Often people who make a covered carport in their yard for protection from the sun and atmospheric precipitation, the cross-section of the posts on which the canopy will rest is not calculated, but the cross-section is selected by eye or by consulting a neighbor.
You can understand them, the loads on the racks, in in this case being columns, not so big, the volume of work performed is also not enormous, and appearance columns are sometimes much more important than their load-bearing capacity, so even if the columns are made with a multiple safety margin, there is no big problem in this. Moreover, you can spend an endless amount of time searching for simple and clear information about the calculation of solid columns without any result - understand the examples of calculating columns for industrial buildings applying a load at several levels without good knowledge of strength materials is almost impossible, and ordering a column calculation from an engineering organization can reduce all expected savings to zero.
This article was written with the goal of at least slightly changing the current state of affairs and is an attempt to outline the main stages of the calculation as simply as possible metal column, no more. All basic calculation requirements metal columns can be found in SNiP II-23-81 (1990).
General provisions
From a theoretical point of view, the calculation of a centrally compressed element, such as a column or rack in a truss, is so simple that it is even inconvenient to talk about it. It is enough to divide the load into design resistance the steel from which the column will be made - that's all. In mathematical expression it looks like this:
F = N/Ry (1.1)
F- required cross-sectional area of the column, cm²
N- concentrated load applied to the center of gravity of the cross section of the column, kg;
Ry- the calculated resistance of the metal to tension, compression and bending at the yield point, kg/cm². The value of the design resistance can be determined from the corresponding table.
As you can see, the level of complexity of the task belongs to the second, maximum to the third class primary school. However, in practice everything is not as simple as in theory, for a number of reasons:
1. Applying a concentrated load exactly to the center of gravity of the cross-section of a column is only possible theoretically. In reality, the load will always be distributed and there will still be some eccentricity in the application of the reduced concentrated load. And since there is eccentricity, it means there is a longitudinal bending moment acting in the cross section of the column.
2. The centers of gravity of the cross sections of the column are located on one straight line - the central axis, also only theoretically. In practice, due to the heterogeneity of the metal and various defects, the centers of gravity of the cross sections can be shifted relative to the central axis. This means that the calculation must be made along a section whose center of gravity is as far away from the central axis as possible, which is why the eccentricity of the force for this section is maximum.
3. The column may not have a rectilinear shape, but be slightly curved as a result of factory or installation deformation, which means that the cross sections in the middle part of the column will have the greatest eccentricity of load application.
4. The column can be installed with deviations from the vertical, which means that it is vertical effective load can create an additional bending moment, maximum in the lower part of the column, or more precisely, at the point of attachment to the foundation, however, this is relevant only for free-standing columns.
5. Under the influence of loads applied to it, the column can deform, which means that the eccentricity of the load application will again appear and, as a consequence, an additional bending moment.
6. Depending on how exactly the column is fixed, the value of the additional bending moment at the bottom and in the middle part of the column depends.
All this leads to the appearance of longitudinal bending and the influence of this bending must be taken into account somehow in the calculations.
Naturally, it is almost impossible to calculate the above deviations for a structure that is still being designed - the calculation will be very long, complex, and the result is still doubtful. But it is very possible to introduce a certain coefficient into formula (1.1) that would take into account the above factors. This coefficient is φ - buckling coefficient. The formula that uses this coefficient looks like this:
F = N/φR (1.2)
Meaning φ is always less than one, this means that the cross section of the column will always be larger than if you simply calculate using formula (1.1), what I mean is that now the fun begins and remember that φ always less than one - it won't hurt. For preliminary calculations you can use the value φ within 0.5-0.8. Meaning φ depends on the steel grade and column flexibility λ :
λ = l ef/ i (1.3)
l ef- design length of the column. The calculated and actual length of a column are different concepts. The estimated length of the column depends on the method of securing the ends of the column and is determined using the coefficient μ :
l ef = μ l (1.4)
l - actual length of the column, cm;
μ - coefficient taking into account the method of securing the ends of the column. The coefficient value can be determined from the following table:
Table 1. Coefficients μ for determining the design lengths of columns and racks of constant cross-section (according to SNiP II-23-81 (1990))
As we can see, the coefficient value μ changes several times depending on the method of fastening the column, and the main difficulty here is which design scheme to choose. If you don’t know which fastening scheme suits your conditions, then take the value of the coefficient μ=2. The value of the coefficient μ=2 is accepted mainly for free-standing columns, clear example a free-standing column - a lamppost. The coefficient value μ=1-2 can be taken for canopy columns on which beams rest without rigid attachment to the column. This design scheme can be adopted when the canopy beams are not rigidly attached to the columns and when the beams have a relatively large deflection. If the column will be supported by trusses rigidly attached to the column by welding, then the value of the coefficient μ=0.5-1 can be taken. If there are diagonal connections between the columns, then you can take the value of the coefficient μ = 0.7 for non-rigid fastening of diagonal connections or 0.5 for rigid fastening. However, such stiffness diaphragms do not always exist in 2 planes and therefore such coefficient values must be used carefully. When calculating the truss posts, the coefficient μ=0.5-1 is used, depending on the method of securing the posts.
