\mathrm{2,3427}}=0.7922$$

N <F = γ · φ _min · R _y · A⇒N _1-3 = 6582.4 <F = 0.8 · 0.7922 · 240 · 52.69 = 8013.9 gN.

Conclusion: the stability of the support brace from an equally stable I-beam profile is provided. The margin of stability is 17.1%.

Find the required cross-sectional area from the pipe, round in cross section

$$A_{\mathrm{one}-3}=\frac{{{\rm N}}_{\mathrm{one}-3}}{\gamma \cdot \varphi \cdot R_y}=\frac{6582,\mathrm{four}}{0,95\cdot 0,8\cdot 2\mathrm{four}0}=36,\mathrm{one}\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

Assign a pipe round in cross section

- Ø273 · 4.5, A = 38 cm ² , i _x = 9.5 cm, m = 29.8 kg / m.

We check the stability of the tubular support brace round in cross section relative to any axis. Effective compressive force should be less than the limit value, limited by the norms.

The flexibility of the tubular rod relative to any axis with a length of 4.02 m, µ = 1:

$${\lambda }_x=\frac{{\ell }_{xef}}{i_x}=\frac{402}{9.5}=42.32⊲60\Rightarrow \gamma =0.95$$

The stability coefficient φ _{min is} found by the given flexibility [8, p. 9]: N _1-3 = 6582.4 gN

$$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=42.32\sqrt{\frac{240}{206000}}=\mathrm{1,4443}⊲\mathrm{2,5}$$

then according to the formula [8, p. 9]

$${\phi }_{\min }=\mathrm{one}-\mathrm{0,066}\cdot \overline{\lambda }\cdot \sqrt{\overline{\lambda }}=\mathrm{one}-\mathrm{0,066}\cdot \mathrm{1,4443}\cdot \sqrt{\mathrm{1,4443}}=0.8854⊳0.8$$

N <F = γ · φ _min · R _y · A⇒N _1-3 = 6582.4 <F = 0.95 · 0.8854 · 240 · 38 = 7671.4 gN.

The stability of the cylindrical support brace ⌀273 · 4,5 provided. The margin of stability is 16.5%.

We turn the tubular section of the compressed support brace into a concrete element.

Concrete grade 250 300 350 400 450 500 550 k _bya 1.92 1.83 1.73 1.66 1,59 1.55 1,50 R _PR , MPa 10.79 13.24 15,2 17.17 19.13 21.09 R _PR , kgf / cm ² 110 135 155 175 195 215

The bearing capacity of pipe-concrete columns N <(A R _{os os os} + A k _Tr R _y) φ,

where A _b and A _Tr - the area of the concrete core and steel pipe;

k _ba - coefficient of increase of concrete strength in the pipe;

R _ba = R _CR and R _y - the design resistance of concrete and steel;

φ is the coefficient of stability of the rod of the concrete pipe.

Reduced flexibility $${\lambda }_{PR}=\frac{{\ell }_{ef}}{i_{b\mathrm{I\; am}}}\sqrt{\frac{k+\mu }{0.25k+0.5\mu }},$$

$${\ell }_{ef}$$

- расчетная длина стержня;Where $$k=\frac{k_{b\mathrm{I\; am}}\cdot R_{b\mathrm{I\; am}}}{R_y};\mu =\frac{A_{TR}}{A_{b\mathrm{I\; am}}};$$

$${\ell }_{ef}$$

- estimated length of the rod;

- бетонного ядра. $$i_{b\mathrm{I\; am}}=\sqrt{\frac{J_{b\mathrm{I\; am}}}{{\mathrm{BUT}}_{b\mathrm{I\; am}}}}$$

- concrete core.

Stability coefficient φ of a compressed concrete pipe

Reduced flexibility λ _Pr 10 twenty thirty 40 fifty 60 Concrete grade 250 0.988 0.963 0.931 0.888 0.850 0.791 Concrete grade 500 0.988 0.974 0.950 0.922 0.893 0.852 Reduced flexibility λ _Pr 70 80 90 one hundred 110 120 Concrete grade 250 0.728 0.654 0.591 0.527 0.461 0.400 Concrete grade 500 0.80 0.731 0.663 0.588 0.518 0.450

We fill the pipe Ø273 · 4.5, (A = 38 cm ² , i _x = 9.5 cm, m = 29.8 kg / m) of the support brace with fine-grained expanding concrete of the M250 grade with prismatic strength R _Pr = 10.79 MPa.

