Calculation of vertical struts of the metal frame. Calculation of a metal beam online (calculator). Center pillar calculation

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

Figure 1 - Strut loading diagram and cross-section.

The total mass 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 per strut.

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

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

Consider a box-section strut welded from two plates

Determination of eccentricities:

If the eccentricity t x has a value from 0.1 to 5 - eccentrically compressed (stretched) rack; if T from 5 to 20, the tension or compression of the beam must be taken into account in the calculation.

t x= 2.5 - eccentrically compressed (stretched) stance.

Determining the size of the cross-section of the rack:

The main load on the strut is the longitudinal force. Therefore, to select a section, a tensile (compression) strength calculation is used:

(9)

From this equation, find the required cross-sectional area

, mm 2 (10)

The permissible stress [σ] during endurance work depends on the steel grade, 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 on the yield strength of the material. Stress concentration in welded joints is most often caused by welded seams. The concentration factor value depends on the shape, size and location of the seams. The higher the stress concentration, the lower the allowable stress.

The most loaded section of the bar structure designed in operation is located near the place of its attachment to the wall. Attachment with frontal fillet seams corresponds to the 6th group, 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 asymmetry index of the cycle p, equal to the ratio of the minimum voltage per cycle to the maximum, i.e.

-1≤ρ<1,

and also from the sign of stresses. Stretching promotes and compression prevents cracking, therefore the value γ for equal ρ depends on the sign of σ max. In the case of pulsating loading, when σ min= 0, ρ = 0 in compression γ = 2 in tension γ = 1,67.

As ρ → ∞ γ → ∞. In this case, the allowable stress [σ] becomes very large. This means that the risk of fatigue failure is reduced, but does not mean that the strength is ensured, since failure is possible during the first loading. Therefore, when determining [σ], it is necessary to take into account the conditions of static strength and stability.

Static tension (no bending)

[σ] = R y. (12)

The value of the design resistance R y according to the yield point is determined by the formula

(13)

where γ m is the material safety factor.

For 09G2S σ T = 325 MPa, γ t = 1,25

In static compression, the permissible stress is reduced due to the danger of loss of stability:

where 0< φ < 1. Коэффициент φ зависит от гибкости и относительного эксцентриситета. Его точное значение может быть найдено только после определения размеров сечения. Для ориентировочного выбора Атрпо формуле следует задаться значением φ. With a small eccentricity of the load application, we can take φ = 0.6. This factor means that the compressive strength of the bar due to buckling decreases to 60% of the tensile strength.

We substitute the data into the formula:

We choose the smallest of the two values [σ]. And in the future, it will be used for calculation.

Allowable voltage

We supply data to the formula:

Since 295.8 mm 2 is an extremely small cross-sectional area, based on the structural dimensions and the magnitude of the moment, we increase to

We will select the channel number according to the area.

The minimum area of the channel should be - 60 cm 2

Channel number - 40P. Has parameters:

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

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

Substitute the data into formula 12 and calculate the voltages again:

= 146.7 MPa

The acting stresses in the section are less than the limiting stresses for the metal. This means that the material of construction can withstand the applied load.

Checking calculation of the overall stability of the racks.

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

The flexibility of the rack with respect to the material axis (i.e., the axis intersecting the section elements) is determined by the formula:

(15)

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

μ is the coefficient depending on the fixing condition; at console = 2;

i min - radius of gyration, is found by the formula:

(16)

We substitute the data into the formula 20 and 21:

The calculation of stability is carried out according to 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 y (λ yo) when bending around the y-axis. Coefficient with takes into account the decrease in stability from the action of the moment M NS.

Metal structures is a complex and extremely responsible topic. Even a small mistake can cost hundreds of thousands and millions of rubles. In some cases, the cost of a mistake can be the life of people at the construction site, as well as during operation. So, it is necessary and important to check and double-check the calculations.

Using Excel to solve computational problems is, on the one hand, not new, but not entirely familiar. However, Excel calculations have a number of undeniable advantages:

Openness- each such calculation can be disassembled by the bones.
Availability- the files themselves exist in the public domain, they are written by MK developers according to 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, moreover, require serious efforts to master.

They should not be considered a panacea. Such calculations allow solving 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:

Beam calculation for bending
Beam bending calculation online
Check the calculation of the strength and stability of the column.
Check the selection of the cross-section of the bar.

Universal calculation file MK (EXCEL)

Table for the selection of sections of metal structures, for 5 different points of 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 stretched elements.
calculation of eccentrically compressed or compressed-bending elements.

