Often, people who make the indoor canopy in the yard for a car or to protect against the sun and atmospheric precipitation, the cross-section of the racks, which will be relying a canopy, do not count, but choose the cross section on the eye or consulting the neighbor.

You can understand them, the loads on the racks, in this case are columns, not so what large, the volume of work performed is also not huge, and the appearance of the columns is sometimes much more important than their carrying ability, so even if the columns are made with a multiple margin by strength - There is no big trouble in this. Moreover, in search of simple and intelligible information on the calculation of solid columns, you can spend an infinite amount of time without any result - to understand the examples of calculating the columns for production buildings with the load application in several levels without good knowledge of the conversion is almost impossible, and ordering the calculation of the column The engineering organization can reduce all the expected savings to zero.

This article is written in order to further change the existing state of affairs and is an attempt to maximally state the main stages of the calculation of the metal column, no more. All major requirements for the calculation of metal columns can be found in SNiP II-23-81 (1990).

General provisions

From theoretical point of view, the calculation of the central-compressed element, which is a column, or a rack in a farm, so simple that it is even uncomfortable to talk about it. It is enough to divide the load on the calculated resistance of the steel, from which the column will be manufactured - all. In mathematical expression it looks like this:

F \u003d N / R y. (1.1)

F. - Required cross section area, see

N. - a concentrated load attached to the center of gravity of the cross section of the column, kg;

R. y. - Estimated metal resistance to stretching, compression and bending over the yield strength, kg / cm & sup2. The value of the calculated resistance can be determined by the corresponding table.

As you can see, the level of complexity of the task refers to the second, the maximum of the third grade of primary school. However, in practice, everything is not as simple as in theory, for a number of reasons:

1. Attach the concentrated load exactly to the center of gravity of the cross section of the column can be only theoretically. In reality, the load will always be distributed and there will still be some eccentricity of the application of the above concentrated load. And since there is an eccentricity, it means there is a longitudinal bending moment acting in the cross section of the column.

2. The centers of gravity of the cross sections of the columns are located on one straight line - the central axis is also only theoretically. In practice, due to the heterogeneity of the metal and various defects, the centers of the severity of the cross sections can be shifted relative to the central axis. This means that the calculation should be done in a section, the center of severity of which is removed as much as possible from the central axis, which is why the eccentricity of the force for this cross section is maximum.

3. The column may not have a straightforward form, but be a bit curved as a result of the factory or installation deformation, which means that the cross sections in the middle part of the column will have the highest eccentricity of the load application.

4. The column can be installed with the deviations from the vertical, which means that the vertically active load can create an additional bending moment, maximum in the lower part of the column, and if more precisely, at the fastening site to the foundation, however, it is relevant, only for separate columns. .

5. Under the action of the loads applied to it, the column can be deformed, which means that the eccentricity of the load application appears again and as a result, an additional bending moment.

6. Depending on how the column is fixed, the value of the additional bending moment below and in the middle part of the column depends.

All this leads to the appearance of longitudinal bending and the influence of this bend in the calculations should be somehow taken into account.

Naturally, it is almost impossible to calculate the above deviations for the design, which is still designed, it will be practically impossible - the calculation will be very long, difficult, and the result is still dubious. But to introduce a certain coefficient in formula (1.1), which would take into account the foregoing factors, it is very possible. This coefficient is φ - The coefficient of longitudinal bending. The formula in which this coefficient is used, it looks like this:

F \u003d n / φR (1.2)

Value φ Always less than one, it means that the cross section of the column will always be greater than if you simply calculate the formula (1.1), it is for the fact that now the most interesting thing will begin and remember that φ Always less than one - it does not hurt. For preliminary calculations, you can use the value φ in the range of 0.5-0.8. Value φ Depends on the steel grade and flexibility of the column λ :

λ = l. EF / i. (1.3)

l. EF. - The estimated length of the column. The estimated and real length of the column is different concepts. The estimated length of the column depends on the method of fixing the ends of the column and is determined using the coefficient μ :

l. EF \u003d μ. l. (1.4)

l. - the real length of the column, cm;

μ - coefficient, taking into account the method of fixing the ends of the column. The value of the coefficient can be determined by the following table:

Table 1.The coefficients μ to determine the calculated lengths of the columns and stands of constant cross section (according to SNiP II-23-81 (1990))