The diameter of the concrete core d _ba = D-2t = 27.3-2 · 0.9 = 25.5 cm.

The area of the concrete core in the steel pipe $$A_{b\mathrm{I\; am}}=\frac{\pi \cdot d^2}{\mathrm{four}}=\frac{\pi \cdot {25.5}^2}{\mathrm{four}}=510.71\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

The moment of inertia of the concrete core $$J_{b\mathrm{I\; am}}=\frac{\pi \cdot d^{\mathrm{four}}}{64}=\frac{\pi \cdot {25.5}^{\mathrm{four}}}{64}=20755.4\mathrm{from}{m}^{\mathrm{four}}$$

Radius of inertia of concrete core $$i_{b\mathrm{I\; am}}=\sqrt{\frac{J_{b\mathrm{I\; am}}}{{\mathrm{BUT}}_{b\mathrm{I\; am}}}}=\sqrt{\frac{20755.4}{510.71}}=\mathrm{6,375}\mathrm{from}m$$

$$\mu =\frac{A_{Тр}}{A_{бя}}=\frac{38}{\mathrm{510,71}}=\mathrm{0,074406}$$

Odds $$k=\frac{k_{b\mathrm{I\; am}}\cdot R_{b\mathrm{I\; am}}}{R_y}=\frac{1.92\cdot 10.79}{240}=\mathrm{0,08632};$$

$$\mu =\frac{A_{TR}}{A_{b\mathrm{I\; am}}}=\frac{38}{510.71}=\mathrm{0,074406}$$

Reduced flexibility

, $${\lambda }_{PR}=\frac{{\ell }_{ef}}{i_{b\mathrm{I\; am}}}\sqrt{\frac{k+\mathrm{<figure-callout\; id="2"\; label="\mu "\; filenames="00000122.png"\; state="\{\{state\}\}">\mu </figure-callout>}}{0,25k+0,5\mu }}=\frac{\mathrm{four}02}{6,375}\sqrt{\frac{0,08632+0,069\mathrm{four}2}{0,25\cdot 0,08632+0,5\cdot 0,07\mathrm{four}\mathrm{four}06}}=\mathrm{one}0\mathrm{four},3\Rightarrow \varphi =0,553$$

,

Bearing capacity of a concrete bar

(510,71·10,79·1,92+38·240)0,553=10894,3 гНN≤ (A R _{os os os} + A k _Tr R _y) φ⇒6582,4 $$⊲$$

(510.71 · 10.79 · 1.92 + 38 · 240) 0.553 = 10894.3 gN

The stability of the concrete pipe rod increased in comparison with a steel rod by 10894.3 / 6971.3 = 1.563 times.

Assign a support brace from an oval pipe section

The oval rod was obtained from a pipe Ø273 · 4.5, A = 38 cm ² , i _x = 9.5 cm, m = 29.8 kg / m.

The vertical and horizontal dimensions of the oval profile (along the axis passing through the middle of the wall thickness)

$$2a=\frac{\mathrm{1,5}\cdot A}{\pi \cdot t_0}=\frac{\mathrm{1,5}\cdot 38}{\pi \cdot 0.45}=\mathrm{40,319}\mathrm{from}m\Rightarrow \mathrm{but}=20.16\Rightarrow b=\frac{a}{3}=6.72\mathrm{from}m$$

2b = 12.144 cm.