The Excel version must be at least 2010. To see the instructions, click on the plus in the upper left corner of the screen.

METALLIC

The program is an EXCEL book with macro support.
And it is intended for the calculation of steel structures according to
SP16 13330.2013 "Steel structures"

Selection and calculation of runs

The selection of a run is only a trivial task at first glance. The step of the runs and their size depend on many parameters. And it would be nice to have an appropriate calculation at hand. Actually, this is what the obligatory article tells about:

calculation of a run without strands
calculation of a single strand run
calculation of a run with two strands
calculation of the run taking into account the bimoment:

But there is a small fly in the ointment - apparently the file contains errors in the calculation part.

Calculation of the moments of inertia of a section in excel tables

If you need to quickly calculate the moment of inertia of a composite section, or it is not possible to determine the GOST by which the metal structures are made, then this calculator will come to your aid. There is a small explanation at the bottom of the table. In general, the work is simple - we select a suitable section, set the sizes of these sections, we obtain the main parameters of the section:

Moments of inertia of the section
Section resistance moments
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 tube
triangle

A column is a vertical member of a building's supporting structure 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, a composite section of channels, corners, sheets are usually used.

For centrally compressed columns, it is optimal to use a pipe or a square profile - they are economical in terms of metal weight and have a beautiful aesthetic appearance, however, the internal cavities cannot be painted, therefore this profile must be airtight.

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

The method of fixing the column in the foundation is of great importance. The column can be hinged, 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 calculated length of the column depends on the method of fastening.

It is also necessary to take into account the method of fastening the purlins, wall panels, beams or trusses to the column, if the load is transferred to the side of the column, then the 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, it is usually considered 0.7l in the calculation, since the beam bends under the action of the load and there is no complete pinching.

In practice, the column is not considered separately, but a frame or a three-dimensional model of a building is modeled in the program, it is loaded and the column in the assembly is calculated and the required profile is selected, but it can be difficult to take into account the weakening of the section by the bolt holes in the programs, so it is sometimes necessary to check the section manually ...

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

The column is calculated according to the following parameters:

1. Strength at central tensile / compressive

2. Stability under central compression (in 2 planes)

3. Strength under the combined action of longitudinal force and bending moments

4. Checking the ultimate flexibility of the rod (in 2 planes)

1. Strength at central tensile / compressive

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 performed according to the formula

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

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

γ с - coefficient of working conditions (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 the verification calculations, the selection of the section of the column can be done only by the method of selection of the section, so here we can set the starting point, less than which the section cannot be.

2. Stability under central compression

The calculation for stability is carried out in accordance with SP 16.13330 clause 7.1.3 according to the formula

A- the cross-sectional area of the gross profile, i.e. excluding the weakening of its holes;

φ - coefficient of stability at 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 bar λ (indicated with a bar above).

where R y is the calculated steel resistance;

E- elastic modulus;

λ - the flexibility of the bar, calculated by the formula:

where l ef is the calculated length of the bar;

i- radius of gyration of the section.

Estimated lengths l ef columns (posts) of constant cross-section or individual sections of stepped columns in accordance with SP 16.13330 clause 10.3.1 should be determined by the formula

where l- the length of the column;

μ - coefficient of the calculated length.

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

The radius of gyration of the section can be found in the corresponding GOST for the profile, i.e. the profile must have already been specified in advance, and the calculation is reduced to an enumeration of sections.

Because the radius of gyration in 2 planes for most of the 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 therefore the calculated lengths can also be different, then the stability calculation must be made for 2 planes.

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

If the limiting 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 the table

Coefficient values φ calculated by this formula should be taken no more than (7.6 / λ 2) when the values of the conditional slenderness are over 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 calculate 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 model, for example, by changing the hinged termination to rigid or by tying a column in the span, you can reduce the calculated length of the bar.

Compressed elements with solid walls of an open U-shaped section are recommended to be reinforced with strips or lattice. If there are no strips, then the stability should be checked for stability in the bending-torsional form of 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 by an axial compressive load, but also by a bending moment, for example, from the wind. The moment is also formed if the vertical load is applied not along 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 of SP 16.13330 according to the formula

where N- longitudinal compressive force;

A n - net cross-sectional area (taking into account weakening by holes);

R y - design steel resistance;

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

n, Сx and Сy- coefficients taken 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 - moments of resistance of the section relative to the X-X and Y-Y axes (can be found in GOST on 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 the warping;

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

If there should be no questions with the first 3 components, then accounting for the bimoment causes some difficulties.