As we see, the value of the coefficient μ varies several times depending on the method of fixing the column and then the main complexity is exactly what the calculation scheme is to choose. If you do not know which attachment scheme matches your conditions, then take the value of the coefficient μ \u003d 2. The value of the coefficient μ \u003d 2 is made mainly for separately standing columns, a visual example of a separate column - a lamppost. The value of the coefficient μ \u003d 1-2 can be taken for the shed columns, which are based on the beams without rigid mounting to the column. This calculation scheme can be taken when the canopy beams will not be fastened to columns and when the beams will have a relatively big deflection. If a farm is rigged on the column, rigidly attached to the welding column, then you can take the value of the coefficient μ \u003d 0.5-1. If diagonal connections are between the columns, you can take the value of the coefficient μ \u003d 0.7 with a non-rigid fastening of diagonal bonds or 0.5 with a hard mount. However, such rigidity diaphragms are not always in 2 planes and therefore it is necessary to use such the values \u200b\u200bof the coefficient. When calculating the racks of farms, the coefficient μ \u003d 0.5-1 is used, depending on the attachment method of the racks.

The value of the flexibility coefficient approximately shows the ratio of the calculated length of the column to the height or the width of the cross section. Those. The greater the value λ The less the width or height of the cross section of the column and, accordingly, the larger margin of the cross section will be required at the same length of the column, but it is a little later.

Now when we determined the coefficient μ can calculate the estimated length of the column according to formula (1.4), and in order to find out the value of the flexibility of the column, you need to know the radius of the inertia of the cross section of the column i. :

where I. - the moment of inertia of the cross section relative to one of the axes, and here the most interesting begins, because during the solution of the problem, we must define the necessary area of \u200b\u200bthe cross section of the column F.But this is not enough, it turns out, we still need to know the meaning of the moment of inertia. Since we do not know whether neither the other, the solution of the problem is performed in several stages.

At the preliminary stage, the value is usually taken. λ In the range of 90-60, the column with a relatively small load can be taken λ \u003d 150-120 (the maximum value for the columns - 180, the values \u200b\u200bof the limit flexibility for other elements can be found on table 19 * SNiP II-23-81 (1990). Then Table 2 defines the value of the flexibility coefficient φ :

Table 2. Coefficients of longitudinal bending φ of centrally compressed elements.

Note: The values \u200b\u200bof the coefficient φ The table is equal to 1000 times.

After that, the required radius of the inertia of the cross section is determined by transforming formula (1.3):

i. = l. EF /λ (1.6)

According to the assortment, a rolling profile is selected with the corresponding value of the inertia radius. In contrast to the bending elements, where the cross section is selected only by one axis, since the load is valid only in one plane, in the centrally compressed columns, the longitudinal bend can occur relative to any of the axes and there is the closer the value I z to i y, the better, by others With the words most preferred circular profiles or square sections. Well, now let's try to determine the cross section of the column based on the knowledge gained.

Example of calculating the metal central-compressed column

There is: the desire to make a canopy near the house of about the following type:

In this case, the only central-compressed column under any fixing conditions and with a uniformly distributed load will be a column shown in the figure in red. In addition, the load on this column will be maximum. The columns designated in the figure in blue and green can be considered as centrally compressed, only with the corresponding structural solution and uniform-distributed load, the columns designated by orange will be either centrally compressed or high-centered or frame racks calculated separately. In this example, we calculate the cross section of the column designated in red. For calculations, we will take a constant load from our own weight of the carport 100 kg / m & sup2 and the temporary load of 100 kg / m & sup2 from snow cover.

2.1. Thus, the concentrated load on the column designated in red will be:

N \u003d (100 + 100) · 5 · 3 \u003d 3000 kg

2.2. Take pre-value λ \u003d 100, then Table 2 Bend Coefficient φ \u003d 0.599 (for steel with a computational strength of 200 MPa, this value is made to provide an additional stock by strength), then the required area of \u200b\u200bthe column cross section:

F. \u003d 3000 / (0.599 · 2050) \u003d 2.44 cm & sup2

2.3. Table 1 Take the value μ \u003d 1 (as the roofing coating of the profiled flooring, properly fixed, will provide the stiffness of the structure in the plane parallel to the wall plane, and in the perpendicular plane, the relative immobility of the top point of the column will ensure the attachment of the rafter to the wall), then the inertia radius

i. \u003d 1 · 250/100 \u003d 2.5 cm

2.4. According to the sortiment for square profile tubes, this requirement satisfies the profile with the size of the cross section 70x70 mm with a wall thickness of 2 mm, having an inertia radius of 2.76 cm. The cross section of such a profile is 5.34 cm & sup2. It is much more than required by calculation.