Vertical and horizontal profile oval dimensions

2a + t ₀ = 40.319 + 0.45 = 40.77; 2b + t ₀ = 2 · 6.72-0.45 = 11.694 cm.

$$J_Y=\frac{\pi }{4}\left[a{\left(\frac{a}{3}\right)}^3-\left(a-\frac{t_0}{2}\right){\left(\frac{a}{3}-\frac{t_0}{2}\right)}^3\right]=\frac{\pi }{4}\left[\mathrm{20,16}{\left(\frac{\mathrm{20,16}}{3}\right)}^3-\left(\mathrm{20,16}-\frac{\mathrm{0,45}}{2}\right){\left(\frac{\mathrm{20,16}}{3}-\frac{\mathrm{0,45}}{2}\right)}^3\right]=\mathrm{655,82}с{м}^4$$

Moments of inertia $$J_X=\frac{3A}{16}\left(2a^2+\frac{5}{6}{t}_0\right)=\frac{3\cdot 38}{16}\left(2\cdot {20.16}^2+\frac{5}{6}\cdot 0.45\right)=5792.55\mathrm{from}{m}^{\mathrm{four}}$$

$$J_Y=\frac{\pi }{\mathrm{four}}\left[a{\left(\frac{a}{3}\right)}^3-\left(a-\frac{t_0}{2}\right){\left(\frac{a}{3}-\frac{t_0}{2}\right)}^3\right]=\frac{\pi }{\mathrm{four}}\left[20.16{\left(\frac{20.16}{3}\right)}^3-\left(20.16-\frac{0.45}{2}\right){\left(\frac{20.16}{3}-\frac{0.45}{2}\right)}^3\right]=655.82\mathrm{from}{m}^{\mathrm{four}}$$

Moments of resistance

$$W_X=\frac{J_x}{a+0.5t_0}=\frac{5792.55}{20.16+0.5\cdot 0.45}=284.16\mathrm{from}{m}^3$$

$$W_Y=\frac{J_Y}{b+0.5t_0}=\frac{655.82}{6.72+0.5\cdot 0.45}=94.43\mathrm{from}{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="00000122.png,00000125.png"\; state="\{\{state\}\}">m</figure-callout>}}^3$$

$${\rho }_Y=\frac{W_Y}{A}=\frac{\mathrm{94,43}}{38}=\mathrm{2,485}см,$$

section core radii $${\rho }_X=\frac{W_x}{A}=\frac{284.16}{38}=7.48;$$

$${\rho }_Y=\frac{W_Y}{A}=\frac{94.43}{38}=\mathrm{2,485}\mathrm{from}m,$$

$$i_Y=\sqrt{\frac{J_Y}{А}}=\sqrt{\frac{\mathrm{655,82}}{38}}=\mathrm{4,154}см.$$

radii of inertia $$i_X=\sqrt{\frac{J_x}{\mathrm{BUT}}}=\sqrt{\frac{5792.55}{38}}=\mathrm{12,346};$$

$$i_Y=\sqrt{\frac{J_Y}{\mathrm{BUT}}}=\sqrt{\frac{655.82}{38}}=\mathrm{4,154}\mathrm{from}m.$$

The geometric and calculated length of the support brace from the truss plane is ℓ = 402; ℓ _ef = µ ℓ = 402

The geometric and calculated length of the support brace in the truss plane (two trusses) is equal to µ = 1, ℓ = 402/3 = 134; ℓ _ef = μ ℓ = 134.

Maximum flexibility in the truss plane with a length of ℓ _ef = 134, $${\lambda }_Y=\frac{{\ell }_{ef}}{i_Y}=\frac{134}{\mathrm{4,154}}=32.26⊲60$$

Flexibility from the truss plane with a length of ℓ _ef = 402 cm,

$${\lambda }_X=\frac{{\ell }_{ef}}{i_X}=\frac{402}{\mathrm{12,346}}=32.56⊳32.26$$

We fill the oval profiles with fine-grained expanding concrete and turn the rods into pipe-concrete.

The outer dimensions of the cross section of the concrete core is the largest and smallest 2a-t ₀ = 40.319-0.45 = 39.87; 2b-t ₀ = 2 · 6.72-0.45 = 13.89 cm.

