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 method of fixing the ends of the rack. In structural mechanics, seven methods are considered for securing the ends of the rack. We will consider three main methods:
To ensure a certain margin of stability, it is necessary that the following condition be met:
Where: P - acting force;
A certain stability factor is set
Thus, when calculating elastic systems, it is necessary to be able to determine the value of the critical force Рcr. If we introduce that the force P applied to the rack causes only small deviations from the rectilinear shape of the rack with length ι, then it can be determined from the equation
where: E - modulus of elasticity;
J_min - minimum moment of inertia of the section;
M(z) - bending moment equal to M(z) = -P ω;
ω - the magnitude of the deviation from the rectilinear shape of the rack;
Solving this differential equation 
A and B constants of integration are determined by the boundary conditions.
Having performed 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 rack will look like:
where: z - current ordinate, at the maximum value z=l;
The admissible expression for the critical force is called L. Euler's formula. It can be seen that the value of the critical force depends on the rigidity of the rack EJ min in direct proportion and on the length of the rack l - inversely proportional.
As mentioned, the stability of the elastic rack depends on how it is fixed.
The recommended safety margin for steel studs 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 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 - rack length;
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 fixing the ends of the rack and is given in the tables of the handbook on strength materials (G.S. Pisarenko and S.P. Fesik)
Let us give an example of calculating the critical force for a rod of a solid section of a rectangular shape - 6 × 1 cm, the length of the rod ι = 2m. Fixing the ends according to scheme III.
Calculation:
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:
It is obvious that 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), however, this force will cause the rod to bend, which means loss of stability.
Consider another example of calculating a wooden rack of circular cross section, pinched at the lower end and hinged at the upper end (S.P. Fesik). Stand length 4m, compression force N=6tf. Permissible stress [σ]=100kgf/cm 2 . We accept the reduction factor of the allowable stress for compression φ=0.5. We calculate the sectional area of the rack:
Determine the diameter of the rack: 
Moment of inertia of section 
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 stress in the rack is 100kgf/cm 2 and it is exactly the allowable stress [σ]=100kgf/cm 2
Let's consider the third example of calculation of a steel rack from an I-profile, 1.5 m long, compression force 50 tf, allowable stress [σ]=1600 kgf/cm 2 . The lower end of the rack is pinched, and the upper end is free (I method).
To select the section, we use the formula and set the coefficient ϕ=0.5, then:
We select from the range I-beam No. 36 and its data: F = 61.9 cm 2, i min = 2.89 cm.
Determine the flexibility of the rack:
where: μ from the table, equal to 2, taking into account the way the rack is pinched;
The design voltage in the rack will be: 
5kgf, which is approximately equal to the allowable voltage, and 0.97% more, which is acceptable in engineering calculations.
The cross section of the rods working in compression will be rational with the largest radius of inertia. 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 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 value of the critical force is obtained from the differential equation of the curved axial line of the rack (L. Euler).
To take into account all the factors characterizing the loaded rack, the concept of rack flexibility - λ, provided length factor - μ, stress reduction factor - ϕ, critical load factor - ϑ. Their values are taken from reference tables (G.S. Pisarentko and S.P. Fesik).
Approximate calculations of struts are given to determine the critical force - Рcr, critical stress - σcr, strut diameter - d, strut flexibility - λ and other characteristics.
The optimal section for racks and columns is tubular thin-walled profiles with the same principal moments of inertia.
Used Books:
G.S Pisarenko "Handbook on the strength of materials."
S.P. Fesik "Handbook of Strength of Materials".
IN AND. Anuryev "Handbook of the designer-machine builder".
SNiP II-6-74 "Loads and impacts, design standards".
P 
the apron of the building (Fig. 5) is once statically indeterminate. 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. Calculation scheme of the frame
5.1. Definition of geometric characteristics
1. Rack section height 
. Accept 
.
2. The width of the section of the rack is taken according to the assortment, taking into account the sharpness 
mm .
3. Cross-sectional area 
.
section modulus 
.
Static moment 
.
Moment of inertia of section 
.
Radius of gyration of section 
.
5.2. Load collection
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 accept 
;
);
- load safety factor;
- normative value of wind pressure (according to the task).
