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

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

Where: p - active effort;

Sets a certain stability factor

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

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

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

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

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

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

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

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

and the critical tension will be:

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

Determine the diameter of the rack:

Moment of inertia section

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

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

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

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

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

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

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

Calculation of the central rack

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

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

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

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

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

N \u003d R 2 G,. (3.9)

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

R is an internal radius of the tank, m.

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

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

And the total weight of the overlap N, kg:

The effort perceived by the resistant is calculated:

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

CM 2, (3.12)

where: N-full overlap weight, kg;

1600 kgf / cm 2, for Steel InstallationSp;

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

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

Pipe cross sections, F:

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

J \u003d cm4, (3.14)

where - the geometric characteristics of the section.

Inertia Radius:

r \u003d, cm, (3.15)

where J is the moment of profile inertia;

F-area of \u200b\u200bthe required section.

Flexibility:

The stance in the rack is determined by the formula:

Kgf / cm (3.17)

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

Calculation of the Stand Base Strength

The calculated pressure P on the foundation is determined:

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

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

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

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

P st - Ints, kg;

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

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

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

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

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

Estimated bending moment, m:

The estimated moment of resistance, w:

Plate thickness D:

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

The height of the rack and the length of the shoulder of the power of the power P is selected constructively, according to the drawing. Take a cross section of the rack as 2sh. Based on the ratio H 0 / L \u003d 10 and H / B \u003d 1.5-2, select the cross section is not greater than H \u003d 450mm and B \u003d 300mm.

Figure 1 - Rack loading diagram and cross-section.

The total mass of the design is:

m \u003d 20,1 + 5 + 0.43 + 3 + 3,2 + 3 \u003d 34.73 tons

Weight coming on one of 8 racks is:

P \u003d 34.73 / 8 \u003d 4.34 tons \u003d 43400n - Pressure per rack.

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

MX \u003d P * L; MX \u003d 43400 * 5000 \u003d 217000000 (H * mm)

Consider a box of a box cross section, cooked from two plates

Definition of eccentricity:

If eccentricity t. H.it matters from 0.1 to 5 - an echocently compressed (stretched) resistant; if a t.from 5 to 20, the stretching or compression of the beam must be taken into account in the calculation.

t. H. \u003d 2.5 - an echocently compressed (stretched) rack.

Determine the size of the cross section of the rack:

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

From this equation find the required cross-sectional area

, mm 2 (10)

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

(11)

Estimated resistance R U.depends on the concentration of voltage and on the yield strength of the material. The concentration of voltage in welded connections is most often due to 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 voltage.

The most loaded cross section of the designed rod design is located near the place of its attachment to the wall. Attaching winding angular seams corresponds to the 6th group, therefore, R U \u003d 45MPa.

For the 6th group, with n \u003d 10 -6, α \u003d 1.63;

Coefficient w.reflects the dependence of the allowable stresses from the indicator of the asymmetry of the cycle p equal to the ratio of the minimum voltage per cycle to the maximum, i.e.

-1≤ρ.<1,

as well as from stress sign. The stretching contributes, and the compression prevents the occurrence of cracks, so the value γ at the same ρ depends on the sign Σ MAX. In case of pulsating load when Σ min\u003d 0, ρ \u003d 0 with compression γ \u003d 2 when tensile γ = 1,67.

For ρ → ∞ Γ → ∞. In this case, the allowable voltage [σ] becomes very large. This means that the danger of fatigue destruction decreases, but does not mean that the strength is ensured, since it is possible to destruction during the first loading. Therefore, when determining [σ], it is necessary to take into account the conditions of static strength and stability.

With static tension (without bending)

[Σ] \u003d R y. (12)

The value of the calculated resistance R y on the yield strength is determined by the formula

(13)

where γ M is the reliability coefficient by material.

For 09g2s Σ T \u003d.325 MPa, Γ T \u003d.1,25

With static compression, the allowable voltage is reduced due to the danger of loss of stability:

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 application application, you can accept φ = 0.6. Such a coefficient means that the strength of the rod in compression due to the loss of stability is reduced to 60% of tensile strength.

We substitute the data in the formula:

Of the two values \u200b\u200b[σ], choose the smallest. And in the future it will be calculated.

Allowable voltage

We supply data in the formula:

Since 295,8mm 2 is an extremely small cross section area, based on the design sizes and the magnitude of the moment increase to

Schawler's number will be published in the area.

The minimum slicer of the sewller must be 60 cm 2

Schawler number - 40p. It has parameters:

h \u003d 400 mm; B \u003d 115mm; S \u003d 8mm; T \u003d 13.5 mm; F \u003d 18.1 cm 2;

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

We substitute the data in the formula 12 and calculate the voltages again:

\u003d 146.7 MPa

The active stresses in the cross section are less limiting voltages for the metal. This means that the design material is withstanding the applied load.

