Calculation of metal columns. Excel Calculators for steel structures

A column is a vertical element of a building's load-bearing structure that transfers loads from higher structures to the foundation.

When calculating steel columns, it is necessary to be guided by SP 16.13330 "Steel structures".

For a steel column, an I-beam, a pipe, a square profile, a composite section of channels, corners, sheets are usually used.

For centrally compressed columns, it is optimal to use a pipe or a square profile - they are economical in terms of metal mass and have a beautiful aesthetic appearance, however, the internal cavities cannot be painted, so this profile must be airtight.

The use of a wide-shelf I-beam for columns is widespread - when the column is pinched in one plane, this type of profile is optimal.

Of great importance is the method of fixing the column in the foundation. The column can be hinged, rigid in one plane and hinged in another, or rigid in 2 planes. The choice of fastening depends on the structure of the building and is more important in the calculation, because. the estimated length of the column depends on the method of fastening.

It is also necessary to take into account the method of attaching purlins, wall panels, beams or trusses to the column, if the load is transferred from the side of the column, then the eccentricity must be taken into account.

When the column is pinched in the foundation and the beam is rigidly attached to the column, the calculated length is 0.5l, but 0.7l is usually considered in the calculation. the beam bends under the action of the load and there is no complete pinching.

In practice, the column is not considered separately, but a frame or a 3-dimensional building model is modeled in the program, it is loaded and the column in the assembly is calculated and the required profile is selected, but in programs it can be difficult to take into account the weakening of the section by bolt holes, so it may be necessary to check the section manually .

To calculate the column, we need to know the maximum compressive / tensile stresses and moments that occur in key sections, for this we build stress diagrams. In this review, we will consider only the strength calculation of the column without plotting.

We calculate the column according to the following parameters:

1. Tensile/compressive strength

2. Stability under central compression (in 2 planes)

3. Strength under the combined action of longitudinal force and bending moments

4. Checking the ultimate flexibility of the rod (in 2 planes)

1. Tensile/compressive strength

According to SP 16.13330 p. 7.1.1 strength calculation of steel elements with standard resistance R yn ≤ 440 N/mm2 in case of central tension or compression by force N should be carried out according to the formula

A n is the cross-sectional area of the net profile, i.e. taking into account the weakening of its holes;

R y is the design resistance of rolled steel (depends on the steel grade, see Table B.5 of SP 16.13330);

γ c is the coefficient of working conditions (see Table 1 of SP 16.13330).

Using this formula, you can calculate the minimum required cross-sectional area of \u200b\u200bthe profile and set the profile. In the future, in the verification calculations, the selection of the section of the column can be done only by the method of selection of the section, so here we can set the starting point, which the section cannot be less than.

2. Stability under central compression

Calculation for stability is carried out in accordance with SP 16.13330 clause 7.1.3 according to the formula

A- the cross-sectional area of the gross profile, i.e. without taking into account the weakening of its holes;

φ is the coefficient of stability under central compression.

As you can see, this formula is very similar to the previous one, but here the coefficient appears φ , in order to calculate it, we first need to calculate the conditional flexibility of the rod λ (denoted with a dash above).

where R y is the design resistance of steel;

E- elastic modulus;

λ - the flexibility of the rod, calculated by the formula:

where l ef is the calculated length of the rod;

i is the radius of inertia of the section.

Effective lengths l ef columns (pillars) of constant cross section or individual sections of stepped columns in accordance with SP 16.13330 clause 10.3.1 should be determined by the formula

where l is the length of the column;

μ - effective length coefficient.

Effective length factors μ columns (pillars) of constant cross section should be determined depending on the conditions for fixing their ends and the type of load. For some cases of fixing the ends and the type of load, the values μ are shown in the following table:

The radius of gyration of the section can be found in the corresponding GOST for the profile, i.e. the profile must be pre-specified and the calculation is reduced to enumerating the sections.

