Calculation of metal columns. Calculation of racks Online rack calculation program
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 longitudinal force. Therefore, to select a cross-section, tensile (compressive) strength calculations are used:
(9)
From this equation the required area is found cross section
,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)
We 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.
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 way the ends of the rack are secured. In structural mechanics, there are seven ways to secure the ends of a strut. We will consider three main methods:
To ensure a certain margin of stability, it is necessary that the following condition be met:
Where: P - effective force;
A certain stability factor is established
Thus, when calculating elastic systems, it is necessary to be able to determine the value of the critical force Pcr. If we take into account that the force P applied to the rack causes only small deviations from the rectilinear shape of the rack of length ι, then it can be determined from the equation
where: E - elastic modulus;
J_min - minimum moment of inertia of the section;
M(z) - bending moment equal to M(z) = -P ω;
ω - the amount of deviation from the rectilinear shape of the rack;
Solving this differential equation 
A and B are constants of integration, determined by the boundary conditions.
After performing certain actions and substitutions, we obtain the final expression for the critical force P 
The minimum value of the critical force will be for n = 1 (integer) and
The equation of the elastic line of the rack will be:
where: z - current ordinate, with maximum value z=l;
An acceptable 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 strut EJ min in direct proportion and on the length of the strut l - in inverse proportion.
As was said, the stability of the elastic strut depends on the method of its fastening.
The recommended safety factor for steel racks 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 securing the ends of the rack, the coefficient of the ends of the reduced flexibility of the rack is introduced.
where: μ - reduced length coefficient (Table);
i min - the smallest radius of gyration of the cross section of the rack (table);
ι - length of the stand;
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, the value of which depends on the method of securing the ends of the rack and is given in the tables of the strength of materials reference book (G.S. Pisarenko and S.P. Fesik)
Let us give an example of calculating the critical force for a solid rectangular cross-section rod - 6 × 1 cm, rod length ι = 2 m. Fastening the ends according to scheme III.
Calculation:
From the table we find the coefficient ϑ = 9.97, μ = 1. The moment of inertia of the section will be:
and the critical voltage will be:
Obviously, the critical force P cr = 247 kgf will cause a stress in the rod of only 41 kgf/cm 2, which is significantly less than the flow limit (1600 kgf/cm 2), however, this force will cause bending of the rod, and therefore loss of stability.
Let's consider another example of calculating a wooden post with a circular cross-section, clamped at the lower end and hinged at the upper end (S.P. Fesik). Rack length 4m, compression force N=6t. Allowable stress [σ]=100kgf/cm2. We accept the reduction factor for the permissible compressive stress φ=0.5. We calculate the cross-sectional area of the rack:
Determine the diameter of the stand: 
Section moment of inertia 
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: 
Obviously, the voltage in the rack is 100 kgf/cm 2 and it is equal to the permissible voltage [σ] = 100 kgf/cm 2
Let's consider the third example of calculating a steel rack made of an I-profile, 1.5 m long, compression force 50 tf, permissible stress [σ] = 1600 kgf/cm 2. The lower end of the rack is pinched, and the upper end is free (method I).
To select the cross section, we use the formula and set the coefficient ϕ=0.5, then:
We select I-beam No. 36 from the assortment and its data: F = 61.9 cm 2, i min = 2.89 cm.
Determining the flexibility of the rack:
where: μ from the table, equal to 2, taking into account the method of pinching the rack;
The calculated voltage in the rack will be: 
5 kgf, which is approximately equal to the permissible voltage, and 0.97% more, which is acceptable in engineering calculations.
The cross-section of rods working in compression will be rational at the largest radius of gyration. When calculating the specific radius of gyration
the most optimal is tubular sections, thin-walled; for which the value is ξ=1÷2.25, and for solid or rolled profiles ξ=0.204÷0.5
conclusions
When calculating the strength and stability of racks and columns, it is necessary to take into account the method of securing the ends of the racks and apply the recommended safety factor.
The value of the critical force is obtained from the differential equation of the curved centerline of the strut (L. Euler).
To take into account all the factors characterizing a loaded rack, the concept of rack flexibility - λ, provided length coefficient - μ, voltage reduction coefficient - ϕ, critical load coefficient - ϑ - was introduced. Their values are taken from reference tables (G.S. Pisarentko and S.P. Fesik).
Approximate calculations of racks are given to determine the critical force - Pcr, critical stress - σcr, diameter of racks - d, flexibility of racks - λ and other characteristics.
The optimal cross-section for racks and columns is tubular thin-walled profiles with the same main moments of inertia.
Used Books:
G.S. Pisarenko “Handbook on the strength of materials.”
S.P. Fesik “Handbook on the strength of materials.”
IN AND. Anuriev “Handbook of mechanical engineering designer”.
SNiP II-6-74 “Loads and impacts, design standards.”
Metal structures are a complex and extremely important topic. Even a small mistake can cost hundreds of thousands and millions of rubles. In some cases, the cost of an error may be the lives of people at a construction site, as well as during operation. So, checking and double-checking calculations is necessary and important.
Using Excel to solve calculation problems is, on the one hand, not new, but at the same time not entirely familiar. However, Excel calculations have a number of undeniable advantages:
- Openness— each such calculation can be disassembled piece by piece.
- Availability— the files themselves exist in the public domain, written by MK developers to suit their needs.