Verification calculation of the overall stability of racks.

Such an inspection is required only under the action of compressing longitudinal forces. If the forces are applied to the centered center (MX \u003d MU \u003d 0), the static string strength to the static strength due to the stability loss is evaluated by the coefficient φ depending on the flexibility of the rack.

The flexibility of the rack relative to the material axis (i.e. the axes crossing the elements of the section) are determined by the formula:

(15)

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

μ - coefficient-dependent consolidation; with console \u003d 2;

i min - inertia radius, located by the formula:

(16)

We substitute the data in the formula 20 and 21:

The calculation of stability is carried out by the formula:

(17)

The coefficient φ is determined as well as under the central compression, in the table. 6 Depending on the flexibility of the rack λ y (λ О), when the axis is bent around the axis. Coefficient fromtakes into account the moment of resistance from the moment M. x.

1. Harvesting loads

Before starting the calculation of the steel beam, it is necessary to assemble the load acting on the metal beam. Depending on the duration of the load, the load is divided into permanent and temporary.

• own weight of the metal beam;
• own overlap weight, etc.;

• long-term load (payload is made depending on the designation of the building);
• short-term load (snow load, is adopted depending on the geographical location of the building);
• special load (seismic, explosive, etc. As part of this calculator is not taken into account);

Loads on the beam are separated into two types: settlement and regulatory. Estimated loads are used to calculate the beams for strength and stability (1 limit state). Regulatory loads are set by norms and is used to calculate the beam to the deflection (2 limit state). Calculated loads are determined by multiplying the regulatory load on the reliability load coefficient. As part of this calculator, the estimated load is used when determining the beam brack definition.

After assembled the surface load on the overlap measured in kg / m2, it is necessary to calculate how much of this surface load takes the beam. To do this, you need to multiply the surface load onto the beam step (the so-called cargo strip).

For example: We counted that the total load turned out to the check. \u003d 500kg / m2, and the beam step is 2.5m. Then the distributed load on the metal beam will be: QSPR. \u003d 500kg / m2 * 2,5m \u003d 1250kg / m. This load is entered into the calculator

2. Building Epur

Next, the construction of the moments, transverse force is made. The step depends on the loading scheme of the beam, the type of supporting beams. Epur is being built according to the rules of construction mechanics. For the most part used loading and support schemes, there are ready-made tables with derived formulas of EPUR and deflection.

3. Calculation of strength and deflection

After constructing the EPUR, it is calculated by strength (1 limit state) and a deflection (2 limit state). In order to pick up the beam, it is necessary to find the required moment of inertia Wetr and from the sorting table to choose a suitable metal plate. The vertical limit of FULT deflection is accepted according to Table 19 of Snip 2.01.07-85 * (load and exposure). Paragraph2.a, depending on the span. For example, the limit of deflection FULT \u003d L / 200 with a span l \u003d 6m. This means that the calculator will select the cross section of the rolling profile (of a 2-hander, chapeller or two channels in the box), the limit of which will not exceed FULT \u003d 6M / 200 \u003d 0.03m \u003d 30mm. For the selection of metal profiles by deflection, the required moment of IT inertia is found, which is obtained from the formula for finding the limit deflection. And also from the spreadsheet table pick up suitable metal products.

4. Selection of a metal beam from the sorting table

Of the two results of the selection (1 and 2 limit state), metal photographs are selected with a large cross section.

The column is a vertical element of the supporting structure of a building that transmits loads from the above-described structures on the foundation.

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

For the steel column, two-way, pipe, square profile, 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 by metal mass and have a beautiful aesthetic appearance, but the internal cavities can not be painted, so this profile should be tightly.

The use of a broad-boiler for columns is widespread - when pinching the column in one plane, this type of profile is optimal.

The method of fixing the column in the foundation is of great importance. The column may have a hinge fastening, rigid in one plane and hinged to another or rigid in 2 planes. The selection of the mount depends on the building of the building and has more value when calculating because The calculated length of the column depends on the fastening method.

It is also necessary to take into account the method of fastening runs, wall panels, beams or farms on the column, if the load is transmitted from the side of the column, the eccentricity must be taken into account.

When pinching the column in the foundation and rigid fastening of the beam to the column, the calculated length is 0.5L, but 0.7L is usually considered in the calculation. The beam under the action of the load is bent and no complete pinch.

In practice, the column is not considered separately, and model the building or a 3-dimensional building model in the program, load it and calculate the column in the assembly and select the necessary profile, but in the programs it is difficult to consider the weakening of the cross section of the bolts from the bolts, therefore it is 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, the voltage plots are built for this. In this review, we will consider only the strength calculation of the column without constructing the EPUR.

