Calculation of metal racks. The procedure for performing stability calculations. Calculation of the central pillar

The forces in the racks are calculated taking into account the loads applied to the rack.

B-pillars

The middle pillars of the building frame work and are calculated as centrally compressed elements to the action of the greatest compressive force N from the own weight of all coating structures (G) and snow load and snow load (P sn).

Figure 8 – Loads on the middle pillar

Calculation of centrally compressed middle pillars is carried out:

a) for strength

Where - design resistance wood is compressed along the grain;

Net area cross section element;

b) for stability

where is the coefficient longitudinal bending;

– calculated cross-sectional area of ​​the element;

Loads are collected from the coverage area according to the plan, per one middle pillar ().

Figure 9 – Loading areas of the middle and outer columns

End posts

The outermost post is under the influence of longitudinal loads relative to the axis of the post (G and P sn), which are collected from the area and transverse, and X. In addition, longitudinal force arises from the action of wind.

Figure 10 – Loads on the outer pillar

G – load from the dead weight of the coating structures;

X – horizontal concentrated force applied at the point of contact of the crossbar with the rack.

In the case of rigid embedding of racks for a single-span frame:

Figure 11 – Scheme of loads during rigid pinching of racks in the foundation

where are the horizontal wind loads, respectively, from the wind on the left and right, applied to the post at the point where the crossbar adjoins it.

where is the height of the supporting section of the crossbar or beam.

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

In the case of hinged support of the rack on the foundation for a single-span frame:

Figure 12 – Load diagram for hinged support of racks on the foundation

For multi-span frame structures, when there is wind from the left, p 2 and w 2, and when there is wind from the right, p 1 and w 2 will be equal to zero.

The outer pillars are calculated as compressed-bending elements. The values ​​of the longitudinal force N and the bending moment M are taken for the combination of loads at which the greatest compressive stresses occur.

1) 0.9(G + P c + wind from left)

2) 0.9(G + P c + wind from the right)

For a post included in the frame, the maximum bending moment is taken as max from those calculated for the case of wind on the left M l and on the right M in:

where e is the eccentricity of the application of longitudinal force N, which includes the most unfavorable combination of loads G, P c, P b - each with its own sign.

The eccentricity for racks with a constant section height is zero (e = 0), and for racks with a variable section height it is taken as the difference between the geometric axis of the supporting section and the axis of application of the longitudinal force.

Calculation of compressed - curved outer pillars is carried out:

a) for strength:

b) for the stability of a flat bending shape in the absence of fastening or with an estimated length between the fastening points l p > 70b 2 /n according to the formula:

The geometric characteristics included in the formulas are calculated in the reference section. From the plane of the frame, the struts are calculated as a centrally compressed element.

Calculation of compressed and compressed-bent composite sections is carried out according to the above formulas, however, when calculating the coefficients φ and ξ, these formulas take into account the increase in the flexibility of the rack due to the compliance of the connections connecting the branches. This increased flexibility is called reduced flexibility λn.

Calculation of lattice racks can be reduced to the calculation of trusses. In this case, the uniformly distributed wind load is reduced to concentrated loads in the nodes of the truss. It is believed that vertical forces G, P c, P b are perceived only by the strut belts.

P the building frame (Fig. 5) is once statically indeterminate. We reveal the indeterminacy based on the condition of equal rigidity of the left and right struts and the same magnitude of horizontal displacements of the hinged end of the struts.

Rice. 5. Design diagram of the frame

5.1. Determination of geometric characteristics

1. Rack section height
. Let's accept
.

2. The width of the rack section is taken according to the assortment, taking into account the shank
mm .

3. Sectional area
.

Section moment of resistance
.

Static moment
.

Section moment of inertia
.

Section radius of gyration
.

5.2. Load collection

a) horizontal loads

Linear wind loads

, (N/m)

,

Where - coefficient taking into account the value of wind pressure in height (Appendix Table 8);

- aerodynamic coefficients (at
m accept
;
);

- load reliability factor;

- standard value of wind pressure (as specified).