The cross-sectional area of the concrete core

$$A_{b\mathrm{I\; am}}=\frac{\pi }{\mathrm{four}}(2a-t_0)(2b-t_0)=\frac{\pi }{\mathrm{four}}(\mathrm{40,319}-0.45)(6.72-0.45)=366.17\mathrm{<figure-callout\; id="2"\; label="c\; m"\; filenames="00000122.png"\; state="\{\{state\}\}">c\; </figure-callout>}{m}^{}{}^2$$

The moment of inertia of the concrete core

$$J_{b\mathrm{I\; am}Y}=\frac{A_{b\mathrm{I\; am}}}{\mathrm{four}}{\left(b-\frac{t_0}{2}\right)}^2=\frac{366.17}{\mathrm{four}}{\left(3.36-\frac{0.45}{2}\right)}^2=3129.64\mathrm{from}{m}^{\mathrm{four}}$$

Radius of inertia of concrete core $$i_{b\mathrm{I\; am}Y}=\sqrt{\frac{J_{b\mathrm{I\; am}Y}}{{\mathrm{BUT}}_{b\mathrm{I\; am}}}}=\sqrt{\frac{3129.64}{366.17}}=\mathrm{3,924}\mathrm{from}m$$

$$k=\frac{k_b\cdot R_b}{R_{\mathrm{at}}}=\frac{1.92\cdot 10.79}{240}=\mathrm{0,08632};$$

$$\mu =\frac{A_{Tp}}{A_b}=\frac{38}{366.17}=\mathrm{0,103776}$$

Reduced flexibility

, $${\lambda }_{PR}=\frac{{\ell }_{ef}}{i_b}\sqrt{\frac{k+\mu }{0.25k+0.5\mu }}=\frac{134}{\mathrm{3,924}}\sqrt{\frac{\mathrm{0,08632}+\mathrm{0,103776}}{0.25\cdot \mathrm{0,08632}+0.5\cdot \mathrm{0,103776}}}=54.94\Rightarrow \varphi =0.879$$

,

Bearing capacity of a concrete bar

(366,17·10,79·1,92+38·240)0,879=14684,6 гНN≤ (A R _{os os os} + A k _Tr R _y) φ⇒6582,4 $$⊲$$

(366.17 · 10.79 · 1.92 + 38 · 240) 0.879 = 14684.6 gN

The stability of the concrete pipe rod increased compared to a steel rod by 14,684.6 / 6971.3 = 2.106 times.

Support brace from two corners (for comparison)

The required cross-sectional area at φ = 0.8:

$$A=\frac{{{\rm N}}_{\mathrm{one}-3}}{\gamma \cdot \varphi \cdot R_y}=\frac{\mathrm{6582,4}}{0.8\cdot 0.8\cdot 230}=44.7\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

Assign a symmetrical section from a pair of corners 2L160 · 10,

A = 2 · 31.4 = 62.8 cm ² , radii of inertia i _x = 4.96, i _y = 6.97 cm,

m = 2 · 24.7 = 49.4 kg / m

The flexibility of the support brace relative to the x axis with a length of 402.2 cm, µ = 1:

$${\lambda }_x=\frac{{\ell }_{xef}}{i_x}=\frac{402}{4.96}=81.09⊳60\Rightarrow \gamma =0.8$$

The stability coefficient φ is found by the reduced flexibility:

$$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=81.09\sqrt{\frac{240}{206000}}=2.7095⊳\mathrm{2,5}$$

then according to the formula [8, p. 9]

$${\varphi }_{\min }=1.46-0.34\overline{\lambda }+\mathrm{0,021}{\overline{\lambda }}^2=1.46-0.34\cdot 2.7095+\mathrm{0,021}\cdot {2.7095}^2=0.6929$$

N <F = γ · φ _min · R _y · A⇒N _1-3 = 6582, $$\mathrm{four}⊲F=0.95\cdot 0.6929\cdot 230\cdot 62.8=9507.8gN.$$

Conclusion: the stability of the support brace from a pair of corners 2L 160 · 10 is provided. The compressive force in the support brace N _1-3 = -6582.4 gN.

Compare the material consumption of the compressed support brace from an equally stable I-beam profile PI20K1 and from tubular profiles Ø245 · 8 and Ø200 · 5 mm, that is, the first, second and third rows of the table.