2.5.1. We can increase the flexibility of the column, while the required radius of inertia will decrease. For example, for λ \u003d 130 bend coefficient φ \u003d 0.425, then the required cross-sectional area of \u200b\u200bthe column:

F \u003d 3000 / (0,425 · 2050) \u003d 3.44 cm & sup2

2.5.2. Then

i. \u003d 1 · 250/130 \u003d 1.92 cm

2.5.3. According to the grangle for square profile tubes, these requirements satisfies the profile with a cross-sectional size of 50x50 mm with a wall thickness of 2 mm having a radius of inertia 1.95 cm. The cross section of such a profile of 3.74 cm & sup2, the moment of resistance for this profile is 5.66 cm & sup3.

Instead of square profile pipes, you can use an equilibrium corner, a channel, a 2-way, a conventional pipe. If the calculated resistance of the steel profile has become more than 220 MPa, then you can recalculate the cross section of the column. Here in principle, and everything related to the calculation of metal centrally compressed columns.

Calculation of an echocent-compressed column

Here, of course, the question arises: how to calculate the remaining columns? The answer to this question is highly dependent on the method of fastening a canopy to columns. If the beams of the canopy are hard to fasten to the columns, then a rather complicated statically undefined frame will be formed and then the columns should be considered as part of this frame and calculate the cross section of the columns additionally on the transverse bending moment, we will further consider the situation when the columns shown in the figure , connected with a shed manner (a column marked with red, we no longer consider). For example, the headband of the column has a supporting platform - a metal plate with holes for bolted beams of canopy. For various reasons, the load on such columns can be transmitted with a large enough eccentricity:

The beam shown in the figure, beige, under the influence of the load will slightly bend and will lead to the fact that the load on the column will not be transmitted in the center of severity of the cross section of the column, but with the eccentricity e. And when calculating the extreme columns, this eccentricity must be taken into account. Cases of the extracentrate loading of columns and possible transverse sections of the columns There are a large set described by the corresponding formulas for the calculation. In our case, to check the cross section of the eccentric-compressed column, we will use one of the simplest:

(N / φf) + (m z / w z) ≤ r y (3.1)

In this case, when we have already defined the cross section of the loaded column itself, it is enough to check whether such a section is suitable for the rest of the columns for the reason that the tasks of building the steel plant we do not have, and we simply expect columns for a canopy that will be all the same cross section For reasons of unification.

What N., φ and R. Y we already know.

Formula (3.1) after the simplest transformations, will take the following form:

F \u003d (N / R Y) (1 / φ + E z · f / w z) (3.2)

as M z \u003d n · e zWhy the moment the moment is exactly what is the moment of resistance W, is explained in detail in a separate article.

The columns indicated in the figure blue and green will be 1,500 kg. Check the desired section with such a load and previously defined φ = 0,425

F \u003d (1500/2050) (1 / 0.425 + 2.5 · 3.74 / 5.66) \u003d 0.7317 · (2,353 + 1,652) \u003d 2.93 cm & sup2

In addition, formula (3.2) allows you to determine the maximum eccentricity, which will solve the already calculated column, in this case the maximum eccentricity will be 4.17 cm.

The required cross section of 2.93 cm & sup2 is less than the received 3.74 cm & sup2, and therefore a square profile pipe with a cross-sectional size of 50x50 mm with a wall thickness of 2 mm can also be used for extreme columns.

Calculation of an echocent-compressed conditional flexibility column

Oddly enough, but for the selection of a cross-sectional squeezed column - a solid rod is an even more simple formula:

F \u003d n / φ e. R. (4.1)

φ E. - the coefficient of longitudinal bend, depending on the eccentricity, it would be possible to call the eccentricity coefficient of the longitudinal deflection, so as not to be confused with the coefficient of the longitudinal deflection φ . However, the calculation for this formula may be longer than by formula (3.2). To determine the coefficient φ E. It is necessary to know the value of the expression anyway e z · f / w z - Which we met in formula (3.2). This expression is called the relative eccentricity and is indicated. m.:

m \u003d e z · f / w z (4.2)

After that, the reduced relative eccentricity is determined:

m. eF. \u003d HM. (4.3)

h. - This is not the height of the section, but the coefficient determined by Table 73 of SNIPA II-23-81. Just say that the value of the coefficient h. It vary in the range from 1 to 1.4, for most simple calculations, it is possible to use H \u003d 1.1-1.2.

