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

The height of the rack and the length of the arm of the application of the force P is selected constructively, according to the drawing. Let's take the section of the rack as 2Sh. Based on the ratio h 0 /l=10 and h/b=1.5-2, we select a section no more than h=450mm and b=300mm.

Figure 1 - Scheme of loading the rack and cross section.

The total weight of the structure is:

m= 20.1+5+0.43+3+3.2+3 = 34.73 tons

The weight coming to one of the 8 racks is:

P \u003d 34.73 / 8 \u003d 4.34 tons \u003d 43400N - pressure per rack.

The force does not act in the center of the section, so it causes a moment equal to:

Mx \u003d P * L; Mx = 43400 * 5000 = 217000000 (N*mm)

Consider a box-section strut welded from two plates

Definition of eccentricities:

If the eccentricity t x has a value from 0.1 to 5 - eccentrically compressed (stretched) rack; if t from 5 to 20, then the tension or compression of the beam must be taken into account in the calculation.

t x\u003d 2.5 - eccentrically compressed (stretched) rack.

Determining the size of the section of the rack:

The main load for the rack is the longitudinal force. Therefore, to select the section, the calculation for tensile (compressive) strength is used:

(9)

From this equation find the required cross-sectional area

,mm 2 (10)

Permissible stress [σ] during endurance work depends on the steel grade, stress concentration in the section, number of loading cycles and cycle asymmetry. In SNiP, the allowable stress during endurance work is determined by the formula

(11)

Design resistance R U depends on the stress concentration and on the yield strength of the material. Stress concentration in welded joints is most often caused by welds. The value of the concentration coefficient depends on the shape, size and location of the seams. The higher the stress concentration, the lower the allowable stress.

The most loaded section of the rod structure designed in the work is located near the place of its attachment to the wall. Attachment with frontal fillet welds corresponds to the 6th group, therefore, RU = 45 MPa.

For the 6th group, with n = 10 -6, α = 1.63;

Coefficient at reflects the dependence of allowable stresses on the cycle asymmetry index p, equal to the ratio of the minimum stress per cycle to the maximum, i.e.

-1≤ρ<1,

as well as from the sign of stresses. Tension promotes, and compression prevents cracking, so the value γ for the same ρ depends on the sign of σ max. In the case of pulsating loading, when σmin= 0, ρ=0 in compression γ=2 in tension γ = 1,67.

As ρ→ ∞ γ→∞. In this case, the allowable stress [σ] becomes very large. This means that the risk of fatigue failure is reduced, but does not mean that strength is ensured, since failure during the first loading is possible. Therefore, when determining [σ], it is necessary to take into account the conditions of static strength and stability.

Under static tension (no bending)

[σ] = R y. (12)

The value of the design resistance R y according to the yield strength is determined by the formula

(13)

where γ m is the reliability factor for the material.

For 09G2S σ Т = 325 MPa, γ t = 1,25

In static compression, the allowable stress is reduced due to the risk of buckling:

where 0< φ < 1. Коэффициент φ зависит от гибкости и относительного эксцентриситета. Его точное значение может быть найдено только после определения размеров сечения. Для ориентировочного выбора Атрпо формуле следует задаться значением φ. With a small eccentricity of the load application, φ can be taken = 0.6. This coefficient means that the compressive strength of the rod is reduced to 60% of the tensile strength due to buckling.

We substitute the data in the formula:

Of the two values ​​of [ σ] choose the smallest. And in the future, it will be calculated.

Allowable voltage

Putting the data into the formula:

Since 295.8 mm 2 is an extremely small cross-sectional area, based on the design dimensions and the magnitude of the moment, we increase it to

We will select the channel number according to the area.

The minimum area of ​​​​the channel should be - 60 cm 2

Channel number - 40P. Has options:

h=400 mm; b=115mm; s=8mm; t=13.5mm; F=18.1 cm 2 ;

We get the cross-sectional area of ​​\u200b\u200bthe rack, consisting of 2 channels - 61.5 cm 2.

Substitute the data in formula 12 and calculate the stresses again:

=146.7 MPa

The effective stresses in the section are less than the limiting stresses for the metal. This means that the material of construction can withstand the applied load.

Verification calculation of the overall stability of the racks.

Such a check is required only under the action of compressive longitudinal forces. If forces are applied to the center of the section (Mx=Mu=0), then the reduction in the static strength of the rack due to the loss of stability is estimated by the coefficient φ, which depends on the flexibility of the rack.