Comparison of the material intensity of the compressed support brace A cm ² i _{x min} m kg λ _x γ

$${\overline{\lambda }}_x$$

φ _xmin F, gN N / F PI20K1 52.69 6.39 41.36 62.94 0.8 2,1483 0.7222 8013.9 0.821 Ø245.5 37.12 6.98 29.14 57.62 0.95 1.9254 0.8237 6971.27 0.944 2∠16010 62.8 4.96 49.4 81.09 0.8 2.7095 0.6929 9507.8 0.724

The bearing capacity of the support brace from an equidistant I-beam profile EI20K2 has a greater margin of stability, but inferior to the tubular profile ⌀351 · 8. However, the construction of nodes in the first case is facilitated. The material consumption of the compressed support brace from the corners 2 L 160 · 14 is the largest of all options.

The purpose of the section of the compressed struts and braces

The assignment of the sections of the compressed braces and racks is carried out in the same way as the section of the compressed upper belt of the farm, the extended braces - the same as the stretched lower belt. In this case, the calculated lengths of the elements determine l ₀ = µ · l , where l is the distance between the fastened points, and µ is the coefficient of reduction of length.

For the upper and lower truss belts and for supporting braces in the plane and from the plane of the truss, the coefficient of reduction of the length is μ = 1.

For compressed braces and struts, the lattices in the truss plane µ = 0.8, and from the truss plane µ = 1.

The purpose of the cross section of the compressed rack (rod 4-17)

- We check the stability of the tubular square section of the rack relative to the y axis (from the plane). Effective compressive force should be less than the limit value, limited by the norms.

Find the required cross-sectional area (Fig. 4) from a pipe square in cross-section:

$$A_{\mathrm{four}-17}=\frac{{{\rm N}}_{\mathrm{four}-17}}{\gamma \cdot \varphi \cdot R_y}=\frac{1019.5}{0.8\cdot 0.7\cdot 240}=7.59\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

Assign a pipe square in cross section

□ 90 · 3, A = 10.1 cm ² , i _x = 3.92 cm, m = 8.83 kg / m.

The flexibility of the tubular rod with respect to any axis with length

300 cm, µ = 1: $${\lambda }_x=\frac{{\ell }_{xef}}{i_x}=\frac{300}{3.92}=76.53⊳60\Rightarrow \gamma =0.8$$

The stability coefficient φ is found from the reduced flexibility

$$\overline{\lambda }=\lambda \sqrt{\frac{{\text{R}}_y}{\text{E}}}=76.53\sqrt{\frac{240}{206000}}=\mathrm{2,612}⊳\mathrm{2,5};\mathrm{2,612}⊲\mathrm{4,5}$$

then according to the formula [8, p. 9]

$${\phi }_{\min }=1.46-0.34\cdot \overline{\lambda }+\mathrm{0,021}\cdot {\overline{\lambda }}^2=1.46-0.34\cdot \mathrm{2,612}+\mathrm{0,021}\cdot {\mathrm{2,612}}^2=0.7152⊳0.7$$

N <F = γ · φ _min · R _y · A⇒N _1-3 = 1019.5 <F = 0.8 · 0.7152 · 240 · 10.1 = 1386.8 gN.

Conclusion: the stability of the tubular rack is provided with a margin.

The purpose of the cross section of the compressed concrete pipe rack (rod 4-17)

We fill the pipe □ 90 · 3 (A = 10.1 cm ² , i _x = 3.92 cm, m = 8.83 kg / m) with fine-grained expanding concrete of the M250 grade with prismatic strength R _Pr = 10.79 MPa.

The area of the concrete core in the steel pipe

$$A_{b\mathrm{I\; am}}=a\times b=8.4\times 8.4=70.56\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

The moment of inertia of the concrete core $$J_{b\mathrm{I\; am}}=\frac{b\cdot {\mathrm{<figure-callout\; id="3"\; label="h"\; filenames="00000122.png,00000125.png"\; state="\{\{state\}\}">h</figure-callout>}}^3}{12}=\frac{8.4\cdot {8.4}^3}{12}=414.9\mathrm{from}{m}^{\mathrm{four}}$$

