1. Obtaining information about the material of the rod to determine the maximum flexibility of the rod by calculation or according to the table:

2. Obtaining information about the geometric dimensions of the cross section, length and methods of securing the ends to determine the category of the rod depending on the flexibility:

where A is the cross-sectional area; J m i n - minimum moment of inertia (from axial ones);

μ - coefficient of reduced length.

3. Selection of calculation formulas for determining the critical force and critical stress.

4. Verification and sustainability.

When calculating using the Euler formula, the stability condition is:

F- effective compressive force;

- permissible safety factor.

When calculated using the Yasinsky formula Where a, b

- design coefficients depending on the material (the values ​​of the coefficients are given in Table 36.1)

If the stability conditions are not met, it is necessary to increase the cross-sectional area.

Sometimes it is necessary to determine the stability margin at a given load:

When checking stability, the calculated endurance margin is compared with the permissible one:

Examples of problem solving

Solution

1. The flexibility of the rod is determined by the formula

2. Determine the minimum radius of gyration for the circle. Substituting expressions for Jmin And A

1. (section circle) μ = 0,5.
2. Length reduction factor for a given fastening scheme

The flexibility of the rod will be equal to Example 2.

Examples of problem solving

How will the critical force for the rod change if the method of securing the ends is changed? Compare the presented diagrams (Fig. 37.2)

Critical force will increase 4 times. Example 3. How will the critical force change when calculating stability if an I-section rod (Fig. 37.3a, I-beam No. 12) is replaced with a rectangular section rod of the same area (Fig. 37.3 ) b

Examples of problem solving

? Other design parameters do not change. Perform the calculation using Euler's formula.

1. Determine the width of the section of the rectangle, the height of the section is equal to the height of the section of the I-beam. The geometric parameters of I-beam No. 12 according to GOST 8239-89 are as follows: cross-sectional area A 1 =

14.7 cm 2;

By condition, the area of ​​the rectangular cross-section is equal to the cross-sectional area of ​​the I-beam. Determine the width of the strip at a height of 12 cm.

2. Let us determine the minimum of the axial moments of inertia.

3. The critical force is determined by Euler’s formula:

4. Other things being equal, the ratio of the critical forces is equal to the ratio of the minimum moments of inertia:

5. Thus, the stability of a rod with an I-section No. 12 is 15 times higher than the stability of a rod of the selected rectangular cross-section.

Example 4. Check the stability of the rod. A rod 1 m long is clamped at one end, the cross-section is channel No. 16, the material is StZ, the stability margin is threefold. The rod is loaded with a compressive force of 82 kN (Fig. 37.4).

Examples of problem solving

1. Determine the main geometric parameters of the rod section according to GOST 8240-89. Channel No. 16: cross-sectional area 18.1 cm 2; minimum axial section moment 63.3 cm 4 ; minimum radius of gyration of the section r t; n = 1.87 cm.

Ultimate flexibility for material StZ λpre = 100.

Design flexibility of the rod at length l = 1m = 1000mm

The rod being calculated is a highly flexible rod; the calculation is carried out using the Euler formula.

4. Stability condition

82kN< 105,5кН. Устойчивость стержня обеспечена.

Example 5. In Fig. Figure 2.83 shows the design diagram of a tubular strut of an aircraft structure. Check the stand for stability at [ n y] = 2.5, if it is made of chromium-nickel steel, for which E = 2.1*10 5 and σ pts = 450 N/mm 2.

Examples of problem solving

To calculate stability, the critical force for a given rack must be known. It is necessary to establish by what formula the critical force should be calculated, i.e. it is necessary to compare the flexibility of the rack with the maximum flexibility for its material.

We calculate the value of the maximum flexibility, since there is no tabular data on λ, pre for the rack material:

To determine the flexibility of the calculated rack, we calculate the geometric characteristics of its cross section:

Determining the flexibility of the rack:

and make sure that λ< λ пред, т. е. критическую силу можно опреде­лить ею формуле Эйлера:

We calculate the calculated (real) stability factor:

Thus, n y > [ n y] by 5.2%.

Example 2.87. Check the strength and stability of the specified rod system (Fig. 2.86). The material of the rods is St5 steel (σ t = 280 N/mm 2). Required safety factors: strength [n]= 1.8; sustainability = 2.2. The rods have a circular cross-section d 1 = d 2= 20 mm, d 3 = 28 mm.

Examples of problem solving

By cutting out the node where the rods meet and composing equilibrium equations for the forces acting on it (Fig. 2.86)

we establish that the given system is statically indeterminate (three unknown forces and two static equations). It is clear that in order to calculate rods for strength and stability, it is necessary to know the magnitude of the longitudinal forces arising in their cross sections, i.e., it is necessary to reveal static indetermination.

We create a displacement equation based on the displacement diagram (Fig. 2.87):

or, substituting the values ​​of changes in the lengths of the rods, we get

Having solved this equation together with the equations of statics, we find:

Stresses in cross sections of rods 1 Jmin 2 (see Fig. 2.86):

Their safety factor

To determine the stability safety factor of the rod 3 it is necessary to calculate the critical force, and this requires determining the flexibility of the rod in order to decide what formula to find N Kp should be used.