Concentrated forces from wind load at the level of the top of the rack:

,
,

Where - supporting part of the farm.

b) vertical loads

We will collect the loads in tabular form.

Table 5

Collection of load on the rack, N

Note:

1. The load from the covering panel is determined according to table 1

,
.

2. The load from the beam is determined

.

3. Own weight of the arch
defined:

Upper belt
;

Bottom belt
;

Racks.

To obtain the design load, the arch elements are multiplied by , corresponding to metal or wood.

,
,
.

Unknown
:
.

Bending moment at the base of the post
.

Lateral force
.

5.3. Verification calculation

In the bending plane

1. Check for normal voltages

,

Where - coefficient taking into account the additional moment from the longitudinal force.

;
,

Where - consolidation coefficient (assume 2.2);
.

Undervoltage should not exceed 20%. However, if accepted minimum dimensions racks and
, then the undervoltage can exceed 20%.

2. Checking the supporting part for chipping during bending

.

3. Checking the stability of plane deformation:

,

Where
;
(Table 2 app. 4).

From the bending plane

4. Stability test

,

Where
, If
,
;

- the distance between the connections along the length of the rack. In the absence of connections between the racks, the total length of the rack is taken as the estimated length
.

5.4. Calculation of attaching the rack to the foundation

Let's write out the loads
And
from Table 5. The design of attaching the rack to the foundation is shown in Fig. 6.

Where
.

Rice. 6. Design of attaching the rack to the foundation

2. Compressive stress
, (Pa)

Where
.

3. Dimensions of compressed and stretched zones
.

4. Dimensions And :

;
.

5. Maximum tensile force in anchors

, (N)

6. Required area of ​​anchor bolts

,

Where
- coefficient taking into account thread weakening;

- coefficient taking into account stress concentration in the thread;

- coefficient taking into account the uneven operation of two anchors.

7. Required anchor diameter
.

We accept the diameter according to the assortment (Appendix Table 9).

8. For the accepted diameter of the anchor, a hole in the traverse will be required
mm.

9. Width of traverse (angle) fig. 4 must be at least
, i.e.
.

Let's take an isosceles angle according to the assortment (Appendix Table 10).

11. The magnitude of the distribution load along the width of the rack (Fig. 7 b).

.

12. Bending moment
,

Where
.

13. Required moment of resistance
,

Where - the design resistance of steel is assumed to be 240 MPa.

14. For a pre-adopted corner
.

If this condition is met, we proceed to checking the voltage; if not, we return to step 10 and accept a larger angle.

15. Normal stresses
,

Where
- coefficient of working conditions.

16. Traverse deflection
,

Where
Pa – modulus of elasticity of steel;

- maximum deflection (accept ).

17. Select the diameter of the horizontal bolts from the condition of their placement across the fibers in two rows along the width of the rack
, Where
- distance between bolt axes. If we accept metal bolts, then
,
.

Let us take the diameter of the horizontal bolts according to the appendix table. 10.

18. The smallest load-bearing capacity of a bolt:

a) according to the condition of collapse of the outermost element
.

b) according to the bending condition
,

Where
- application table. eleven.

19. Number of horizontal bolts
,

Where
- the smallest load-bearing capacity from clause 18;
- number of slices.

Let's take the number of bolts even number, because We arrange them in two rows.

20. Overlay length
,

Where - the distance between the axes of the bolts along the fibers. If the bolts are metal
;

- number of distances along the length of the overlay.

In practice, it often becomes necessary to calculate a rack or column for the maximum axial (longitudinal) load. The force at which the rack loses its stable state (bearing capacity) is critical. The stability of the rack is influenced by the way the ends of the rack are secured. In structural mechanics, seven methods are considered for securing the ends of a strut. We will consider three main methods:

To ensure a certain margin of stability, it is necessary that the following condition be met:

Where: P - effective force;