Calculate columns We carry out the following parameters:

1. Strength with central tension / compression

2. Stability under central compression (in 2 planes)

3. Strength with the joint action of longitudinal force and bending moments

4. Check the limit flexibility of the rod (in 2 planes)

1. Strength with central tension / compression

According to SP 16.13330 p. 7.1.1 Calculation of the strength of elements from steel with regulatory resistance R.yn ≤ 440 N / mm2 with central tension or compression by force N should be performed by the formula

A.n is the cross section of the net profile, i.e. taking into account the weakening of it by holes;

R.y - the calculated resistance of steel rolled (depends on the brand of steel, see Table V.5 SP 16.13330);

γ c - coefficient of working conditions (see table 1 SP 16.13330).

According to this formula, you can calculate the minimum-needed area of \u200b\u200bthe profile cross section and set a profile. In the future, in verification calculations, the selection of the cross section of the column can be made only by selecting the section, so here we can set a starting point, less which can not be a cross section.

2. Sustainability under the central compression

Calculation of stability is made according to SP 16.13330 p. 7.1.3 by formula

A. - Cross profile cross section area, i.e. take into account the weakening of it with holes;

R.

γ

φ - Sustainability coefficient in central compression.

As you can see this formula, it is very similar to the previous one, but the coefficient appears here. φ To calculate it at first, it will be necessary to calculate the conditional flexibility of the rod. λ (denoted with a feature from above).

where R.y - settling the resistance of steel;

E. - elastic modulus;

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

where l.eF - the estimated length of the rod;

i. - radius of inertia section.

Estimated lengths L.eF columns (racks) of constant cross section or individual sections of stepped columns according to SP 16.13330 p. 10.3.1 should be determined by the formula

where l. - length of the column;

μ - Coefficient of the calculated length.

Coefficients of settlement length μ Columns (racks) of permanent 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 value μ List in the following table:

The radius of the inertia can be found in the corresponding Gulling on the profile, i.e. A pre-profile must be set and the calculation is reduced to the cross sections.

Because The inertia radius in 2 planes for most profiles has different values \u200b\u200bon 2 planes (only the pipe and the square profile) and the fixation can be different, and therefore, and the calculated lengths can also be different, then the calculation of stability must be made For 2 planes.

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

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

the value of the coefficient δ It should be calculated by the formula:

factors α and β see Table

The values \u200b\u200bof the coefficient φ calculated by this formula should be taken no more (7.6 / λ 2) with the values \u200b\u200bof the conditional flexibility above 3.8; 4.4 and 5.8 for types of cross sections, respectively, A, B and C.

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

The values \u200b\u200bof the coefficient φ Led in Appendix D SP 16.13330.

Now when all source data is known to make a calculation by the formula presented at first:

As mentioned above, it is necessary to make 2-a calculation for 2 planes. If the calculation does not satisfy the condition, then select a new profile with a larger value of the inertia of the cross section. You can also change the calculated scheme, for example, changing the hinge embelling on rigid or fixing the column in the span, you can reduce the calculated rod length.

Compressed elements with solid walls of the open P-shaped section are recommended to strengthen the straps or grille. If the planks are absent, then stability should be checked for stability in the flexible-twist form of stability loss in accordance with P.7.1.5 SP 16.13330.

3. Strength with the joint action of longitudinal force and bending moments

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

where N. - longitudinal compressive force;

A.n is the area of \u200b\u200bthe net (taking into account the weakening of the holes);

R.y - the calculated resistance of steel;

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

n, CXand SY. - coefficients accepted according to Table E.1 SP 16.13330

MX. and MY. - moments relative to x-x and y-y axes;

W.xn, min and W.yn, min - the moments of the section resistance relative to the X-X and Y-Y axes (can be found in the GOST on the profile or in the directory);

B. - Bimoment, in SNIP II-23-81 * this parameter was not in the calculations, this parameter was introduced to account for deploitation;

W.ω, min - sectoral moment of section resistance.

If there should be no first to be issues with the first 3 components, then the Bimome's accounting causes some difficulties.

Bimoment characterizes the changes made to the linear distribution zones of the separation of the cross section and, in fact, is a pair of moments aimed at opposite parties

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

4. Check the limit flexibility of the rod

Flexibility of compressed elements λ \u003d LEF / I, as a rule, should not exceed the limit values λ u shown in the table

The coefficient α in this formula is the coefficient of use of the profile, according to the calculation of consistency in central compression.

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

In the event that the profile does not suit, it is necessary to change the cross section by increasing the radius of the inertia of the cross section or changing the calculated scheme (change the consolidation or consolidate with the connections to reduce the calculated length).

If the critical factor is the limit flexibility, the steel brand can be taken the smallest because On the limit flexibility, the steel brand does not affect. The optimal option can be calculated by the selection method.

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