The flexibility of the rack relative to the material axis (i.e., the axis intersecting the section elements) is determined by the formula:

(15)

where - the length of the half-wave of the curved axis of the rack,

μ - coefficient depending on the condition of fixing; at console = 2;

i min - radius of inertia, is found by the formula:

(16)

We substitute the data in the formula 20 and 21:

Calculation of stability is carried out according to the formula:

(17)

The coefficient φ y is determined in the same way as with central compression, according to table. 6 depending on the flexibility of the rack λ y (λ yo) when bending around the y axis. Coefficient with takes into account the decrease in stability due to the action of the moment M X.

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 method of fixing the ends of the rack. In structural mechanics, seven methods are considered for securing the ends of the rack. We will consider three main methods:

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

Where: P - acting force;

A certain stability factor is set

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

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

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

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

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

where: z - current ordinate, at the maximum value z=l;
The admissible 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 rack EJ min in direct proportion and on the length of the rack l - inversely proportional.
As mentioned, the stability of the elastic rack depends on how it is fixed.
The recommended safety margin for steel studs 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 fixing the ends of the rack, the coefficient of the ends of the reduced flexibility of the rack is introduced.

where: μ - coefficient of reduced length (Table) ;
i min - the smallest radius of gyration of the cross section of the rack (table);
ι - rack length;
Enter the critical load factor:

, (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 fixing the ends of the rack and is given in the tables of the handbook on strength materials (G.S. Pisarenko and S.P. Fesik)
Let us give an example of calculating the critical force for a rod of solid section of a rectangular shape - 6 × 1 cm, the length of the rod ι = 2m. Fixing the ends according to scheme III.
Calculation:
According to the table, we find the coefficient ϑ = 9.97, μ = 1. The moment of inertia of the section will be:

and the critical stress will be:

It is obvious that the critical force P cr = 247 kgf will cause a stress in the rod of only 41 kgf / cm 2, which is much less than the flow limit (1600 kgf / cm 2), however, this force will cause the rod to bend, which means loss of stability.
Consider another example of calculating a wooden rack of circular cross section, pinched at the lower end and hinged at the upper end (S.P. Fesik). Stand length 4m, compression force N=6tf. Permissible stress [σ]=100kgf/cm 2 . We accept the reduction factor of the allowable stress for compression φ=0.5. We calculate the sectional area of ​​​​the rack:

Determine the diameter of the rack:

Moment of inertia of section

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:

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

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

where: μ from the table, equal to 2, taking into account the way the rack is pinched;
The design voltage in the rack will be:

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

findings
When calculating the strength and stability of racks, columns, it is necessary to take into account the method of fixing the ends of the racks, apply the recommended margin of safety.
The value of the critical force is obtained from the differential equation of the curved axial line of the rack (L. Euler).
To take into account all the factors characterizing the loaded rack, the concept of rack flexibility - λ, provided length factor - μ, stress reduction factor - ϕ, critical load factor - ϑ. Their values ​​are taken from reference tables (G.S. Pisarentko and S.P. Fesik).
Approximate calculations of struts are given to determine the critical force - Рcr, critical stress - σcr, strut diameter - d, strut flexibility - λ and other characteristics.
The optimal section for racks and columns is tubular thin-walled profiles with the same principal moments of inertia.

Used Books:
G.S Pisarenko "Handbook on the strength of materials."
S.P. Fesik "Handbook of Strength of Materials".
IN AND. Anuryev "Handbook of the designer-machine builder".
SNiP II-6-74 "Loads and impacts, design standards".

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.

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:

kgf/cm (3.17)

At the same time, according to the tables of Appendix 17 (A.N. Serenko) = 0.34

Rack Base Strength Calculation

The design pressure P on the foundation is determined by:

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

R st \u003d F L g, kg, (3.19)

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

where: P "-force of the vertical rack P" \u003d 5885.6 kg;

R st - weight racks, kg;

g - specific gravity of steel.g \u003d 7.85 * 10 -3 kg /.

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

g b - specific gravity of concrete grade. g b \u003d 2.4 * 10 -3 kg /.

The required area of ​​​​the shoe plate at the allowable pressure on the sandy base [y] f \u003d 2 kg / cm 2:

A slab with sides is accepted: aChb \u003d 0.65×0.65 m. The distributed load, q per 1 cm of the slab is determined:

Estimated bending moment, M:

Estimated moment of resistance, W:

Plate thickness d:

The plate thickness d = 20 mm is taken.

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;

R

γ

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 the pipe and the 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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