A certain stability factor is established

Thus, when calculating elastic systems, it is necessary to be able to determine the value of the critical force Pcr. If we take into account that the force P applied to the rack causes only small deviations from the rectilinear shape of the rack of length ι, then it can be determined from the equation

where: E - elastic modulus;
J_min - minimum moment of inertia of the section;
M(z) - bending moment equal to M(z) = -P ω;
ω - the amount of deviation from the rectilinear shape of the rack;
Solving this differential equation

A and B are constants of integration, determined by the boundary conditions.
After performing certain actions and substitutions, we obtain the final expression for the critical force P

The minimum value of the critical force will be for n = 1 (integer) and

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

where: z - current ordinate, with maximum value z=l;
An acceptable expression for the critical force is called L. Euler's formula. It can be seen that the magnitude of the critical force depends on the rigidity of the strut EJ min in direct proportion and on the length of the strut l - in inverse proportion.
As mentioned, the stability of the elastic strut depends on the method of its fastening.
The recommended safety factor for steel racks is
n y =1.5÷3.0; for wooden n y =2.5÷3.5; for cast iron n y =4.5÷5.5
To take into account the method of securing the ends of the rack, the coefficient of the ends of the reduced flexibility of the rack is introduced.

where: μ - reduced length coefficient (Table);
i min - the smallest radius of gyration of the cross section of the rack (table);
ι - length of the stand;
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, the value of which depends on the method of securing the ends of the rack and is given in the tables of the strength of materials reference book (G.S. Pisarenko and S.P. Fesik)
Let us give an example of calculating the critical force for a solid cross-section rod rectangular shape- 6×1 cm, rod length ι = 2 m. Fastening the ends according to scheme III.
Calculation:
From the table we find the coefficient ϑ = 9.97, μ = 1. The moment of inertia of the section will be:

and the critical voltage will be:

Obviously, the critical force P cr = 247 kgf will cause a stress in the rod of only 41 kgf/cm 2, which is significantly less than the flow limit (1600 kgf/cm 2), however, this force will cause bending of the rod, and therefore loss of stability.
Let's look at another example of calculating a wooden stand round section pinched at the lower end and hinged at the upper (S.P. Fesik). Rack length 4m, compression force N=6t. Allowable stress [σ]=100kgf/cm2. We accept the reduction factor for the permissible compressive stress φ=0.5. We calculate the cross-sectional area of ​​the rack:

Determine the diameter of the stand:

Section moment of inertia

We calculate the flexibility of the rack:
where: μ=0.7, based on the method of pinching the ends of the rack;
Determine the voltage in the rack:

Obviously, the voltage in the rack is 100 kgf/cm 2 and it is equal to the permissible voltage [σ] = 100 kgf/cm 2
Let's consider the third example of calculating a steel rack made of an I-profile, 1.5 m long, compression force 50 tf, permissible stress [σ] = 1600 kgf/cm 2. The lower end of the rack is pinched, and the upper end is free (method I).
To select the cross section, we use the formula and set the coefficient ϕ=0.5, then:

We select I-beam No. 36 from the assortment and its data: F = 61.9 cm 2, i min = 2.89 cm.
Determining the flexibility of the rack:

where: μ from the table, equal to 2, taking into account the method of pinching the rack;
The calculated voltage in the rack will be:

5 kgf, which is approximately equal to the permissible voltage, and 0.97% more, which is acceptable in engineering calculations.
The cross-section of rods working in compression will be rational at the largest radius of gyration. When calculating the specific radius of gyration
the most optimal is tubular sections, thin-walled; for which the value is ξ=1÷2.25, and for solid or rolled profiles ξ=0.204÷0.5

conclusions
When calculating the strength and stability of racks and columns, it is necessary to take into account the method of securing the ends of the racks and apply the recommended safety factor.
The critical force value is obtained from differential equation curved center line racks (L. Euler).
To take into account all the factors characterizing a loaded rack, the concept of rack flexibility - λ, provided length coefficient - μ, voltage reduction coefficient - ϕ, critical load coefficient - ϑ - was introduced. Their values ​​are taken from reference tables (G.S. Pisarentko and S.P. Fesik).
Given approximate calculations racks, to determine the critical force - Pcr, critical stress - σcr, diameter of racks - d, flexibility of racks - λ and other characteristics.
The optimal cross-section for racks and columns is tubular thin-walled profiles with the same main moments of inertia.