The bimoment characterizes the changes made to the linear stress distribution zones of the section warping and, in fact, is a pair of moments directed in opposite directions

It should be noted that many programs cannot calculate the bimoment, including SCAD does not take it into account.

4. Checking the ultimate flexibility of the bar

Slenderness of compressed members λ = lef / i, as a rule, should not exceed the limit values λ u given in the table

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

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

If the profile does not fit, it is necessary to change the section by increasing the radius of inertia of the section or by changing the design model (change the fixings or fix with ties to reduce the calculated length).

If ultimate flexibility is a critical factor, the smallest steel grade can be taken. the steel grade does not affect the ultimate flexibility. The best option can be calculated by the fitting method.

Posted in Tagged,

In practice, it is often necessary to calculate the rack or column for the maximum axial (longitudinal) load. The force at which the strut loses its steady state (load-bearing capacity) is critical. The stability of the post is influenced by the way the ends of the post are secured. In structural mechanics, seven methods are considered for securing the ends of a rack. We will consider three main ways:

To ensure a certain margin of stability, it is necessary that the following condition be met:

Where: P - acting effort;

A certain safety factor of stability is established

Thus, when calculating elastic systems, it is necessary to be able to determine the value of the critical force Pcr. If we have to introduce that the force P applied to the rack causes only small deviations from the rectilinear form of the rack of length v, then it can be determined from the equation

where: E is the modulus of elasticity;
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 smallest value of the critical force will be at n = 1 (integer) and

The equation of the elastic line of the strut will look like:

where: z is the current ordinate, at the maximum value z = l;
The admissible expression for the critical force is called Euler's formula. It can be seen that the magnitude of the critical force depends on the stiffness of the strut EJ min in direct proportion and on the strut length l - inversely.
As it was said, the stability of the elastic strut depends on the method of its fastening.
The recommended safety factor for steel posts is even
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 fixing the ends of the rack, the coefficient of the ends of the reduced flexibility of the rack is introduced.

where: μ - coefficient of reduced length (Table);
i min - the smallest radius of gyration of the cross-section of the rack (table);
ι is the length of the rack;
The critical load factor is introduced:

, (table);
Thus, when calculating the cross-section of the rack, it is necessary to take into account the coefficients μ and ϑ whose value depends on the method of fixing the ends of the rack and is given in the tables of the reference book on strength materials (G.S. Pisarenko and S.P. Fesik)
Let us give an example of calculating the critical force for a solid rectangular bar - 6 × 1 cm, bar length ι = 2m. Fastening the ends according to scheme III.
Payment:
According to the table, we find the coefficient ϑ = 9.97, μ = 1. The moment of inertia of the section will be:

and the critical stress 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 much less than the flow limit (1600 kgf / cm 2), but this force will cause the rod to bend, and hence the loss of stability.
Let us consider another example of calculating a wooden rack with a circular cross-section clamped at the lower end and hinged at the upper end (S.P. Fesik). The length of the rack is 4m, the compression force is N = 6tf. Allowable stress [σ] = 100kgf / cm 2. We take the coefficient of decreasing the allowable compressive stress φ = 0.5. We calculate the cross-sectional area of the rack:

Determine the diameter of the rack:

Section moment of inertia

Calculating 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 stress in the rack is 100 kgf / cm 2 and it is exactly the permissible stress [σ] = 100 kgf / cm 2
Let us consider the third example of calculating a steel rack made of an I-profile, 1.5 m long, compressive force 50 tf, allowable 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 section, we use the formula and set the coefficient ϕ = 0.5, then:

We select I-beams No. 36 and its data from the assortment: F = 61.9 cm 2, i min = 2.89 cm.
Determine the flexibility of the rack:

where: μ from the table, even 2, taking into account the way the rack is pinched;
The calculated rack stress will be:

5kgs, which is approximately exactly the permissible voltage, and 0.97% more, which is permissible in engineering calculations.
The cross-section of the rods working in compression will be rational with the largest radius of gyration. When calculating the specific radius of gyration
the most optimal is tubular sections, thin-walled; for which the value ξ = 1 ÷ 2.25, and for solid or rolled profiles ξ = 0.204 ÷ 0.5

conclusions
When calculating the strength and stability of the racks, columns, it is necessary to take into account the method of fixing the ends of the racks, apply the recommended margin of safety.
The critical force value is obtained from the differential equation of the curved centerline of the strut (L. Euler).
To take into account all the factors characterizing the loaded rack, the concept of rack flexibility - λ, the provided length factor - μ, the voltage reduction factor - ϕ, and the critical load factor - ϑ have been introduced. Their values are taken from reference tables (G.S. Pisarentko and S.P. Fesik).
The approximate calculations of the racks are given for determining the critical force - Ркр, critical stress - σкр, diameter of the racks - d, flexibility of the racks - λ and other characteristics.
The optimal cross-section for posts and columns is tubular thin-walled profiles with the same principal moments of inertia.

