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

B-pillars

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

Figure 8 – Loads on the middle pillar

Calculation of centrally compressed middle pillars is carried out:

a) for strength

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

Net cross-sectional area of ​​the element;

b) for stability

where is the buckling coefficient;

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

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

Figure 9 – Loading areas of the middle and outer columns

End posts

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

Figure 10 – Loads on the end post

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Calculation of compressed - curved outer pillars is carried out:

a) for strength:

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

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

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

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

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

Figure 1 – Rack loading diagram 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 arriving at one of the 8 racks is:

P = 34.73 / 8 = 4.34 tons = 43400N – pressure on one rack.

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

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

Let's consider a box-section rack welded from two plates

Definition of eccentricities:

If 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=2.5 - eccentrically compressed (stretched) stand.

Determining the size of the rack section:

The main load for the rack is the longitudinal force. Therefore, to select a cross-section, tensile (compressive) strength calculations are used:

From this equation the required cross-sectional area is found

,mm 2 (10)

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

(11)

Design resistance R U depends on the stress concentration and the yield strength of the material. Stress concentrations in welded joints are most often caused by weld seams. The value of the concentration coefficient depends on the shape, size and location of the seams. The higher the stress concentration, the lower the permissible 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 group 6, therefore, R U = 45 MPa.

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

Coefficient at reflects the dependence of the permissible 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,

and also on the sign of the stresses. Tension promotes, and compression prevents the occurrence of cracks, so the value γ at the same ρ depends on the sign of σ max. In the case of pulsating loading, when σ min= 0, ρ=0 for compression γ=2 for tension γ = 1,67.

For ρ→ ∞ γ→∞. In this case, the permissible 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 is possible during the first load. Therefore, when determining [σ], it is necessary to take into account the conditions of static strength and stability.

With static stretching (without bending)

[σ] = R y. (12)

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

(13)

where γ m is the reliability coefficient for the material.

For 09G2S σ T = 325 MPa, γ t = 1,25

During static compression, the permissible stress is reduced due to the risk 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 load application, you can take φ = 0.6. This coefficient means that the compressive strength of the rod due to loss of stability is reduced to 60% of the tensile strength.

Substitute the data into the formula:

Of the two values ​​[σ], we choose the smallest. And in the future, calculations will be made based on it.

Allowable voltage

We put 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 cm2

Channel number – 40P. Has parameters:

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

We obtain the cross-sectional area of ​​the rack, consisting of 2 channels - 61.5 cm 2.

Let's substitute the data into formula 12 and calculate the voltages 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 the structure can withstand the applied load.

Verification calculation of the overall stability of the racks.

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

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 – half-wave length of the curved axis of the stand,

μ – coefficient depending on the fastening condition; at console = 2;

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

(16)

Substitute the data into formula 20 and 21:

Stability calculations are carried out using the formula:

(17)

The coefficient φ y is determined in the same way as for central compression, according to table. 6 depending on the flexibility of the strut λ у (λ уо) when bending around the y axis. Coefficient With takes into account the decrease in stability due to torque M X.

A column is a vertical element of the supporting structure of a building that transfers loads from the structures above 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, or a composite section of channels, angles, and 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 weight and have a beautiful aesthetic appearance, however, the internal cavities cannot be painted, so this profile must be hermetically sealed.

The use of wide-flange I-beams for columns is widespread - when pinching a column in one plane, this type of profile is optimal.

The method of securing the column in the foundation is of great importance. The column can have a hinged fastening, rigid in one plane and hinged in the other, 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 design length of the column depends on the method of fastening.

It is also necessary to consider the method of fastening purlins, wall panels, beams or trusses to the column; if the load is transmitted from the side of the column, then 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, however, in the calculation it is usually considered 0.7l because the beam bends under the influence of the load and there is no complete pinching.

In practice, the column is not considered separately, but a frame or a 3-dimensional model of the building 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 holes from bolts, so it is sometimes necessary to check the section manually .