The slenderness coefficient value approximately shows the ratio of the design length of the column to the height or width of the cross section. Those. the higher the value λ , the smaller the width or height of the cross-section of the column and, accordingly, the greater the cross-sectional margin required for the same column length, but more on that a little later.
Now that we have determined the coefficient μ , you can calculate the design length of the column using formula (1.4), and in order to find out the flexibility value of the column, you need to know the radius of gyration of the column section i :
Where I- moment of inertia of the cross section relative to one of the axes, and here the most interesting thing begins, because in the course of solving the problem we must determine required area column sections F, but this is not enough; it turns out that we still need to know the value of the moment of inertia. Since we do not know either one or the other, the solution to the problem is carried out in several stages.
On preliminary stage usually the value is taken λ within 90-60, for columns with a relatively small load you can take λ = 150-120 (maximum value for columns - 180, values extreme flexibility for other elements can be found in table 19* SNiP II-23-81 (1990). Then, according to Table 2, the value of the flexibility coefficient is determined φ :
Table 2. Buckling coefficients φ of centrally compressed elements.
Note: coefficient values φ in the table are magnified 1000 times.
After this, the required radius of gyration of the cross section is determined by transforming formula (1.3):
i = l ef/λ (1.6)
A rolled profile with a corresponding radius of gyration value is selected according to the assortment. Unlike bending elements, where the section is selected along only one axis, since the load acts only in one plane, in centrally compressed columns longitudinal bending can occur relative to any of the axes and therefore the closer the value of I z to I y, the better, in other words In other words, the most preferable profiles are round or square section. Well, now let’s try to determine the cross-section of the column based on the knowledge gained.
Example of calculation of a metal centrally compressed column
There is: a desire to make a canopy near the house approximately as follows:
In this case, the only centrally compressed column under any fastening conditions and under a uniformly distributed load will be the column shown in red in the figure. In addition, the load on this column will be maximum. Columns marked in blue and green, can be considered as centrally compressed only with appropriate constructive solution and uniformly distributed load, columns marked orange, will be either centrally compressed or eccentrically compressed or frame racks calculated separately. In this example, we will calculate the cross section of the column indicated in red. For calculations, we will assume a permanent load from the canopy’s own weight of 100 kg/m² and a temporary load of 100 kg/m² from the snow cover.
2.1. Thus, the concentrated load on the column, indicated in red, will be:
N = (100+100) 5 3 = 3000 kg
2.2. We accept the preliminary value λ = 100, then according to table 2 the bending coefficient φ = 0.599 (for steel with a design strength of 200 MPa, given value adopted to provide an additional safety margin), then the required cross-sectional area of the column is:
F= 3000/(0.599 2050) = 2.44 cm²
2.3. According to table 1 we take the value μ = 1 (since roof covering made of profiled flooring, properly fixed, will ensure rigidity of the structure in a plane parallel to the plane of the wall, and in a perpendicular plane, the relative immobility of the top point of the column will be ensured by fastening the rafters to the wall), then the radius of inertia
i= 1·250/100 = 2.5 cm
2.4. According to the assortment for square profile pipes, these requirements are satisfied by a profile with cross-sectional dimensions of 70x70 mm with a wall thickness of 2 mm, having a radius of gyration of 2.76 cm. The cross-sectional area of such a profile is 5.34 cm². This is much more than is required by calculation.