Used Books:
GS Pisarenko "Handbook on the strength of materials."
SP Fesik "Handbook on the strength of materials."
IN AND. Anuryev "Handbook of the constructor-mechanical engineer".
SNiP II-6-74 “Loads and Impacts, Design Standards”.

NS the operek of the building (Fig. 5) is statically undefined once. We reveal the indeterminacy based on the condition of the same rigidity of the left and right struts and the same magnitude of horizontal displacements of the hinged end of the struts.

Rice. 5. Design scheme of the frame

5.1. Determination of geometric characteristics

1. Height of the section of the rack
... We will accept
.

2. The width of the section of the rack is taken according to the assortment, taking into account the clipping
mm .

3. Sectional area
.

Cross section moment of resistance
.

Static moment
.

Section moment of inertia
.

Section radius of gyration
.

5.2. Collecting load

a) horizontal loads

Linear wind loads

, (N / m)

where - coefficient taking into account the value of wind pressure along the height (Appendix Table 8);

- aerodynamic coefficients (at
m to accept
;
);

- load safety factor;

- standard value of wind pressure (on assignment).

Concentrated forces from wind load at the level of the top of the strut:

,
,

where - the supporting part of the farm.

b) vertical loads

Let's collect the loads in tabular form.

Table 5

Collecting the load on the rack, N

Name
Constant
1. From the cover panel
2. From the supporting structure
3. Self-weight of the rack (approximate)
Total:
Temporary
4. Snowy

Note:

1. The load from the coating panel is determined according to table 1

,
.

2. The load from the beam is determined

3. Self-weight of the arch
is determined by:

Upper belt
;

Lower belt
;

Racks.

To obtain the design load, the elements of the arch are multiplied by corresponding to metal or wood.

,
,
.

Unknown
:
.

Bending moment at the base of the rack
.

Transverse force
.

5.3. Check calculation

In the plane of bending

1. Checking for normal voltages

where - coefficient that takes into account the additional moment from the longitudinal force.

;
,

where - the coefficient of fastening (take 2.2);
.

Undervoltage should not exceed 20%. However, if the minimum rack dimensions and
, then the undervoltage may exceed 20%.

2. Checking the bearing for spalling when bending

3. Checking the stability of a flat deformation:

where
;
(Table 2 Appendix 4).

From the plane of bending

4. Test for stability

where
, if
,
;

- the distance between the ties along the length of the rack. In the absence of connections between the posts, the total length of the post is taken as the calculated length.
.

5.4. Calculation of the attachment of the rack to the foundation

Let's write out the loads
and
from table 5. The structure of attaching the rack to the foundation is shown in fig. 6.

where
.

Rice. 6. The structure of attaching the rack to the foundation

2. Compression stress
, (Pa)

where
.

3. Sizes of the compressed and stretched zones
.

4. Dimensions and :

;
.

5. Maximum tensile force in anchors

, (H)

6. Required area of anchor bolts

where
- coefficient taking into account the weakening of the thread;

- coefficient taking into account the concentration of stresses in the thread;

- coefficient taking into account the unevenness of the two anchors.

7. Required anchor diameter
.

We accept the diameter according to the assortment (Appendix Table 9).

8. For the accepted anchor diameter, a hole in the traverse is required.
mm.

9. Width of traverse (angle) fig. 4 must be at least
, i.e.
.

Let's take an isosceles corner according to the assortment (Appendix Table 10).

11. The value of the distribution load in the section of the width of the rack (Fig. 7 b).

12. Bending moment
,

where
.

13. Required moment of resistance
,

where - the design resistance of steel is taken to be 240 MPa.

14. For pre-accepted corner
.

If this condition is met, we proceed to checking the voltage, if not, we return to step 10 and take a larger corner.

15. Normal voltages
,

where
- coefficient of working conditions.

16. Beam deflection
,

where
Pa is the modulus of elasticity of steel;

- ultimate deflection (accept ).

17. Choose the diameter of the horizontal bolts from the condition of their arrangement across the fibers in two rows along the width of the rack
, where
- distance between bolt axles. If we accept metal bolts, then
,
.