So λ 0< λ < λ пред и крити­ческую силу следует определять по эмпирической формуле:

Safety factor

Thus, the calculation shows that the stability safety factor is close to the required one, and the safety factor is significantly higher than the required one, i.e., when the system load increases, the rod loses stability 3 more likely than the occurrence of yield in the rods 1 Jmin 2.

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 stability is achieved by correctly selecting the section of the rack.

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. It is 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 of 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 of concrete grade. g b = 2.4 * 10 -3 kg/.

Required area of ​​the shoe plate with permissible pressure on the sand 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.

Metal structures are a complex and extremely important topic. Even a small mistake can cost hundreds of thousands and millions of rubles. In some cases, the cost of an error may be the lives of people at a construction site, as well as during operation. So, checking and double-checking calculations is necessary and important.

Using Excel to solve calculation problems is, on the one hand, not new, but at the same time not entirely familiar. However, Excel calculations have a number of undeniable advantages:

• Openness— each such calculation can be disassembled piece by piece.
• Availability— the files themselves exist in the public domain, written by MK developers to suit their needs.
• Convenience- almost any PC user is able to work with programs from the MS Office package, while specialized design solutions are expensive and, in addition, require serious effort to master.

They should not be considered a panacea. Such calculations make it possible to solve narrow and relatively simple design problems. But they do not take into account the work of the structure as a whole. In a number of simple cases they can save a lot of time:

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

Universal calculation file MK (EXCEL)

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

• calculation of a single-span hinged beam.
• calculation of centrally compressed elements (columns).
• calculation of tensile elements.
• calculation of eccentrically compressed or compressed-bending elements.

The Excel version must be at least 2010. To see instructions, click on the plus sign in the upper left corner of the screen.

METALLICA

The program is an EXCEL workbook with macro support.
And is intended for the calculation of steel structures according to
SP16 13330.2013 “Steel structures”

Selection and calculation of runs

Selecting a run is only a trivial task at first glance. The pitch of the purlins and their size depend on many parameters. And it would be nice to have the corresponding calculation at hand. This is what this must-read article talks about:

• calculation of the run without strands
• calculation of a run with one strand
• calculation of a purlin with two strands
• calculation of the run taking into account the bi-moment:

But there is a small fly in the ointment - apparently the file contains errors in the calculation part.

Calculation of moments of inertia of a section in excel tables

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

• Section moments of inertia
• Section moments of resistance
• Section radius of gyration
• Cross-sectional area
• Static moment
• Distances to the center of gravity of the section.

The table contains calculations for the following types of sections:

• pipe
• rectangle
• I-beam
• channel
• rectangular pipe
• triangle

Often people who make a covered carport in their yard or for protection from the sun and precipitation do not calculate the cross-section of the posts on which the canopy will rest, but select the cross-section by eye or by consulting a neighbor.

You can understand them, the loads on the racks, which in this case are columns, are not so great, the volume of work performed is also not enormous, and the appearance of the columns is sometimes much more important than their load-bearing capacity, so even if the columns are made with a multiple safety margin - There is no big problem in this. Moreover, you can spend an infinite amount of time searching for simple and intelligible information about the calculation of solid columns without any result - it is almost impossible to understand examples of the calculation of columns for industrial buildings with the application of loads at several levels without good knowledge of strength of materials, and ordering a column calculation in 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 relates to the second, maximum to the third grade of 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 a vertically acting load can create an additional bending moment, maximum at the bottom of the column, or more precisely, at the point of attachment to the foundation, however, this is only relevant 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 greater 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; a clear example of a free-standing column is 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 :

When calculated using the Yasinsky formula 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.

At the preliminary stage, the value is usually 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, round or square profiles are most preferable. 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 fastening conditions and under a uniformly distributed load will be the column shown in red in the figure. In addition, the load on this column will be maximum. Columns marked in blue and green in the figure can be considered as centrally compressed, only with an appropriate design solution and a uniformly distributed load, columns marked in orange will be either centrally compressed or eccentrically compressed or frame racks calculated separately. In this example, we will calculate the cross section of the column indicated in red. For calculations, we will assume a permanent 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, this value is taken to provide an additional safety margin), then the required cross-sectional area of ​​the column:

F= 3000/(0.599 2050) = 2.44 cm²

2.3. According to table 1 we take the value μ = 1 (since a roof covering made of profiled decking, 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 - a metal plate with holes for bolting the canopy beams. For various reasons, the load on such columns can be transmitted with a fairly large eccentricity:

The beam shown in the figure, in beige color, will bend slightly under the influence of the load 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.

for the columns indicated in blue and green in the figure will be 1500 kg. We check the required cross-section at such a load and previously determined φ = 0,425

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 a square profile pipe with cross-sectional dimensions of 50x50 mm with a wall thickness of 2 mm can also be used for the 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 h = 1.1-1.2 can be used.

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.