Concentrated forces from wind load at the level of the top of the rack:
,
,
where 
- the supporting part of the farm.
b) vertical loads
We will collect the loads in tabular form.
Table 5
Collecting the load on the rack, N
|
Name |
|
|
|
|
Constant |
|||
|
1. Off panel cover |
|
|
|
|
2. From the supporting structure |
|
|
|
|
3. Net weight of rack (approximately) |
|
|
|
|
Total: |
|
|
|
|
Temporary |
|||
|
4. Snowy |
|
|
|
|
|
|
||
Note:
1. The load from the cover panel is determined from table 1
,
.
2. The load from the beam is determined
.
3. Own weight of the arch 
defined:
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 column 
.
Shear force 
.
5.3. Check calculation
In the plane of the bend
1. Normal stress test
,
where 
- coefficient taking into account the additional moment from the longitudinal force.
;
,
where 
- fixing coefficient (accept 2.2); 
.
Undervoltage should not exceed 20%. However, if minimum rack dimensions are accepted and 
, then the undervoltage can exceed 20%.
2. Checking the supporting part for chipping when bending
.
3. Checking the stability of a flat deformation form:
,
where 
;
(Table 2 appendix 4).
From the plane of the bend
4. Stability test
,
where 
, if 
,
;
- the distance between the bonds along the length of the rack. In the absence of connections between the racks, the full length of the rack is taken as the estimated length 
.
5.4. Calculation of attaching the rack to the foundation
Let's write out the loads 
and 
from table 5. The design of attaching the rack to the foundation is shown in fig. 6.
where 
.
Rice. 6. The design of attaching the rack to the foundation
2. Compressive stresses 
, (Pa)
where 
.
3. Dimensions of the compressed and stretched zones 
.
4. Dimensions 
and 
:
; 
.
5. Maximum tensile force in anchors
, (N)
6. Required area of anchor bolts
,
where 
- coefficient taking into account the weakening of the thread;
- coefficient taking into account the stress concentration in the thread;
- coefficient taking into account the uneven operation of two anchors.
7. Required anchor diameter 
.
We accept the diameter according to the assortment (Appendix Table 9).
8. Accepted anchor diameter will require a hole in the traverse 
mm.
9. Width of the traverse (corner) fig. 4 must be at least 
, i.e. 
.
Let's take an equilateral 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 assumed to be 240 MPa.
14. For pre-accepted corner 
.
If this condition is met, we proceed to the voltage test, if not, we return to step 10 and accept a larger angle.
15. Normal stresses 
,
where 
- coefficient of working conditions.
16. Traverse deflection 
,
where 
Pa is the modulus of elasticity of steel;
- ultimate deflection (accept 
).
17. We choose the diameter of the horizontal bolts from the condition of their placement across the fibers in two rows along the width of the rack 
, where 
- distance between the axes of the bolts. If we accept metal bolts, then 
,
.
Let's take the diameter of the horizontal bolts according to the application table. ten.
18. The smallest bearing capacity of the bolt:
a) by the condition of the collapse of the extreme element 
.
b) according to the condition of bending 
,
where 
- appendix table. eleven.
19. Number of horizontal bolts 
,
where 
- the smallest bearing capacity from clause 18; 
- the number of cuts.
Let's take the number of bolts as an even number, because arrange them in two rows.
20. Lining length 
,
where 
- the distance between the axes of the bolts along the fibers. If the bolts are metal 
;
- number of distances 
along the length of the patch.
The height of the rack and the length of the arm of the application of the force P is selected constructively, according to the drawing. Let's take the section of the rack as 2Sh. 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 - Scheme of loading the rack 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 coming to one of the 8 racks is:
P \u003d 34.73 / 8 \u003d 4.34 tons \u003d 43400N - pressure per rack.
The force does not act in the center of the section, so it causes a moment equal to:
Mx \u003d P * L; Mx = 43400 * 5000 = 217000000 (N*mm)
Consider a box-section strut welded from two plates
Definition 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, then the tension or compression of the beam must be taken into account in the calculation.
t x\u003d 2.5 - eccentrically compressed (stretched) rack.