After that, it is necessary to determine the conditional flexibility of the column λ¯ :

λ¯ \u003d λ√~ (R y / e) (4.4)

and only after that Table 3 determine the value φ e. :

Table 3. The coefficients φ e to test the stability of non-centrane-compressed (compressed-bending) of solid researched rods in the point of action plane coinciding with the symmetry plane.

Notes:

1. The values \u200b\u200bof the coefficient φ E enlarged 1000 times.
2. Meaning φ e should be taken no more φ .

Now, for clarity, check the cross section of the columns loaded with the eccentricity, according to formula (4.1):

4.1. The concentrated load on the columns indicated by blue and green will be:

N \u003d (100 + 100) · 5 · 3/2 \u003d 1500 kg

Eccentricity of the application of the load e. \u003d 2.5 cm, coefficient of longitudinal bending φ = 0,425.

4.2. The value of the relative eccentricity we already determined:

m \u003d 2.5 · 3.74 / 5.66 \u003d 1,652

4.3. Now we define the value of the reduced coefficient m. eF. :

m. eF. \u003d 1,652 · 1.2 \u003d 1.984 ≈ 2

4.4. Conditional flexibility with the flexibility coefficient adopted by us. λ \u003d 130, steel strength R. Y \u003d 200 MPa and elastic module E. \u003d 200000 MPa will be:

λ¯ \u003d 130√~ (200/200000) \u003d 4.11

4.5. Table 3 define the value of the coefficient φ E ≈ 0.249.

4.6. Determine the desired cross section of the column:

F \u003d 1500 / (0.249 · 2050) \u003d 2.94 cm & sup2

Let me remind you that when determining the cross-sectional area of \u200b\u200bthe column according to formula (3.1), we obtained almost the same result.

Tip: So that the load from the canopy is transmitted with the minimum eccentricity, a special platform is made in the support part of the beam. If the beam is metallic, from the rolling profile, then it is usually enough to weld to the bottom shelf of the beam piece of fittings.

Metal designs Theme is complex, extremely responsible. Even a small mistake can cost hundreds of thousands and millions of rubles. In some cases, the time of the error can be the life of people at a construction site, as well as during operation. So what to check and recheck calculations is necessary and important.

Using Excel to solve settlement tasks - the case on one side is not new, but not quite familiar. However, Excel has a number of indisputable advantages:

• Openness - Each such calculation can be disassembled around the bones.
• Availability - The files themselves exist in general access, written by the developers of the MK under their needs.
• Convenience- Almost any PC user is able to work with programs from the MS Office package, while specialized design solutions - roads, and also require serious efforts for their development.

Do not count them with panacea. Such calculations allow you to solve narrow and relatively simple design tasks. But they do not take into account the work of the design as a whole. In a number of ordinary cases, a lot of time can save:

• Calculation of bending beams
• Calculation of bending beam online
• Check the calculation of the strength and stability of the column.
• Check the selection of the cross section of the rod.

Universal Settlement File MK (Excel)

Table for sections of metal structures, 5 different points of the SP 16.13330.2011
Actually, using this program, you can perform the following calculations:

• calculation of a single-break hinge beam.
• the calculation is centrally compressed elements (columns).
• calculation of stretched elements.
• calculation of nonsense-compressed or compressed-bending elements.

Excel version should not be lower than 2010. To see the instructions, click on the plus in the upper left corner of the screen.

Metallica

The program is an Excel book with support for macros.
And is intended to calculate steel structures according to
SP16 13330.2013 "Steel structures"

Selection and calculation of runs

Selection of a run - the task only at first glance is trivial. The step of the run and their size depends on many parameters. And it would be nice to have an appropriate calculation at hand. Actually about this and tells the article Mandatory to familiarize:

• calculation of the run without heavy
• the calculation of the run of one gravity
• calculation of the run with two heavy
• bimmer calculation of the Bimome:

But there is a small spoon of tar - apparently in the file there are errors in the settlement part.