Radius of inertia of concrete core $$i_{b\mathrm{I\; am}}=\sqrt{\frac{J_{b\mathrm{I\; am}}}{{\mathrm{BUT}}_{b\mathrm{I\; am}}}}=\sqrt{\frac{414.9}{70.56}}=2.42\mathrm{from}m$$

$$\mu =\frac{A_{Тр}}{A_{бя}}=\frac{\mathrm{10,1}}{\mathrm{70,56}}=\mathrm{0,143}$$

Odds $$k=\frac{k_{b\mathrm{I\; am}}\cdot R_{b\mathrm{I\; am}}}{R_y}=\frac{1.92\cdot 10.79}{240}=\mathrm{0,08632};$$

$$\mu =\frac{A_{TR}}{A_{b\mathrm{I\; am}}}=\frac{10.1}{70.56}=0.143$$

Reduced flexibility

$${\lambda }_{PR}=\frac{{\ell }_{ef}}{i_{b\mathrm{I\; am}}}\sqrt{\frac{k+\mu }{0.25k+0.5\mu }}=\frac{300}{2.42}\sqrt{\frac{\mathrm{0,08632}+0.143}{0.25\cdot \mathrm{0,08632}+0.5\cdot 0.143}}=194.62\Rightarrow \varphi =0.553$$

Bearing capacity of a concrete pillar

(70,56·10,79·1,92+10,1·240)0,553=2336,33 гНN≤ (A R _{os os os} + A k _Tr R _y) φ⇒1019,5 $$⊲$$

(70.56 · 10.79 · 1.92 + 10.1 · 240) 0.553 = 2336.33 gN

The stability of the concrete bar increased in comparison with a steel bar in 2336.33 / 1019.5 = 2.291 times.

The purpose of the cross section of the compressed concrete pipe rack (rod 5-17)

The oval rod was obtained from a pipe ⌀273 · 4.5, A = 38 cm ² , i _x = 9.5 cm, m = 29.8 kg / m.

The vertical and horizontal dimensions of the oval profile (along the axis passing through the middle of the wall thickness)

$$2a=\frac{\mathrm{1,5}\cdot A}{\pi \cdot t_0}=\frac{\mathrm{1,5}\cdot 38}{\pi \cdot 0.45}=\mathrm{40,319}\mathrm{from}m\Rightarrow \mathrm{but}=20.16\Rightarrow b=\frac{a}{3}=6.72\mathrm{from}m;2b=\mathrm{12,144}\mathrm{from}m.$$

Vertical and horizontal profile oval dimensions

2a + t ₀ = 40.319 + 0.45 = 40.77; 2b + t ₀ = 2 · 6.72-0.45 = 11.694 cm.

Moments of inertia $$J_X=\frac{3A}{16}\left(2a^2+\frac{5}{6}{t}_0\right)=\frac{3\cdot 38}{16}\left(2\cdot {20.16}^2+\frac{5}{6}\cdot 0.45\right)=5792.55\mathrm{from}{m}^{\mathrm{four}}$$

$$J_Y=\frac{\pi }{\mathrm{four}}\left[a{\left(\frac{a}{3}\right)}^3-\left(a-\frac{t_0}{2}\right){\left(\frac{a}{3}-\frac{t_0}{2}\right)}^3\right]=\frac{\pi }{\mathrm{four}}\left[20.16{\left(\frac{20.16}{3}\right)}^3-\left(20.16-\frac{0.45}{2}\right){\left(\frac{20.16}{3}-\frac{0.45}{2}\right)}^3\right]=655.82\mathrm{from}{m}^{\mathrm{four}}$$

Moments of resistance

$$W_X=\frac{J_X}{a+0.5t_0}=\frac{5792.55}{20.16+0.5\cdot 0.45}=284.16\mathrm{from}{m}^3$$

$$W_Y=\frac{J_Y}{b+0.5t_0}=\frac{655.82}{6.72+0.5\cdot 0.45}=94.43\mathrm{from}{\mathrm{<figure-callout\; id="3"\; label="m"\; filenames="00000122.png,00000125.png"\; state="\{\{state\}\}">m</figure-callout>}}^3$$

$${\rho }_Y=\frac{W_Y}{A}=\frac{\mathrm{94,43}}{38}=\mathrm{2,485}см$$

section core radii $${\rho }_X=\frac{W_X}{A}=\frac{284.16}{38}=7.48;$$

$${\rho }_Y=\frac{W_Y}{A}=\frac{94.43}{38}=\mathrm{2,485}\mathrm{from}m$$

radii of inertia

$$i_X=\sqrt{\frac{J_X}{\mathrm{BUT}}}=\sqrt{\frac{5792.55}{38}}=\mathrm{12,346};$$

$$i_Y=\sqrt{\frac{J_Y}{\mathrm{BUT}}}=\sqrt{\frac{655.82}{38}}=\mathrm{4,154}\mathrm{from}m$$

