Calculation of brick load-bearing walls for stability. How to calculate masonry walls for stability

In case of independent design brick house there is an urgent need to calculate whether the brickwork can withstand the loads that are laid down in the project. A particularly serious situation develops in masonry areas weakened by window and doorways. In the event of a heavy load, these areas may not withstand and be destroyed.

The exact calculation of the resistance of the wall to compression by the overlying floors is quite complicated and is determined by the formulas laid down in the regulatory document SNiP-2-22-81 (hereinafter referred to as<1>). In engineering calculations of the compressive strength of a wall, many factors are taken into account, including the configuration of the wall, the compressive strength, the strength of a given type of material, and more. However, approximately, "by eye", you can estimate the resistance of the wall to compression, using the indicative tables, in which the strength (in tons) is linked depending on the width of the wall, as well as brands of brick and mortar. The table is compiled for a wall height of 2.8 m.

Brick wall strength table, tons (example)

If the value of the width of the wall is in the range between the indicated ones, it is necessary to focus on the minimum number. At the same time, it should be remembered that the tables do not take into account all the factors that can correct the stability, structural strength and resistance of the brick wall to compression in a fairly wide range.

In terms of time, loads are temporary and permanent.

Permanent:

• weight of elements of structures (weight of fences, load-bearing and other structures);
• soil and rock pressure;
• hydrostatic pressure.

Temporary:

• weight of temporary structures;
• loads from stationary systems and equipment;
• pressure in pipelines;
• loads from stored products and materials;
• climatic loads (snow, ice, wind, etc.);
• and many others.

When analyzing the loading of structures, it is necessary to take into account the total effects. Below is an example of calculating the main loads on the walls of the first floor of a building.

Loading brickwork

To take into account the force acting on the designed section of the wall, it is necessary to sum up the loads:

In the case of low-rise construction, the task is greatly simplified, and many live load factors can be neglected by setting a certain margin of safety at the design stage.

However, in the case of the construction of 3 or more storey structures, a thorough analysis is necessary using special formulas that take into account the addition of loads from each floor, the angle of application of force, and much more. In some cases, the strength of the pier is achieved by reinforcement.

Load Calculation Example

This example shows the analysis of the existing loads on the walls of the 1st floor. Here, only permanent loads from various structural elements of the building are taken into account, taking into account the uneven weight of the structure and the angle of application of forces.

Initial data for analysis:

• number of floors - 4 floors;
• brick wall thickness T = 64 cm (0.64 m);
• specific weight of masonry (brick, mortar, plaster) M = 18 kN / m3 (the indicator is taken from the reference data, Table 19<1>);
• the width of the window openings is: W1=1.5 m;
• height of window openings - B1 = 3 m;
• section of the wall 0.64 * 1.42 m (loaded area, where the weight of the overlying structural elements is applied);
• floor height Vet=4.2 m (4200 mm):
• pressure is distributed at an angle of 45 degrees.

1. Example of determining the load from the wall (plaster layer 2 cm)

Hst \u003d (3-4SH1V1) (h + 0.02) Myf \u003d (* 3-4 * 3 * 1.5) * (0.02 + 0.64) * 1.1 * 18 \u003d 0, 447 MN.

The width of the loaded area П=Вет*В1/2-Ш/2=3*4.2/2.0-0.64/2.0=6 m

Np \u003d (30 + 3 * 215) * 6 \u003d 4.072 MN

Nd \u003d (30 + 1.26 + 215 * 3) * 6 \u003d 4.094 MN

H2 \u003d 215 * 6 \u003d 1.290 MN,

including H2l=(1.26+215*3)*6= 3.878MN

1. Self weight of piers

Npr \u003d (0.02 + 0.64) * (1.42 + 0.08) * 3 * 1.1 * 18 \u003d 0.0588 MN

The total load will be the result of a combination of the indicated loads on the walls of the building, to calculate it, the summation of the loads from the wall, from the floors of the 2nd floor and the weight of the projected area is performed).

Scheme of analysis of load and structural strength

To calculate the pier of a brick wall, you will need:

• the length of the floor (it is also the height of the site) (Wat);
• number of floors (Chat);
• wall thickness (T);
• width brick wall(Sh);
• masonry parameters (type of brick, brand of brick, brand of mortar);

1. Wall area (P)

1. According to table 15<1>it is necessary to determine the coefficient a (elasticity characteristic). The coefficient depends on the type, brand of brick and mortar.
2. Flexibility index (G)

1. Depending on indicators a and D, according to table 18<1>you need to look at the bending factor f.
2. Finding the height of the compressed part

where е0 is the extensibility index.