Calculation of moments of cross section in Excel tables

If you need to quickly calculate the moment of the inertia of the composite section, or there is no possibility to determine the GOST for which the metal structures are made, then this calculator will come to the rescue. At the bottom of the table a small explanation. In general, the work is simple - choose a suitable section, set the dimensions of these sections, we obtain the main parameters of the section:

• Moments of inertia section
• Moments of resistance cross section
• Radius of inertia section
• Cross-sectional area
• Static moment
• Distances to the center of severity.

The table implements calculations for the following types of sections:

• trumpet
• rectangle
• iTOVR
• channel
• rectangular trumpet
• triangle

Calculation of efforts in racks are made taking into account the loads applied to the rack.

Middle racks

The middle racks of the building frame work and are calculated as the centrally compressed elements on the action of the greatest compressive force n from its own weight of all coating structures (G) and snow load and snow load (p sN).

Figure 8 - Loads on the middle rack

The calculation of the centrally compressed average racks produced:

a) for strength

where - the calculated resistance of wood compression along the fibers;

Nette-cross section area of \u200b\u200bthe element;

b) on stability

where is the longitudinal bend coefficient;

- the calculated cross-sectional area of \u200b\u200bthe element;

The loads are collected from the area of \u200b\u200bthe coating according to the plan coming on one middle rack ().

Figure 9 - Freight areas of the middle and extreme columns

Extreme racks

The extreme stand is under the action of longitudinal axes of the load rack (G and R sN) who are collected from the square and transverse, and H. In addition, a longitudinal force arises from the wind action.

Figure 10 - Loads on the extreme rack

G is the load from the own weight of the coating designs;

X is a horizontal concentrated force attached at the point of the beelel adjoining to the rack.

In the case of tight sealing racks for a single-spare frame:

Figure 11 - Load Scheme when rigid pinched racks in the foundation

where - horizontal wind loads, respectively, from the wind to the left and right, attached to the rack in the place of the brightness of the riglel.

where is the height of the reference cross section of the beegel or beam.

The influence of the forces will be significantly if the rigle on the support has a significant height.

In the case of a hinge based on the foundation for a single-spare frame:

Figure 12 - Load Scheme with Hinged Packing Racks on the foundation

For multiplet frame structures under the wind on the left P 2 and W 2, and in the wind to the right P 1 and W 2 will be zero.

Extreme racks are calculated as compressed-bending elements. The values \u200b\u200bof the longitudinal force N and the bending moment M are accepted for such a combination of loads in which the largest compressive stresses occur.

1) 0.9 (G + P C + wind on the left)

2) 0.9 (G + P C + wind on the right)

For a rack included in the frame, the maximum bending moment is taken as Max from the wind calculated for the case of the left M L and on the right m:

where e is the eccentricity of the application of the longitudinal force N, which includes the most unfavorable combination of loads G, P C, P B - each with his sign.

The eccentricity for racks with a constant height of section is zero (E \u003d 0), and for the racks with a variable height of section is taken as the difference between the geometric axis of the reference and the axis of the application of the longitudinal force.

Calculation of compressed - curved extreme racks produced:

a) for strength:

b) on the stability of a flat mold of bending in the absence of fixing or at the calculated length between the fixing points L p\u003e 70b 2 / n by the formula:

The geometrical characteristics included in the formulas are calculated in the reference section. From the plane, the rack frame is calculated as a centrally compressed element.

Calculation of compressed and compressed-curved compoundit is produced according to the above formulas, however, when calculating the coefficients φ and ξ in these formulas, an increase in the flexibility of the rack due to the advantage of the bonds connecting branches is taken into account. This enlarged flexibility is called the flexibility λ n.

Calculation of lattice racks You can reduce the calculation of farms. In this case, the wind uniformly distributed load is reduced to focused loads in the farm nodes. It is believed that the vertical forces G, P C, P B are perceived only by the belts of the rack.

Calculation of the central rack

Stands are called structural elements that work mainly on compression and longitudinal bending.

When calculating the rack, it is necessary to ensure the strength and stability. Supporting stability is achieved by proper selection of the rack section.