Because the radius of gyration in 2 planes for most profiles has different values on 2 planes (only a pipe and a square profile have the same values) and the fastening can be different, and therefore the calculated lengths can also be different, then the calculation for stability must be made for 2 planes.

So now we have all the data to calculate the conditional flexibility.

If the ultimate flexibility is greater than or equal to 0.4, then the stability coefficient φ calculated by the formula:

coefficient value δ should be calculated using the formula:

odds α and β see table

Coefficient values φ , calculated by this formula, should be taken no more than (7.6 / λ 2) at values of conditional flexibility over 3.8; 4.4 and 5.8 for section types a, b and c, respectively.

For values λ < 0,4 для всех типов сечений допускается принимать φ = 1.

Coefficient values φ are given in Appendix D to SP 16.13330.

Now that all the initial data are known, we calculate according to the formula presented at the beginning:

As mentioned above, it is necessary to make 2 calculations for 2 planes. If the calculation does not satisfy the condition, then we select a new profile with a larger value of the radius of gyration of the section. It is also possible to change the design scheme, for example, by changing the hinged attachment to a rigid one or by fixing the column in the span with ties, the estimated length of the rod can be reduced.

Compressed elements with solid walls of an open U-shaped section are recommended to be reinforced with planks or gratings. If there are no straps, then the stability should be checked for stability in the bending-torsional form of buckling in accordance with clause 7.1.5 of SP 16.13330.

3. Strength under the combined action of longitudinal force and bending moments

As a rule, the column is loaded not only with an axial compressive load, but also with a bending moment, for example, from the wind. The moment is also formed if the vertical load is applied not in the center of the column, but from the side. In this case, it is necessary to make a verification calculation in accordance with clause 9.1.1 of SP 16.13330 using the formula

where N- longitudinal compressive force;

A n is the net cross-sectional area (taking into account weakening by holes);

R y is the design resistance of steel;

γ c is the coefficient of working conditions (see Table 1 of SP 16.13330);

n, Сx and Сy- coefficients taken according to table E.1 of SP 16.13330

Mx and My- moments about the axes X-X and Y-Y;

W xn,min and W yn,min - section modulus relative to the X-X and Y-Y axes (can be found in GOST on the profile or in the reference book);

B- bimoment, in SNiP II-23-81 * this parameter was not included in the calculations, this parameter was introduced to account for warping;

Wω,min – sectoral section modulus.

If there should be no questions with the first 3 components, then accounting for the bimoment causes some difficulties.

The bimoment characterizes the changes introduced into the linear zones of the stress distribution of the deformation of the section and, in fact, is a pair of moments directed in opposite directions

It is worth noting that many programs cannot calculate the bimoment, including SCAD does not take it into account.

4. Checking the ultimate flexibility of the rod

Flexibility of compressed elements λ = lef / i, as a rule, should not exceed the limit values λ u given in the table

The coefficient α in this formula is the utilization factor of the profile, according to the calculation of the stability under central compression.

As well as the stability calculation, this calculation must be done for 2 planes.

If the profile does not fit, it is necessary to change the section by increasing the radius of gyration of the section or changing the design scheme (change the fastenings or fix with ties to reduce the estimated length).

If the critical factor is the ultimate flexibility, then the steel grade can be taken as the smallest. the steel grade does not affect the ultimate flexibility. The optimal variant can be calculated by the selection method.

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Calculation of efforts in racks is carried out taking into account the loads applied to the rack.

Middle racks

The average racks of the building frame work and are calculated as centrally compressed elements for the action of the greatest compressive force N from the own weight of all pavement structures (G) and snow load and snow load (P sn).

Figure 8 - Loads on the middle rack

The calculation of the centrally compressed middle racks is carried out:

a) strength

where is the calculated resistance of wood to compression along the fibers;

Net cross-sectional area of the element;

b) stability

where is the buckling coefficient;

is the calculated cross-sectional area of the element;

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

Figure 9 - Cargo areas of the middle and outer columns

Extreme racks

The extreme post is under the action of loads longitudinal with respect to the axis of the post (G and P sn), which are collected from square and transverse, and X. In addition, a longitudinal force arises from the action of the wind.