The geometric and calculated length of the support brace from the truss plane is l = 402; l _ef = μ · l = 402 cm.

The geometric and calculated length of the support brace in the truss plane (two trusses) is µ = 1 l = 402/3 = 134; l _ef = μ · l = 134 cm.

Maximum flexibility in the truss plane with a length l _ef = 134 cm,

$${\lambda }_Y=\frac{{\ell }_{ef}}{i_Y}=\frac{134}{\mathrm{4,154}}=32.26⊲60$$

Flexibility from the truss plane with a length l _ef = 402 cm,

$${\lambda }_X=\frac{{\ell }_{ef}}{i_X}=\frac{402}{\mathrm{12,346}}=32.56⊳32.26$$

We fill the oval profiles with fine-grained expanding concrete and turn the rods into pipe-concrete.

The outer dimensions of the cross section of the concrete core is the largest and smallest 2a-t ₀ = 40.319-0.45 = 39.87; 2b-t ₀ = 2 · 6.72-0.45 = 13.89 cm.

The cross-sectional area of the concrete core

$$A_{b\mathrm{I\; am}}=\frac{\pi }{\mathrm{four}}(2a-t_0)(2b-t_0)=\frac{\pi }{\mathrm{four}}(\mathrm{40,319}-0.45)(6.72-0.45)=366.17\mathrm{<figure-callout\; id="2"\; label="c\; m"\; filenames="00000122.png"\; state="\{\{state\}\}">c\; </figure-callout>}{m}^{}{}^2$$

The moment of inertia of the concrete core

$$J_{b\mathrm{I\; am}Y}=\frac{A_{b\mathrm{I\; am}}}{\mathrm{four}}{\left(b-\frac{t_0}{2}\right)}^2=\frac{366.17}{\mathrm{four}}{\left(3.36-\frac{0.45}{2}\right)}^2=3129.64\mathrm{from}{m}^{\mathrm{four}}$$

Radius of inertia of concrete core $$i_{b\mathrm{I\; am}Y}=\sqrt{\frac{J_{b\mathrm{I\; am}Y}}{{\mathrm{BUT}}_{b\mathrm{I\; am}}}}=\sqrt{\frac{3129.64}{366.17}}=\mathrm{3,924}\mathrm{from}m$$

$$k=\frac{k_b\cdot R_b}{R_y}=\frac{1.92\cdot 10.79}{240}=\mathrm{0,08632};$$

$$\mu =\frac{A_{Tp}}{A_b}=\frac{38}{366.17}=\mathrm{0,103776}$$

Reduced flexibility

$${\lambda }_{PR}=\frac{{\ell }_{ef}}{i_b}\sqrt{\frac{k+\mu }{0.25k+0.5\mu }}=\frac{134}{\mathrm{3,924}}\sqrt{\frac{\mathrm{0,08632}+\mathrm{0,103776}}{0.25\cdot \mathrm{0,08632}+0.5\cdot \mathrm{0,103776}}}=54.94\Rightarrow \varphi =0.879$$

Bearing capacity of a concrete bar

(366,17·10,79·1,92+38·240)0,82=1370,72 гНN≤ (A R _{os os os} + A k _Tr R _y) φ⇒4464,5 $$⊲$$

(366.17 · 10.79 · 1.92 + 38 · 240) 0.82 = 1370.72 gN

The stability of the concrete pipe rack increased compared to a steel rack by 13707.2 / 4464.5 = 3.072 times. The effect is high!

The purpose of the section of the lower stretched belt (rod 15-16)

Estimated tensile force N _15-16 = 11744.2 gN.