Used Books:
G.S. Pisarenko “Handbook on the strength of materials.”
S.P. Fesik “Handbook on the strength of materials.”
IN AND. Anuriev “Handbook of mechanical engineering designer”.
SNiP II-6-74 “Loads and impacts, design standards.”

Calculation of the central pillar

Racks are structural elements that work primarily in compression and longitudinal bending.

When calculating the rack, it is necessary to ensure its strength and stability. Ensuring sustainability is achieved by correct selection rack sections.

When calculating a vertical load, the design diagram of the central pillar is accepted as hinged at the ends, since it is welded at the bottom and top (see Figure 3).

The central post carries 33% of the total weight of the floor.

The total weight of the floor N, kg, will be determined by: including the weight of snow, wind load, load from thermal insulation, load from the weight of the covering frame, load from vacuum.

N = R 2 g,. (3.9)

where g is the total uniformly distributed load, kg/m2;

R - internal radius of the tank, m.

The total weight of the floor consists of the following types of loads:

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

And the total weight of the floor N, kg:

The force perceived by the stand is calculated:

The required cross-sectional area of ​​the rack is determined using the following formula:

See 2, (3.12)

where: N is the total weight of the floor, kg;

1600 kgf/cm 2, for steel VSt3sp;

The buckling coefficient is structurally assumed to be =0.45.

According to GOST 8732-75, a pipe with an outer diameter D h = 21 cm, an inner diameter d b = 18 cm and a wall thickness of 1.5 cm is structurally selected, which is acceptable since the pipe cavity will be filled with concrete.

Pipe cross-sectional area, F:

The moment of inertia of the profile (J) and radius of gyration (r) are determined. Respectively:

J = cm4, (3.14)

where are the geometric characteristics of the section.

Radius of inertia:

r=, cm, (3.15)

where J is the moment of inertia of the profile;

F is the area of ​​the required section.

Flexibility:

The voltage in the rack is determined by the formula:

Kgs/cm (3.17)

In this case, according to the tables in Appendix 17 (A. N. Serenko) it is assumed = 0.34

Calculation of the strength of the rack base

The design pressure P on the foundation is determined:

Р= Р" + Р st + Р bs, kg, (3.18)

Р st =F L g, kg, (3.19)

R bs =L g b, kg, (3.20)

where: P"-force of the vertical stand P"= 5885.6 kg;

R st - weight of the rack, kg;

g - specific gravity of steel. g = 7.85*10 -3 kg/.

R bs - weight concrete poured into the rack, kg;

g b -specific gravity concrete grade.g b =2.4*10 -3 kg/.

The required area of ​​the shoe plate at the permissible pressure on sandy base[y] f =2 kg/cm 2:

A slab with sides is accepted: aChb = 0.65 × 0.65 m. The distributed load, q per 1 cm of the slab will be determined:

Design bending moment, M:

Design moment of resistance, W:

Plate thickness d:

The slab thickness is assumed to be d = 20 mm.

Often people doing in the yard covered canopy for a car or for protection from the sun and precipitation, the cross-section of the posts on which the canopy will rest is not calculated, but the cross-section is selected by eye or by consulting a neighbor.

You can understand them, the loads on the racks, in in this case being columns, not so big, the volume of work performed is also not enormous, and appearance columns are sometimes much more important than them bearing capacity, therefore, even if the columns are made with a multiple margin of strength, there is no big problem in this. Moreover, you can spend an infinite amount of time searching for simple and clear information about the calculation of solid columns without any result - understand the examples of calculating columns for industrial buildings with application of load at several levels without good knowledge strength of strength material is almost impossible, and ordering a column calculation from an engineering organization can reduce all expected savings to zero.

This article was written with the goal of at least slightly changing the current state of affairs and is an attempt to present as simply as possible the main stages of calculating a metal column, nothing more. All basic requirements for the calculation of metal columns can be found in SNiP II-23-81 (1990).