2.5.1. We can increase the flexibility of the column, while the required radius of gyration decreases. For example, when λ = 130 bending factor φ = 0.425, then the required cross-sectional area of the column:
F = 3000/(0.425 2050) = 3.44 cm²
2.5.2. Then
i= 1·250/130 = 1.92 cm
2.5.3. According to the assortment for square profile pipes, these requirements are satisfied by a profile with cross-sectional dimensions of 50x50 mm with a wall thickness of 2 mm, having a radius of gyration of 1.95 cm. The cross-sectional area of such a profile is 3.74 cm², the moment of resistance for this profile is 5.66 cm³.
Instead of square profile pipes, you can use an equal angle angle, a channel, an I-beam, or a regular pipe. If the calculated resistance of the steel of the selected profile is more than 220 MPa, then the cross section of the column can be recalculated. That’s basically all that concerns the calculation of metal centrally compressed columns.
Calculation of an eccentrically compressed column
Here, of course, the question arises: how to calculate the remaining columns? The answer to this question greatly depends on the method of attaching the canopy to the columns. If the canopy beams are rigidly attached to the columns, then a rather complex statically indeterminate frame will be formed, and then the columns should be considered as part of this frame and the cross-section of the columns should be calculated additionally for the action of the transverse bending moment. We will further consider the situation when the columns shown in the figure , are hingedly connected to the canopy (we are no longer considering the column marked in red). For example, the head of the columns has a support platform - metal plate with holes for bolting canopy beams. For various reasons, the load on such columns can be transmitted with a fairly large eccentricity:
The beam shown in the picture is beige color, under the influence of the load it will bend a little and this will lead to the fact that the load on the column will be transmitted not along the center of gravity of the column section, but with eccentricity e and when calculating extreme columns this eccentricity must be taken into account. There are a great many cases of eccentric loading of columns and possible cross sections of columns, described by the corresponding formulas for calculation. In our case, to check the cross-section of an eccentrically compressed column, we will use one of the simplest:
(N/φF) + (M z /W z) ≤ R y (3.1)
In this case, when we have already determined the cross-section of the most loaded column, it is enough for us to check whether such a cross-section is suitable for the remaining columns for the reason that we do not have the task of building a steel plant, but we are simply calculating the columns for the canopy, which will all have the same cross-section for reasons of unification.
What's happened N, φ And R y we already know.
Formula (3.1) after the simplest transformations will take the following form:
F = (N/R y)(1/φ + e z ·F/W z) (3.2)
because M z =N e z, why the value of the moment is exactly what it is and what the moment of resistance W is is explained in sufficient detail in a separate article.
for the columns indicated in blue and green in the figure will be 1500 kg. We check the required cross-section at such a load and previously determined φ = 0,425F = (1500/2050)(1/0.425 + 2.5 3.74/5.66) = 0.7317 (2.353 + 1.652) = 2.93 cm²
In addition, formula (3.2) allows you to determine the maximum eccentricity that the already calculated column will withstand; in this case, the maximum eccentricity will be 4.17 cm.
The required cross-section of 2.93 cm² is less than the accepted 3.74 cm², and therefore square profile pipe with cross-sectional dimensions of 50x50 mm and a wall thickness of 2 mm can also be used for outer columns.
Calculation of an eccentrically compressed column based on conditional flexibility
Oddly enough, to select the cross-section of an eccentrically compressed column - a solid rod - there is an even simpler formula:
F = N/φ e R (4.1)
φ e- buckling coefficient, depending on eccentricity, it could be called the eccentric buckling coefficient, so as not to be confused with the buckling coefficient φ . However, calculations using this formula may turn out to be longer than using formula (3.2). To determine the coefficient φ e you still need to know the meaning of the expression e z ·F/W z- which we met in formula (3.2). This expression is called relative eccentricity and is denoted m:
m = e z ·F/W z (4.2)
After this, the reduced relative eccentricity is determined:
m ef = hm (4.3)
h- this is not the height of the section, but a coefficient determined according to table 73 of SNiPa II-23-81. I'll just say that the coefficient value h varies from 1 to 1.4, for most simple calculations you can use h = 1.1-1.2.
After this, you need to determine the conditional flexibility of the column λ¯ :
λ¯ = λ√‾(R y / E) (4.4)
and only after that, using Table 3, determine the value φ e :
Table 3. Coefficients φ e for checking the stability of eccentrically compressed (compressed-bending) solid-walled rods in the plane of moment action coinciding with the plane of symmetry.