The required sectional area of the lower stretched belt (Fig. 5):

$$A_{TR}=\frac{{{\rm N}}_{\mathrm{fifteen}-16}}{{\gamma }_c{R}_y}=\frac{\mathrm{11744,2}}{0.95\cdot 240}=51.5\mathrm{from}{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="00000122.png"\; state="\{\{state\}\}">m</figure-callout>}}^2$$

We select the section of the lower stretched belt from a new I-beam profile of equal stability with respect to the x axis and y axis. Assign Equidistant I-Beam Profile

PI20K1, A = 52.69 cm ² , i _x = i _y = 6.39 cm, m = 41.36 kg / m.

Flexibility of the lower stretched belt from an equidistant I-beam profile from a plane relative to the y axis:

$${\lambda }_y=\frac{{\ell }_{yef}}{i_x}=\frac{1200}{6.39}=187.8⊲250$$

Flexibility is less than limiting λ _prev = 250. Checking the tensile strength of the lower belt:

$$\sigma =\frac{N_{\mathrm{fifteen}-16}}{A}=\frac{\mathrm{11744,2}}{52.69}=222.9<{\gamma }_c{R}_y=0.95\cdot 240=228mP\mathrm{but}$$

The tensile strength of the lower belt of an equidistant I-beam profile is provided.

After the farm is installed in the design position, the concrete hoses are connected to the pipes of the closed tubular cavities of the farm elements, injected, fine-grained expanding concrete is introduced into the cavity of the elements with the formation of concrete elements after concrete setting.

They increase the survivability of the entire steel truss by 1.5 ... 1.6 times, exclude the possibility of a sudden loss of stability of each of the reinforced compressed elements and transfer the work of the entire truss structure from the dangerous stage of work in the first limit state to more favorable work in the second limit state.

To reduce steel consumption, stretched truss elements are made of low alloy steel.

Figure 1 shows a diagram of a farm in which the survivability is increased by the proposed method, having a compressed upper 1 belt from an equally stable I-beam. Rising compressed support braces, intermediate compressed 3 braces in the truss plane are unfastened with two trusses, this reduces their flexibility in the truss plane by three times. All racks 5 and compressed braces are made of concrete pipes. The descending intermediate stretched 4 braces of an equally stable I-beam profile [9, 14].

The farms are equipped with a system of cross ties along the upper and lower zones, as well as vertical cross ties. In each temperature compartment of the building in the coating system there are spatial communication blocks (in the middle of the compartment and its ends). Intermediate trusses are linked to spatial tie blocks by struts and girders. Consequently, all trusses are unfastened from the plane by ties.

The bearing capacity and survivability of the entire steel truss is increased by 1.5 ... 1.6 times, turning all compressed elements into pipe-concrete and eliminating the possibility of a sudden loss of stability of each of the pipe-concrete compressed elements, and the braces have an oval cross section of 3: 1. The work of the entire farm structure is transferred from the dangerous stage of work in the first limit state to the more favorable stage in the second limit state.

Bibliography

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Claims (1)

A method of increasing the survivability of a steel truss with ascending compressed support and intermediate braces, uprights and descending stretched braces, which consists in the fact that the upper and lower truss belts are made of equally stable I-profiles, in the most stressed zone in the middle of the span, the upper belt is transformed into a tubular closed by welding from the sides to the equally stable I-profile of the trailing sheets, all the compressed elements of the truss lattice are made of oval pipes with a larger to smaller ratio it is equal to three, orienting the large dimension perpendicular to the plane of the farm, and in the plane of the farm the compressed braces are fixed with two trusses, this reduces their flexibility by three times and significantly increases their bearing capacity (stability), after the farm is installed in the design position, the concrete piping hoses are connected to the nozzles closed tubular elements of the farm, fine-grained expanding concrete is injected into the cavity of the elements with the formation of concrete elements after setting concrete, increase the survivability of the entire steel Fairy we 1.5 ... 1.6 times preclude sudden buckling of each fortified compressed struts and translate operation of the entire structure of the truss dangerous work stage at the first limit state more favorable operation in the second limit state.

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