Figure 10 - Loads on the end post

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

X is the horizontal concentrated force applied at the point of junction of the crossbar to the post.

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

Figure 11 - Scheme of loads with rigid pinching of racks in the foundation

where - horizontal wind loads, respectively, from the wind to the left and right, applied to the rack at the junction of the crossbar to 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 - Scheme of loads when the racks are hinged on the foundation

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

The end posts are calculated as compressed-flexible elements. The values of the longitudinal force N and the bending moment M are taken for such a combination of loads at which the greatest compressive stresses occur.

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

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

For a rack that is part of 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 pr:

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 its own sign.

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

The calculation of compressed - curved extreme racks is made:

a) strength:

b) on the stability of the flat shape of the bend in the absence of fastening or with the 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 racks are calculated as a centrally compressed element.

Calculation of Compressed and Compressed-Curved Composite Sections is produced 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 bonds connecting the branches. This increased flexibility is called the reduced flexibility λ n .

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

Metal constructions is a complex and extremely responsible topic. Even a small mistake can cost hundreds of thousands and millions of dollars. In some cases, the price of a mistake can be the lives of people at a construction site, as well as during operation. So, checking and rechecking calculations is necessary and important.

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

openness- each such calculation can be disassembled by bones.
Availability- the files themselves exist in the public domain, are written by the developers of the MK 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, moreover, 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 a beam for bending
Calculation of a beam for bending online
Check the calculation of the strength and stability of the column.
Check the selection of the bar section.

Universal calculation file MK (EXCEL)

Table for the selection of sections of metal structures, according to 5 different points of 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 stretched elements.
calculation of eccentric-compressed or compressed-bent elements.

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

METALLIC

The program is an EXCEL book with macro support.
And it is intended for the calculation of steel structures according to
SP16 13330.2013 "Steel structures"

Selection and calculation of runs

The selection of a run is a trivial task only at first glance. The step of runs and their size depend on many parameters. And it would be nice to have an appropriate calculation at hand. This is what this must-read article is about:

calculation of a run without strands
calculation of a run with one strand
calculation of a run with two strands
calculation of the run taking into account the bimoment:

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

Calculation of the 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 the metal structures are made, then this calculator will come to your aid. A little explanation is at the bottom of the table. In general, the work is simple - we select a suitable section, set the dimensions of these sections, and obtain the main parameters of the section:

Moments of inertia of the section
Section modulus
Radius of gyration of section
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

B-pillar calculation

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

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

The calculation scheme of the central post is adopted when calculating the vertical load, as hinged at the ends, since it is welded at the bottom and top (see Figure 3).

The B-pillar bears 33% of the total floor weight.

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

N \u003d R 2 g,. (3.9)

where g is the total uniformly distributed load, kg / m 2;

R is the inner radius of the tank, m.

The total weight of the floor is made up of the following types of loads:

1. Snow load, g 1 . Accepted g 1 \u003d 100 kg / m 2 .;
2. Load from thermal insulation, g 2. Accepted g 2 \u003d 45 kg / m 2;
3. Wind load, g 3 . Accepted g 3 \u003d 40 kg / m 2;
4. Load from the weight of the cover frame, g 4 . Accepted g 4 \u003d 100 kg / m 2
5. Taking into account the installed equipment, g 5 . Accepted g 5 \u003d 25 kg / m 2
6. Vacuum load, g 6 . Accepted g 6 \u003d 45 kg / m 2.

And the total weight of the overlap N, kg:

The force perceived by the rack is calculated:

The required cross-sectional area of the rack is determined by 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 coefficient of longitudinal bending is structurally accepted = 0.45.

According to GOST 8732-75, a pipe with an outer diameter D h \u003d 21 cm, an inner diameter d b \u003d 18 cm and a wall thickness of 1.5 cm is 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), the radius of inertia (r) is determined. Respectively:

J = cm4, (3.14)

where are the geometric characteristics of the section.