General provisions

From a theoretical point of view, the calculation of a centrally compressed element, such as a column or rack in a truss, is so simple that it is even inconvenient to talk about it. It is enough to divide the load by the design resistance of the steel from which the column will be made - that’s all. In mathematical expression it looks like this:

F = N/Ry (1.1)

F- required cross-sectional area of ​​the column, cm²

N- concentrated load applied to the center of gravity of the cross section of the column, kg;

Ry- the calculated resistance of the metal to tension, compression and bending at the yield point, kg/cm². The value of the design resistance can be determined from the corresponding table.

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

1. Applying a concentrated load exactly to the center of gravity of the cross-section of a column is only possible theoretically. In reality, the load will always be distributed and there will still be some eccentricity in the application of the reduced concentrated load. And since there is 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 column are located on one straight line - the central axis, also only theoretically. In practice, due to the heterogeneity of the metal and various defects the centers of gravity of the cross sections can be shifted relative to the central axis. This means that the calculation must be made along a section whose center of gravity is as far away from the central axis as possible, which is why the eccentricity of the force for this section is maximum.

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

4. The column can be installed with deviations from the vertical, which means that it is vertical effective load can create an additional bending moment, maximum in the lower part of the column, or more precisely, at the point of attachment to the foundation, however, this is relevant only for free-standing columns.

5. Under the influence of loads applied to it, the column can deform, which means that the eccentricity of the load application will again appear and, as a consequence, an additional bending moment.

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

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

Naturally, it is almost impossible to calculate the above deviations for a structure that is still being designed - the calculation will be very long, complex, and the result is still doubtful. But it is very possible to introduce a certain coefficient into formula (1.1) that would take into account the above factors. This coefficient is φ - buckling coefficient. The formula that uses this coefficient looks like this:

F = N/φR (1.2)

Meaning φ is always less than one, this means that the cross section of the column will always be larger than if you simply calculate using formula (1.1), what I mean is that now the fun begins and remember that φ always less than one - it won't hurt. For preliminary calculations you can use the value φ within 0.5-0.8. Meaning φ depends on the steel grade and column flexibility λ :

λ = l ef/ i (1.3)

l ef- design length of the column. The calculated and actual length of a column are different concepts. The estimated length of the column depends on the method of securing the ends of the column and is determined using the coefficient μ :

l ef = μ l (1.4)

l - actual length of the column, cm;

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

Table 1. Coefficients μ for determining the design lengths of columns and racks of constant cross-section (according to SNiP II-23-81 (1990))

As we can see, the coefficient value μ changes several times depending on the method of fastening the column, and the main difficulty here is which design scheme to choose. If you don’t know which fastening scheme suits your conditions, then take the value of the coefficient μ=2. The value of the coefficient μ=2 is accepted mainly for free-standing columns, clear example a free-standing column - a lamppost. The coefficient value μ=1-2 can be taken for canopy columns on which beams rest without rigid attachment to the column. This design scheme can be adopted when the canopy beams are not rigidly attached to the columns and when the beams have a relatively large deflection. If the column will be supported by trusses rigidly attached to the column by welding, then the value of the coefficient μ=0.5-1 can be taken. If there are diagonal connections between the columns, then you can take the value of the coefficient μ = 0.7 for non-rigid fastening of diagonal connections or 0.5 for rigid fastening. However, such stiffness diaphragms do not always exist in 2 planes and therefore such coefficient values ​​must be used carefully. When calculating the truss posts, the coefficient μ=0.5-1 is used, depending on the method of securing the posts.

The slenderness coefficient value approximately shows the ratio of the design length of the column to the height or width of the cross section. Those. the higher the value λ , the smaller the width or height of the cross-section of the column and, accordingly, the greater the cross-sectional margin required for the same column length, but more on that a little later.

Now that we have determined the coefficient μ , you can calculate the design length of the column using formula (1.4), and in order to find out the flexibility value of the column, you need to know the radius of gyration of the column section i :

Where I- moment of inertia of the cross section relative to one of the axes, and here the fun begins, because in the course of solving the problem we must determine the required cross-sectional area of ​​the column F, but this is not enough; it turns out that we still need to know the value of the moment of inertia. Since we do not know either one or the other, the solution to the problem is carried out in several stages.