Determining the size of the section of the rack:
The main load for the rack is the longitudinal force. Therefore, to select the section, the calculation for tensile (compressive) strength is used:
(9)
From this equation find the required cross-sectional area
,mm 2 (10)
Permissible stress [σ] during endurance work depends on the steel grade, stress concentration in the section, number of loading cycles and cycle asymmetry. In SNiP, the allowable 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 welds. The value of the concentration coefficient 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 the work is located near the place of its attachment to the wall. Attachment with frontal fillet welds corresponds to the 6th group, therefore, RU = 45 MPa.
For the 6th group, with n = 10 -6, α = 1.63;
Coefficient at reflects the dependence of allowable 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,
as well as from the sign of stresses. Tension promotes, and compression prevents cracking, so the value γ for the same ρ depends on the sign of σ max. In the case of pulsating loading, when σmin= 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 strength is ensured, since failure during the first loading is possible. Therefore, when determining [σ], it is necessary to take into account the conditions of static strength and stability.
Under static tension (no bending)
[σ] = R y. (12)
The value of the design resistance R y according to the yield strength is determined by the formula
(13)
where γ m is the reliability factor for the material.
For 09G2S σ Т = 325 MPa, γ t = 1,25
In static compression, the allowable stress is reduced due to the risk of buckling:
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, φ can be taken = 0.6. This coefficient means that the compressive strength of the rod is reduced to 60% of the tensile strength due to buckling.
We substitute the data in the formula:
Of the two values of [ σ] choose the smallest. And in the future, it will be calculated.
Allowable voltage 
Putting 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 cm 2
Channel number - 40P. Has options:
h=400 mm; b=115mm; s=8mm; t=13.5mm; F=18.1 cm 2 ;
We get the cross-sectional area of \u200b\u200bthe rack, consisting of 2 channels - 61.5 cm 2.
Substitute the data in formula 12 and calculate the stresses 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 construction can withstand the applied load.
Verification 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=Mu=0), then the reduction 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 relative 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 - coefficient depending on the condition of fixing; at console = 2;
i min - radius of inertia, is found by the formula:
(16)
We substitute the data in the formula 20 and 21:
Calculation of stability is carried out according to the formula:
(17)
The coefficient φ y is determined in the same way as with central compression, according to table. 6 depending on the flexibility of the rack λ y (λ yo) when bending around the y axis. Coefficient With takes into account the decrease in stability due to the action of the moment M X.
1. Collection of loads
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 the action, the load is divided into permanent and temporary.
- own weight of a 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. This calculator does not take into account);
The loads on the beam are divided into two types: design and standard. Design loads are used to calculate the strength and stability of the beam (1 limit state). The normative loads are established by the norms and are used to calculate the beam for deflection (limit state 2). Design loads are determined by multiplying the standard load by the reliability load factor. Within the framework of this calculator, the design load is applied when determining the deflection of the beam to the margin.
After we have collected the surface load on the ceiling, measured in kg / m2, it is necessary to calculate how much of this surface load the beam takes on. To do this, you need to multiply the surface load by the step of the beams (the so-called cargo lane).
For example: We calculated that the total load turned out to be Qsurface = 500kg / m2, and the step of the beams was 2.5m. Then the distributed load on the metal beam will be: Qdistribution = 500kg/m2 * 2.5m = 1250kg/m. This load is entered into the calculator
2. PlottingNext, the diagram of the moments, the transverse force is plotted. The diagram depends on the beam loading scheme, the type of beam support. The plot is built according to the rules of structural mechanics. For the most commonly used loading and support schemes, there are ready-made tables with derived formulas for diagrams and deflections.
3. Calculation of strength and deflectionAfter plotting the diagrams, the strength (1st limit state) and deflection (2nd limit state) are calculated. In order to select a beam for strength, it is necessary to find the required moment of inertia Wtr and select a suitable metal profile from the assortment table. The vertical limit deflection fult is taken according to Table 19 of SNiP 2.01.07-85* (Loads and impacts). Paragraph 2.a depending on the span. For example, the maximum deflection fult=L/200 with a span of L=6m. means that the calculator will select the section of the rolled profile (an I-beam, a 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 according to the deflection, the required moment of inertia Itr is found, which is obtained from the formula for finding the maximum deflection. And also from the assortment table, a suitable metal profile is selected.
