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 diagram of moments and transverse forces is constructed. 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 the cross section 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.
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 this calculator will come to your aid. At the bottom of the table there is a small explanation. In general, the work is simple - choose suitable section, we set the dimensions of these sections, we obtain the main 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
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 area cross section element;
b) for stability
where is the coefficient longitudinal bending;
– calculated cross-sectional area of the element;
Loads are collected from the coverage area according to the plan, per one middle pillar ().
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 with wind from the left, p 2 and w 2, and with 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 greatest 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:
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:
(9)
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 permissible stresses on the cycle asymmetry index p, equal to the ratio minimum voltage per cycle to 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.
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, 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 factor:
, (table);
Thus, when calculating the cross-section of the rack, it is necessary to take into account the coefficients μ and ϑ, the value of which depends on the method of 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 of Strength of Materials.”
IN AND. Anuriev “Handbook of mechanical engineering designer”.
SNiP II-6-74 “Loads and impacts, design standards.”