On preliminary stage usually the value is taken λ within 90-60, for columns with a relatively small load you can take λ = 150-120 (the maximum value for columns is 180, the maximum flexibility values ​​for other elements can be found in table 19* SNiP II-23-81 (1990). Then Table 2 determines the value of the flexibility coefficient φ :

Table 2. Buckling coefficients φ of centrally compressed elements.

Note: coefficient values φ in the table are magnified 1000 times.

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

i = l ef/λ (1.6)

A rolled profile with a corresponding radius of gyration value is selected according to the assortment. Unlike bending elements, where the section is selected along only one axis, since the load acts only in one plane, in centrally compressed columns longitudinal bending can occur relative to any of the axes and therefore the closer the value of I z to I y, the better, in other words In other words, the most preferable profiles are round or square section. Well, now let's try to determine the cross-section of the column based on the knowledge gained.

Example of calculation of a metal centrally compressed column

There is: a desire to make a canopy near the house approximately as follows:

In this case, the only centrally compressed column under any conditions of fastening and at uniform distributed load there will be a column shown in red in the picture. In addition, the load on this column will be maximum. Columns marked in blue and green, can be considered as centrally compressed only with appropriate constructive solution and uniformly distributed load, columns marked orange, will be either centrally compressed or eccentrically compressed or frame struts calculated separately. IN in this example we will calculate the cross section of the column indicated in red. For calculations we will accept constant load from the canopy’s own weight of 100 kg/m² and a temporary load of 100 kg/m² from the snow cover.

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

N = (100+100) 5 3 = 3000 kg

2.2. We accept the preliminary value λ = 100, then according to table 2 the bending coefficient φ = 0.599 (for steel with a design strength of 200 MPa, given value adopted to provide an additional safety margin), then the required cross-sectional area of ​​the column is:

F= 3000/(0.599 2050) = 2.44 cm²

2.3. According to table 1 we take the value μ = 1 (since roofing made of profiled flooring, properly fixed, will ensure rigidity of the structure in a plane parallel to the plane of the wall, and in a perpendicular plane, the relative immobility of the top point of the column will be ensured by fastening the rafters to the wall), then the radius of inertia

i= 1·250/100 = 2.5 cm

2.4. According to the assortment for square profile pipes, these requirements are satisfied by a profile with cross-sectional dimensions of 70x70 mm with a wall thickness of 2 mm, having a radius of gyration of 2.76 cm. The cross-sectional area of ​​such a profile is 5.34 cm². This is much more than is required by calculation.

2.5.1. We can increase the flexibility of the column, while the required radius of gyration decreases. For example, when λ = 130 bending factor φ = 0.425, then the required cross-sectional area of ​​the column:

F = 3000/(0.425 2050) = 3.44 cm²

2.5.2. Then

i= 1·250/130 = 1.92 cm

2.5.3. According to the assortment for square profile pipes, these requirements are satisfied by a profile with cross-sectional dimensions of 50x50 mm with a wall thickness of 2 mm, having a radius of gyration of 1.95 cm. The cross-sectional area of ​​such a profile is 3.74 cm², the moment of resistance for this profile is 5.66 cm³.

Instead of square profile pipes, you can use an equal angle angle, a channel, an I-beam, or a regular pipe. If the calculated resistance of the steel of the selected profile is more than 220 MPa, then the cross section of the column can be recalculated. That’s basically all that concerns the calculation of metal centrally compressed columns.