Notes:
1. Coefficient values φ
e magnified 1000 times.
2. Meaning φ
should not be taken more than φ
.
Now, for clarity, let’s check the cross-section of columns loaded with eccentricity using formula (4.1):
4.1. The concentrated load on the columns indicated in blue and green will be:
N = (100+100) 5 3/2 = 1500 kg
Load application eccentricity e= 2.5 cm, buckling coefficient φ = 0,425.
4.2. We have already determined the value of relative eccentricity:
m = 2.5 3.74/5.66 = 1.652
4.3. Now let’s determine the value of the reduced coefficient m ef :
m ef = 1.652 1.2 = 1.984 ≈ 2
4.4. Conditional flexibility at the flexibility coefficient we adopted λ = 130, steel strength R y = 200 MPa and elastic modulus E= 200000 MPa will be:
λ¯ = 130√‾(200/200000) = 4.11
4.5. Using Table 3, we determine the value of the coefficient φ e ≈ 0.249
4.6. Determine the required column section:
F = 1500/(0.249 2050) = 2.94 cm²
Let me remind you that when determining the cross-sectional area of the column using formula (3.1), we obtained almost the same result.
Advice: To ensure that the load from the canopy is transferred with minimal eccentricity, a special platform is made in the supporting part of the beam. If the beam is metal, made from a rolled profile, then it is usually enough to weld a piece of reinforcement to the bottom flange of the beam.
In practice, it often becomes necessary to calculate a rack or column for the maximum axial (longitudinal) load. The force at which the stand loses steady 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, seven methods are considered for securing 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 look like:
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 mentioned, 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 cross-section rod rectangular shape- 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 look at another calculation example wooden stand round section pinched at the lower end and hinged at the upper (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 critical force value is obtained from differential equation curved center line of the rack (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).
Given approximate calculations racks, 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 of Strength of Materials.”
IN AND. Anuriev “Handbook of mechanical engineering designer”.
SNiP II-6-74 “Loads and impacts, design standards.”
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 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 (1 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. Then the distributed load on the metal beam will be: Qdistributed = 500 kg/m2 * 2.5 m = 1250 kg/m. This load is entered into the calculator
2. Constructing diagramsNext, a moment diagram is constructed, shear force. The diagram depends on the loading pattern of the beam and the type of beam support. The diagram is constructed according to the rules of structural mechanics. For the most frequently used loading and support schemes, there are ready-made tables with derived formulas for diagrams and deflections.
3. Calculation of strength and deflectionAfter 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. The vertical maximum deflection fult is taken according to table 19 from SNiP 2.01.07-85* (Loads and impacts). Point 2.a depending on the span. For example, the maximum deflection is fult=L/200 with a span of L=6m. means that the calculator will select a section of a rolled profile (I-beam, channel or two channels in a box), the maximum deflection of which will not exceed fult=6m/200=0.03m=30mm. To select a metal profile based on deflection, find the required moment of inertia Itr, which is obtained from the formula for finding the maximum deflection. And also a suitable metal profile is selected from the assortment table.
4. Selection of a metal beam from the assortment tableFrom two selection results (limit state 1 and 2), a metal profile with a large section number is selected.
1. Obtaining information about the material of the rod to determine the maximum flexibility of the rod by calculation or according to the table:
2. Obtaining information about geometric dimensions cross-section, length and methods of securing the ends to determine the category of the rod depending on the flexibility:
where A is the cross-sectional area; J m i n - minimum moment of inertia (from axial ones);
μ - coefficient of reduced length.
3. Selection of calculation formulas for determining the critical force and critical stress.
4. Verification and sustainability.
When calculating using the Euler formula, the stability condition is:
F- effective compressive force; - permissible safety factor.
When calculated using the Yasinsky formula
Where a, b- design coefficients depending on the material (the values of the coefficients are given in Table 36.1)
If the stability conditions are not met, it is necessary to increase the cross-sectional area.
Sometimes it is necessary to determine the stability margin at a given load:
When checking stability, the calculated endurance margin is compared with the permissible one:
Examples of problem solving
Solution
1. The flexibility of the rod is determined by the formula
2. Determine the minimum radius of gyration for the circle. 
Substituting expressions for Jmin And A(section circle)
- Length reduction factor for a given fastening scheme μ = 0,5.