- Convenience- almost any PC user is able to work with programs from the MS Office package, while specialized design solutions are expensive and, in addition, require serious effort to master.
They should not be considered a panacea. Such calculations make it possible to solve narrow and relatively simple design problems. But they do not take into account the work of the structure as a whole. In a number of simple cases they can save a lot of time:
- Calculation of beams for bending
- Calculation of beams for bending online
- Check the calculation of the strength and stability of the column.
- Check the selection of the rod cross-section.
Universal calculation file MK (EXCEL)
Table for selecting sections of metal structures, according to 5 different points SP 16.13330.2011
Actually, using this program you can perform the following calculations:
- calculation of a single-span hinged beam.
- calculation of centrally compressed elements (columns).
- calculation of tensile elements.
- calculation of eccentrically compressed or compressed-bending elements.
The Excel version must be at least 2010. To see instructions, click on the plus sign in the upper left corner of the screen.
METALLICA
The program is an EXCEL workbook with macro support.
And is intended for the calculation of steel structures according to
SP16 13330.2013 “Steel structures”
Selection and calculation of runs
Selecting a run is only a trivial task at first glance. The pitch of the purlins and their size depend on many parameters. And it would be nice to have the corresponding calculation at hand. This is what this must-read article talks about:
- calculation of the run without strands
- calculation of a run with one strand
- calculation of a purlin with two strands
- calculation of the run taking into account the bi-moment:
But there is a small fly in the ointment - apparently the file contains errors in the calculation part.
Calculation of moments of inertia of a section in excel tables
If you need to quickly calculate the moment of inertia of a composite section, or there is no way to determine the GOST according to which metal structures are made, then this calculator will come to your aid. At the bottom of the table there is a small explanation. In general, the work is simple - we select a suitable section, set the dimensions of these sections, and obtain the basic parameters of the section:
- Section moments of inertia
- Section moments of resistance
- Section radius of gyration
- Cross-sectional area
- Static moment
- Distances to the center of gravity of the section.
The table contains calculations for the following types of sections:
- pipe
- rectangle
- I-beam
- channel
- rectangular pipe
- triangle
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 coefficient of profile utilization, 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.
Posted in Tagged ,Calculation of the central pillar
Racks are structural elements that work primarily in compression and longitudinal bending.
When calculating the rack, it is necessary to ensure its strength and stability. Ensuring stability is achieved by correctly selecting the section of the rack.
When calculating a vertical load, the design diagram of the central pillar is accepted as hinged at the ends, since it is welded at the bottom and top (see Figure 3).
The central post carries 33% of the total weight of the floor.
The total weight of the floor N, kg, will be determined by: including the weight of snow, wind load, load from thermal insulation, load from the weight of the covering frame, load from vacuum.
N = R 2 g,. (3.9)
where g is the total uniformly distributed load, kg/m2;
R - internal radius of the tank, m.
The total weight of the floor consists of the following types of loads:
- 1. Snow load, g 1. It is accepted g 1 = 100 kg/m 2 .;
- 2. Load from thermal insulation, g 2. Accepted g 2 = 45 kg/m 2;
- 3. Wind load, g 3. It is accepted g 3 = 40 kg/m 2;
- 4. Load from the weight of the coating frame, g 4. Accepted g 4 =100 kg/m 2
- 5. Taking into account the installed equipment, g 5. Accepted g 5 = 25 kg/m 2
- 6. Vacuum load, g 6. Accepted g 6 = 45 kg/m 2.
And the total weight of the floor N, kg:
The force perceived by the stand is calculated:
The required cross-sectional area of the rack is determined using the following formula:
See 2, (3.12)
where: N is the total weight of the floor, kg;
1600 kgf/cm 2, for steel VSt3sp;
The buckling coefficient is structurally assumed to be =0.45.
According to GOST 8732-75, a pipe with an outer diameter D h = 21 cm, an inner diameter d b = 18 cm and a wall thickness of 1.5 cm is structurally selected, which is acceptable since the pipe cavity will be filled with concrete.
Pipe cross-sectional area, F:
The moment of inertia of the profile (J) and radius of gyration (r) are determined. Respectively:
J = cm4, (3.14)
where are the geometric characteristics of the section.
Radius of inertia:
r=, cm, (3.15)
where J is the moment of inertia of the profile;
F is the area of the required section.
Flexibility:
The voltage in the rack is determined by the formula:
Kgs/cm (3.17)
In this case, according to the tables of Appendix 17 (A. N. Serenko) it is assumed = 0.34
Calculation of the strength of the rack base
The design pressure P on the foundation is determined:
Р= Р" + Р st + Р bs, kg, (3.18)
Р st =F L g, kg, (3.19)
R bs =L g b, kg, (3.20)
where: P"-force of the vertical stand P"= 5885.6 kg;
R st - weight of the rack, kg;
g - specific gravity of steel. g = 7.85*10 -3 kg/.
R bs - weight concrete poured into the rack, kg;
g b - specific gravity of concrete grade. g b = 2.4 * 10 -3 kg/.
Required area of the shoe plate with permissible pressure on the sand base [y] f = 2 kg/cm 2:
A slab with sides is accepted: aChb = 0.65 × 0.65 m. The distributed load, q per 1 cm of the slab will be determined:
Design bending moment, M:
Design moment of resistance, W:
Plate thickness d:
The slab thickness is assumed to be d = 20 mm.