4. Selection of a metal beam from the assortment tableFrom the 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 ultimate flexibility of the rod by calculation or according to the table:
2. Obtaining information about the geometric dimensions of the cross section, length and methods of fixing 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);
μ - coefficient of reduced length.
3. The choice of calculation formulas for determining the critical force and critical stress.
4. Verification and sustainability.
When calculating by the Euler formula, the stability condition is:
F- acting compressive force; - allowable stability factor.
When calculating according to the Yasinsky formula
where a, b- design coefficients depending on the material (the values \u200b\u200bof 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 for a given loading:
When checking stability, the calculated endurance is compared with the allowable:
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 BUT(section circle)
- Length reduction factor for a given fastening scheme μ = 0,5.
- The flexibility of the rod will be
Example 2 How will the critical force for the rod change if the method of fixing the ends is changed? Compare the presented schemes (Fig. 37.2)
Solution
Critical power will increase by 4 times. 
Example 3 How will the critical force change when calculating for stability if the I-section rod (Fig. 37.3a, I-beam No. 12) is replaced by a rectangular rod of the same area (Fig. 37.3 b )
? The rest of the design parameters remain unchanged. The calculation is carried out according to the Euler 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 the 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 a rectangular section is equal to the sectional area of an I-beam. We determine the width of the strip at a height of 12 cm.
2. Determine the minimum of the axial moments of inertia.
3. The critical force is determined by the Euler formula:
4. Other things being equal, the ratio of critical forces is equal to the ratio of the minimum moments of inertia:
5. Thus, the stability of a rod with a section of I-beams No. 12 is 15 times higher than the stability of a rod of a selected rectangular section.
Example 4 Check rod stability. A rod 1 m long is pinched at one end, the section is channel No. 16, the material is StZ, the stability margin is three times. The rod is loaded with a compressive force of 82 kN (Fig. 37.4).
Solution
1. We determine the main geometric parameters of the rod section according to GOST 8240-89. Channel No. 16: sectional area 18.1 cm 2; the minimum axial moment of the section is 63.3 cm 4; minimum radius of gyration of the section g t; n = 1.87cm.
Ultimate flexibility for StZ material λ pre = 100.
Calculated bar flexibility at length l = 1m = 1000mm
The calculated rod is a rod of great flexibility, the calculation is carried out according to the Euler formula.
4. Stability condition
82kN< 105,5кН. Устойчивость стержня обеспечена.
Example 5 On fig. 2.83 shows a design diagram of a tubular rack of an aircraft structure. Check the stand for stability when [ n y] \u003d 2.5 if it is made of chromium-nickel steel, for which E \u003d 2.1 * 10 5 and σ pc \u003d 450 N / mm 2.
Solution
For stability analysis, 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 ultimate flexibility for its material.
We calculate the value of ultimate flexibility, since there are no tabular data on λ, prev for the rack material:
To determine the flexibility of the calculated rack, we calculate the geometric characteristics of its cross section:
Determine 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 (actual) stability factor:
In this way, n y > [ n y] by 5.2%.
Example 2.87. Check the given rod system for strength and stability (Fig. 2.86), The rod material is St5 steel (σ t \u003d 280 N / mm 2). Required safety factors: strength [n]= 1.8; sustainability = 2.2. Rods have a round cross section d1 = d2= 20 mm, d 3 = 28 mm.
Solution
Cutting out the node in which the rods converge, and compiling the 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 equations of statics). It is clear that in order to calculate the strength and stability of rods, it is necessary to know the magnitude of the longitudinal forces arising in their cross sections, i.e., it is necessary to reveal the static indeterminacy.
We draw up a displacement equation based on the displacement diagram (Fig. 2.87):
or, substituting the values of changes in the lengths of the rods, we obtain
Solving this equation together with the equations of statics, we find:
Stresses in the cross sections of the rods 1 and 2 (see fig. 2.86):
Their safety factor
To determine the stability 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 which formula to find N Kp should be used.
So, λ 0< λ < λ пред и критическую силу следует определять по эмпирической формуле:
Stability factor
Thus, the calculation shows that the stability factor is close to the required one, and the safety factor is much higher than the required one, i.e., with an increase in the load of the system, the stability loss of the rod 3 more likely than the occurrence of fluidity in the rods 1 and 2.