The design scheme of the central rack is accepted when calculating the vertical load, as it is hinged at the ends, because at the bottom and at the top welded with welding (see Figure 3).

The central stand perceives 33% of the total overlap weight.

The total weight of the overlap N, kg is defined: includes snow weight, wind load, load from thermal insulation, load on the weight of the coating frame, load from vacuum.

N \u003d R 2 G,. (3.9)

where G is the total uniform-distributed load, kg / m 2;

R is an internal radius of the tank, m.

The total weight of the overlap folds from the following types of loads:

• 1. Snow load, G 1. G 1 \u003d 100 kg / m 2 is accepted.;
• 2. Load from thermal insulation, G 2. G 2 \u003d 45 kg / m 2 is accepted;
• 3. Wind load, G 3. G 3 \u003d 40 kg / m 2 is accepted;
• 4. Load from the weight of the coating frame, G 4. G 4 \u003d 100 kg / m 2 is accepted
• 5. Taking into account the installed equipment, G 5. Accepted G 5 \u003d 25 kg / m 2
• 6. Load from vacuum, G 6. Accepted G 6 \u003d 45 kg / m 2.

And the total weight of the overlap N, kg:

The effort perceived by the resistant is calculated:

The required rack cross section is determined by the following formula:

CM 2, (3.12)

where: N-full overlap weight, kg;

1600 kgf / cm 2, for Steel InstallationSp;

The coefficient of longitudinal bending is constructively accepted \u003d 0.45.

According to GOST 8732-75, a pipe with an outer diameter D H \u003d 21cm, the inner diameter D b \u003d 18 cm and the wall thickness of 1.5 cm, which is permissible since the cavity of the pipe will be filled with concrete.

Pipe cross sections, F:

The moment of profile inertia (j) is determined, the inertia radius (R). Respectively:

J \u003d cm4, (3.14)

where - the geometric characteristics of the section.

Inertia Radius:

r \u003d, cm, (3.15)

where J is the moment of profile inertia;

F-area of \u200b\u200bthe required section.

Flexibility:

The stance in the rack is determined by the formula:

Kgf / cm (3.17)

At the same time, according to the tables of Appendix 17 (A. N. Serenko) is accepted \u003d 0.34

Calculation of the Stand Base Strength

The calculated pressure P on the foundation is determined:

P \u003d p "+ p st + r bs, kg, (3.18)

P st \u003d f l g, kg, (3.19)

P bs \u003d L g b, kg, (3.20)

where: p "-usilize vertical rack p" \u003d 5885.6 kg;

P st - Ints, kg;

r - share of steel.g \u003d 7.85 * 10 -3 kg.

R BS - Vesbeton poured in a rack rack, kg;

g, B-tarma of concrete brand.g b \u003d 2.4 * 10 -3 kg.

The required area of \u200b\u200bthe shoe plate with a pressure pressure on the sandy base [y] f \u003d 2 kg / cm 2:

The slab is accepted with the parties: Achb \u003d 0.65h0.65 m. Alloredated load, Q 1 cm plate will be determined:

Estimated bending moment, m:

The estimated moment of resistance, w:

Plate thickness D:

The thickness of the plate d \u003d 20 mm is accepted.

In practice, it often arises the need to calculate the rack or colony to the maximum axial (longitudinal) load. The force in which the rack loses the stable state (carrier) is critical. The resistance of the rack is influenced by the method of fixing the end of the rack. In the construction mechanics we consider seven ways to fix the ends of the rack. MI Consider three main ways:

To ensure a specific sustainability stock, it is necessary that the condition is followed:

Where: p - active effort;

Sets a certain stability factor

Thus, when calculating elastic systems it is necessary to determine the magnitude of the critical power of the RCD. If you have a penalty that the force p is applied to the rack causes only small deviations from the straightforward shape of the rack-length ι, then it can be determined from the equation

where: E is an elastic module;
J_min- minimum moment of inertia;
M (z) - bending moment equal to m (z) \u003d -p ω;
ω is the value of the deviation from the straightforward form of the rack;
Solving it differential equation

A and in constant integration, are determined by boundary conditions.
By producing certain actions and substitutions, we obtain the final expression for the critical force