- The flexibility of the rod will be equal to
Example 2. How will the critical force for the rod change if the method of securing the ends is changed? Compare the presented diagrams (Fig. 37.2)
Solution
Critical force will increase 4 times. 
Example 3. How will the critical force change when calculating stability if the I-section rod (Fig. 37.3a, I-beam No. 12) is replaced by a rod rectangular section the same area (Fig. 37.3 b )
? Other design parameters do not change. Perform the calculation using Euler's formula.
Solution
1. Determine the width of the section of the rectangle, the height of the section is equal to the height of the section of the I-beam. The geometric parameters of I-beam No. 12 according to GOST 8239-89 are as follows:
cross-sectional area A 1 = 14.7 cm 2;
the minimum of the axial moments of inertia.
By condition, the area of the rectangular cross-section is equal to the cross-sectional area of the I-beam. Determine the width of the strip at a height of 12 cm.
2. Let us determine the minimum of the axial moments of inertia.
3. The critical force is determined by Euler’s formula:
4. Other things being equal, the ratio of the critical forces is equal to the ratio of the minimum moments of inertia:
5. Thus, the stability of a rod with an I-section No. 12 is 15 times higher than the stability of a rod of the selected rectangular cross-section.
Example 4. Check the stability of the rod. A rod 1 m long is clamped at one end, the cross-section is channel No. 16, the material is StZ, the stability margin is threefold. The rod is loaded with a compressive force of 82 kN (Fig. 37.4).
Solution
1. Determine the main geometric parameters of the rod section according to GOST 8240-89. Channel No. 16: cross-sectional area 18.1 cm 2; minimum axial section moment 63.3 cm 4 ; minimum radius of gyration of the section r t; n = 1.87 cm.
Ultimate flexibility for material StZ λpre = 100.
Design flexibility of the rod at length l = 1m = 1000mm
The rod being calculated is a highly flexible rod; the calculation is carried out using the Euler formula.
4. Stability condition
82kN< 105,5кН. Устойчивость стержня обеспечена.
Example 5. In Fig. Figure 2.83 shows the design diagram of a tubular strut of an aircraft structure. Check the stand for stability at [ n y] = 2.5, if it is made of chromium-nickel steel, for which E = 2.1*10 5 and σ pts = 450 N/mm 2.
Solution
To calculate stability, the critical force for a given rack must be known. It is necessary to establish by what formula the critical force should be calculated, i.e. it is necessary to compare the flexibility of the rack with the maximum flexibility for its material.
We calculate the value of the maximum flexibility, since there is no tabular data on λ, pre for the rack material:
To determine the flexibility of the calculated rack, we calculate the geometric characteristics of its cross section:
Determining the flexibility of the rack:
and make sure that λ< λ пред, т. е. критическую силу можно определить ею формуле Эйлера:
We calculate the calculated (real) stability factor:
Thus, n y > [ n y] by 5.2%.
Example 2.87. Check the strength and stability of the specified rod system (Fig. 2.86). The material of the rods is St5 steel (σ t = 280 N/mm 2). Required safety factors: strength [n]= 1.8; sustainability = 2.2. The rods have a circular cross-section d 1 = d 2= 20 mm, d 3 = 28 mm.
Solution
By cutting out the node where the rods meet and composing equilibrium equations for the forces acting on it (Fig. 2.86)
we establish that the given system is statically indeterminate (three unknown forces and two static equations). It is clear that in order to calculate rods for strength and stability, it is necessary to know the magnitude of the longitudinal forces arising in their cross sections, i.e., it is necessary to reveal static indetermination.
We create a displacement equation based on the displacement diagram (Fig. 2.87):
or, substituting the values of changes in the lengths of the rods, we get
Having solved this equation together with the equations of statics, we find:
Stresses in cross sections of rods 1 And 2 (see Fig. 2.86):
Their safety factor
To determine the stability safety factor of the rod 3 it is necessary to calculate the critical force, and this requires determining the flexibility of the rod in order to decide what formula to find N Kp should be used.
So λ 0< λ < λ пред и критическую силу следует определять по эмпирической формуле:
Safety factor
Thus, the calculation shows that the stability safety factor is close to the required one, and the safety factor is significantly higher than the required one, i.e., when the system load increases, the rod loses stability 3 more likely than the occurrence of yield in the rods 1 And 2.
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 calculation 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 you can help will come this calculator. 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