Calculation of an eccentrically compressed column

Here, of course, the question arises: how to calculate the remaining columns? The answer to this question greatly depends on the method of attaching the canopy to the columns. If the canopy beams are rigidly attached to the columns, then a rather complex statically indeterminate frame will be formed, and then the columns should be considered as part of this frame and the cross-section of the columns should be calculated additionally for the action of the transverse bending moment. We will further consider the situation when the columns shown in the figure , are hingedly connected to the canopy (we are no longer considering the column marked in red). For example, the head of the columns has a support platform - metal plate with holes for bolting canopy beams. By various reasons the load on such columns can be transmitted with a sufficiently large eccentricity:

The beam shown in the picture is beige color, under the influence of the load it will bend a little and this will lead to the fact that the load on the column will be transmitted not along the center of gravity of the column section, but with eccentricity e and when calculating the outer columns, this eccentricity must be taken into account. There are a great many cases of eccentric loading of columns and possible cross sections of columns, described by the corresponding formulas for calculation. In our case, to check the cross-section of an eccentrically 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 determined the cross-section of the most loaded column, it is enough for us to check whether such a cross-section is suitable for the remaining columns for the reason that we do not have the task of building a steel plant, but we are simply calculating the columns for the canopy, which will all have the same cross-section for reasons of unification.

What's happened N, φ And R y we already know.

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

F = (N/R y)(1/φ + e z ·F/W z) (3.2)

because M z =N e z, why the value of the moment is exactly what it is and what the moment of resistance W is is explained in sufficient detail in a separate article.

F = (1500/2050)(1/0.425 + 2.5 3.74/5.66) = 0.7317 (2.353 + 1.652) = 2.93 cm²

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

The required cross-section of 2.93 cm² is less than the accepted 3.74 cm², and therefore square profile pipe with cross-sectional dimensions of 50x50 mm and a wall thickness of 2 mm can also be used for outer columns.

Calculation of an eccentrically compressed column based on conditional flexibility

Oddly enough, to select the cross-section of an eccentrically compressed column - a solid rod - there is an even simpler formula:

F = N/φ e R (4.1)

φ e- buckling coefficient, depending on eccentricity, it could be called the eccentric buckling coefficient, so as not to be confused with the buckling coefficient φ . However, calculations using this formula may turn out to be longer than using formula (3.2). To determine the coefficient φ e you still need to know the meaning of the expression e z ·F/W z- which we met in formula (3.2). This expression is called relative eccentricity and is denoted m:

m = e z ·F/W z (4.2)

After this, the reduced relative eccentricity is determined:

m ef = hm (4.3)

h- this is not the height of the section, but a coefficient determined according to table 73 of SNiPa II-23-81. I'll just say that the coefficient value h varies from 1 to 1.4, for most simple calculations you can use h = 1.1-1.2.

After this, you need to determine the conditional flexibility of the column λ¯ :

λ¯ = λ√‾(R y / E) (4.4)

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

Table 3. Coefficients φ e for checking the stability of eccentrically compressed (compressed-bending) solid-walled rods in the plane of moment action coinciding with the plane of symmetry.

Notes:

1. Coefficient values φ e magnified 1000 times.
2. Meaning φ should not be taken more than φ .

Now, for clarity, let’s check the cross-section of columns loaded with eccentricity using formula (4.1):

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

N = (100+100) 5 3/2 = 1500 kg

Load application eccentricity e= 2.5 cm, buckling coefficient φ = 0,425.

4.2. We have already determined the value of relative eccentricity:

m = 2.5 3.74/5.66 = 1.652

4.3. Now let’s determine the value of the reduced coefficient m ef :

m ef = 1.652 1.2 = 1.984 ≈ 2

4.4. Conditional flexibility at our accepted flexibility coefficient λ = 130, steel strength R y = 200 MPa and elastic modulus E= 200000 MPa will be:

λ¯ = 130√‾(200/200000) = 4.11

4.5. Using Table 3, we determine the value of the coefficient φ e ≈ 0.249

4.6. Determine the required column section:

F = 1500/(0.249 2050) = 2.94 cm²

Let me remind you that when determining the cross-sectional area of ​​the column using formula (3.1), we obtained almost the same result.

Advice: To ensure that the load from the canopy is transferred with minimal eccentricity, a special platform is made in the supporting part of the beam. If the beam is metal, made from a rolled profile, then it is usually enough to weld a piece of reinforcement to the bottom flange of the beam.