The smallest value of the critical force will be at n \u003d 1 (integer) and

The equation of the elastic line of the rack will look at:

where: z is the current ordinate, with the maximum value z \u003d L;
A permissible expression for critical force is called Formula L. Seiler. It can be seen that the magnitude of the critical force depends on the rigidity of the EJ MIN rack is directly proportional to the length of the l - back proportionally.
As mentioned, the stability of the elastic rack depends on the method of its consolidation.
Recommended Strest Stock For Steel Racks
N y \u003d 1.5 ÷ 3.0; for wooden N y \u003d 2.5 ÷ 3.5; For cast iron N y \u003d 4.5 ÷ 5.5
To account for the method of fixing the ends of the rack, the coefficient of the end of the reduced rack flexibility is introduced.

where: μ is the coefficient of the length (table);
I min - the smallest radius of the inertia of the cross section of the rack (table);
ι - length of the rack;
Enter the critical load coefficient:

, (table);
Thus, when calculating the cross section of the rack, it is necessary to take into account the coefficients μ and θ of which the magnitude of which depends on the method of fixing the ends of the rack and is given in the tables of the reference book on the concomitant (G.S. Parenko and S.P. Fesik)
We give an example of calculating the critical force for the rod of the continuous cross-section of the rectangular shape - 6 × 1 cm., The length of the rod ι \u003d 2m. Fixing ends according to Scheme III.
Payment:
On the table we find the coefficient θ \u003d 9.97, μ \u003d 1. The moment of inertia of the section will be:

and the critical tension will be:

Obviously, the critical force r \u003d 247 kgf will cause a voltage of only 41kc / cm 2 in the rod, which is significantly less than the limit of the flow rate (1600kgs / cm 2), but this force causes the curvature of the rod, which means that the loss of stability.
Consider another example of calculating the wooden rack of the circular section pinched in the lower end and hingedly fixed on the top (S.P. Fesik). Stand length 4m, compression force n \u003d 6TS. Allowable voltage [σ] \u003d 100kgs / cm 2. We accept the coefficient of lowering the allowable voltage to the compression φ \u003d 0.5. Calculate the cross section of the rack:

Determine the diameter of the rack:

Moment of inertia section

Calculate the flexibility of the rack:
where: μ \u003d 0.7, based on the method of pinching the ends of the rack;
Determine the stress in the rack:

Obviously, the stress in the rack is 100kgs / cm 2 and it is exactly allowed voltage [σ] \u003d 100kc / cm 2
Consider the third example of calculating the steel rack from the 2-way profile, a length of 1.5m, a compression force 50tes, allowable voltage [σ] \u003d 1600kgs / cm 2. The lower end of the rack is pinched, and the upper free (i method).
For sections, we use the formula and specify the coefficient φ \u003d 0.5, then:

We select 2All No. 36 from the sorting and its data: f \u003d 61.9cm 2, i min \u003d 2.89cm.
Determine the flexibility of the rack:

where: μ from the table, smooth 2, given the method of pinching the rack;
The calculated stress in the rack will be:

5kgs, which is approximately exactly allowed voltage, and by 0.97% more, which is permissible in engineering calculations.
The cross-section of the squeezing rods will be rational with the largest inertia radius. When calculating the specific inertia radius
The most optimal is tubular sections, thin-walled; For which the value ξ \u003d 1 ÷ 2.25, and for solid or rolling profiles ξ \u003d 0.204 ÷ 0.5

conclusions
When calculating the strength and stability of the racks, the column must be taken into account the method for fastening the ends of the racks, apply the recommended margin of safety.
The value of the critical force was obtained from the differential equation of the curved axial line of the rack (L.Aeler).
For accounting for all factors that characterize the loaded rack, the concept of the rack flexibility - λ was introduced, the coefficient of the valid length - μ, the coefficient of lowering the voltage - φ, the critical load coefficient is θ. Their values \u200b\u200bare taken from the tables of reference books (S.Pisarentko and S.P. Fesik).
Approximate calculations of the racks are given, to determine the critical force - RCR, the critical stress - σkr, the diameter of the racks - d, the flexibility of the racks - λ and other characteristics.
The optimal cross section for racks and columns is tubular thin-walled profile with the same main moments of inertia.

Used Books:
G.S. Pisarenko "Handbook of Material Resistance".
S.P.Fesik "Certificate of Material Resistance".
IN AND. Anuryev "Directory of Designer-Machine Builder".
SNIP II-6-74 "Loads and Impact, Design Norms".