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

We calculate the column using the following parameters:

1. Central 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 maximum flexibility of the rod (in 2 planes)

1. Central tensile/compressive strength

According to SP 16.13330 clause 7.1.1, strength calculation of steel elements with standard resistance R yn ≤ 440 N/mm2 with central tension or compression by force N should be fulfilled according to the formula

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

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

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

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

2. Stability under central compression

Stability calculations are carried out in accordance with SP 16.13330 clause 7.1.3 using the formula

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

R

γ

φ — stability coefficient under central compression.

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

Where R y—calculated resistance of steel;

E- elastic modulus;

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

Where l ef is the design length of the rod;

i— radius of gyration of the section.

Estimated lengths l ef of columns (racks) of constant cross-section or individual sections of stepped columns according to SP 16.13330 clause 10.3.1 should be determined by the formula

Where l— column length;

μ — coefficient of effective length.

Effective length coefficients μ columns (racks) of constant cross-section should be determined depending on the conditions for securing their ends and the type of load. For some cases of fastening the ends and the type of load, the values μ are given in the following table:

The radius of inertia of the section can be found in the corresponding GOST for the profile, i.e. the profile must already be specified in advance 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 consequently the design lengths can also be different, then stability calculations must be made for 2 planes.

So now we have all the data to calculate 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 using this formula, should be taken no more than (7.6/ λ 2) with values ​​of conditional flexibility above 3.8; 4.4 and 5.8 for section types a, b and c, respectively.

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

Coefficient values φ are given in Appendix D SP 16.13330.

Now that all the initial data are known, we perform the calculation using 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. You can also change the design scheme, for example, by changing the hinged seal to a rigid one or by securing the column in the span with ties, you can reduce the design length of the rod.

It is recommended to strengthen compressed elements with solid walls of an open U-shaped section with planks or gratings. If there are no strips, then the stability should be checked for stability in case of flexural-torsional 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. A 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 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—design steel resistance;

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

n, Cx And Сy— coefficients accepted according to table E.1 SP 16.13330

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

W xn,min and W yn,min - sectional moments of resistance relative to the X-X and Y-Y axes (can be found in GOST for 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 take into account deplanation;

Wω,min – sectoral moment of resistance of the section.

If there should be no questions with the first 3 components, then taking into account the bi-moment causes some difficulties.

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

It is worth noting that many programs cannot calculate bi-torque, including SCAD which does not take it into account.

4. Checking the maximum 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

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

Just like the stability calculation, this calculation must be done for 2 planes.

If the profile is not suitable, 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 secure with ties to reduce the design length).

If the critical factor is extreme flexibility, then the lowest grade of steel can be taken because The steel grade does not affect the ultimate flexibility. The optimal option can be calculated using the selection method.

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. It is 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.

1. Load collection

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

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

• long-term load (payload, taken depending on the purpose of the building);
• short-term load (snow load, taken depending on the geographical location of the building);
• special load (seismic, explosive, etc. Not taken into account within this calculator);

Loads on a beam are divided into two types: design and standard. Design loads are used to calculate the beam for strength and stability (1st limit state). Standard loads are established by standards and are used to calculate beams for deflection (2nd limit state). Design loads are determined by multiplying the standard load by the reliability load factor. Within the framework of this calculator, the design load is used to determine the deflection of the beam to reserve.

After you have collected the surface load on the floor, measured in kg/m2, you need to calculate how much of this surface load the beam takes on. To do this, you need to multiply the surface load by the pitch of the beams (the so-called load strip).

For example: We calculated that the total load was Qsurface = 500 kg/m2, and the beam spacing was 2.5 m.

This load is entered into the calculator

After constructing the diagrams, a calculation is made for strength (1st limit state) and deflection (2nd limit state). In order to select a beam based on strength, it is necessary to find the required moment of inertia Wtr and select a suitable metal profile from the assortment table.

4. Selection of a metal beam from the assortment table