1. Finding the area of ​​the compressed part of the section

Pszh \u003d P * (1-2 e0 / T)

1. Determination of the flexibility of the compressed part of the wall

Gszh=Vet/Vszh

1. Definition according to the table. eighteen<1>coefficient fszh, based on Gszh and coefficient a.
2. Calculation of the average coefficient fsr

Fsr=(f+fszh)/2

1. Determination of the coefficient ω (table 19<1>)

ω =1+e/T<1,45

1. Calculation of the force acting on the section
2. Definition of sustainability

Y \u003d Kdv * fsr * R * Pszh * ω

Kdv - long-term exposure coefficient

R - resistance of masonry to compression, can be determined from table 2<1>, in MPa

1. Reconciliation

Masonry Strength Calculation Example

- Wet - 3.3 m

- Chet - 2

- T - 640 mm

– W – 1300 mm

- masonry parameters (clay brick made by plastic pressing, cement-sand mortar, brick grade - 100, mortar grade - 50)

1. Area (P)

P=0.64*1.3=0.832

1. According to table 15<1>determine the coefficient a.

1. Flexibility (G)

G \u003d 3.3 / 0.64 \u003d 5.156

1. Bending factor (table 18<1>).

1. Height of the compressed part

Vszh=0.64-2*0.045=0.55 m

1. The area of ​​the compressed part of the section

Pszh \u003d 0.832 * (1-2 * 0.045 / 0.64) \u003d 0.715

1. Flexibility of the compressed part

Gf=3.3/0.55=6

1. fsf=0.96
2. Calculation of fsr

Fav=(0.98+0.96)/2=0.97

1. According to the table 19<1>

ω=1+0.045/0.64=1.07<1,45

To determine the actual load, it is necessary to calculate the weight of all structural elements that affect the designed section of the building.

1. Definition of sustainability

Y \u003d 1 * 0.97 * 1.5 * 0.715 * 1.07 \u003d 1.113 MN

1. Reconciliation

The condition is met, the strength of the masonry and the strength of its elements is sufficient

Insufficient wall resistance

What to do if the calculated pressure resistance of the walls is not enough? In this case, it is necessary to strengthen the wall with reinforcement. Below is an example of an analysis of the necessary structural modifications in case of insufficient compressive strength.

For convenience, you can use tabular data.

The bottom row shows values ​​for a wall reinforced with 3 mm wire mesh, 3 cm mesh, class B1. Reinforcement of every third row.

The increase in strength is about 40%. Usually this compression resistance is sufficient. It is better to make a detailed analysis by calculating the change in strength characteristics in accordance with the applied method of strengthening the structure.

Below is an example of such a calculation.

An example of calculating the reinforcement of piers

Initial data - see the previous example.

• floor height - 3.3 m;
• wall thickness - 0.640 m;
• masonry width 1,300 m;
• typical masonry characteristics (type of bricks - clay bricks made by pressing, type of mortar - cement with sand, brand of bricks - 100, mortar - 50)

In this case, the condition Y>=H is not satisfied (1.113<1,5).

It is required to increase the compressive strength and structural strength.

Gain

k=Y1/Y=1.5/1.113=1.348,

those. it is necessary to increase the strength of the structure by 34.8%.

Reinforcement of reinforced concrete clip

Reinforcement is made with a clip of concrete B15 with a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 with a step of 0.150 m.

Cross-sectional dimensions of the reinforced structure:

Ш_1=1300+2*60=1.42

Т_1=640+2*60=0.76

With such indicators, the condition Y>=H is fulfilled. Compressive strength and structural strength are sufficient.

Let's check the strength of the brick wall of the bearing wall of a residential building with a variable number of storeys in the city of Vologda.

Initial data:

Floor height - Net=2.8 m;

Number of floors - 8 floors;

Step bearing walls- a=6.3 m;

Dimensions of the window opening - 1.5x1.8 m;

The cross-sectional dimensions of the pier are -1.53x0.68 m;

The thickness of the inner verst is 0.51 m;

Sectional area of ​​the wall-A=1.04m 2 ;

The length of the supporting platform of floor slabs per masonry

Materials: silicate brick thickened front (250CH120CH88) GOST 379-95, grade SUL-125/25, silicate porous stone (250CH120CH138) GOST 379-95, grade SRP -150/25 and hollow silicate brick thickened (250x120x88) GOST 379-95 brand SURP-150/25. For laying 1-5 floors, cement-sand mortar M75 is used, for 6-8 floors, masonry density \u003d 1800 kg / m 3, multilayer masonry, insulation - expanded polystyrene brand PSB-S-35 n \u003d 35 kg / m3 (GOST 15588- 86). With multi-layer masonry, the load will be transferred to the inner verst of the outer wall, therefore, when calculating the thickness of the outer verst and insulation, we do not take into account.

Collection of load from pavement and floors is presented in tables 2.13, 2.14, 2.15. The design wall is shown in fig. 2.5.

Figure 2.12. Settlement wall: a - plan; b - vertical section of the wall; c-calculation scheme; d - plot of moments

Table 2.13. Collection of loads on the coating, kN / m 2

Table 2.14. Collection of loads on the attic floor, kN/m2

Table 2.15. Collection of loads on the interfloor overlap, kN/m2

Table 2.16. Collection of loads per 1 r.m. from the outer wall t=680 mm, kN/m2

We determine the width of the cargo area according to the formula 2.12

where b is the distance between the center axes, m;

a - the value of the support of the floor slab, m.

The length of the loading area of ​​the pier is determined by formula (2.13).

where l is the width of the partition;

l f - width of window openings, m.

The determination of the cargo area (according to Figure 2.6) is carried out according to the formula (2.14)

Figure 2.13. Scheme for determining the cargo area of ​​the pier

The calculation of the force N on the wall from the higher floors at the level of the bottom of the floors of the first floor is based on the cargo area and the existing loads on the floors, coatings and roofs, the load from the weight of the outer wall.

Table 2.17. Collection of loads, kN/m

Load name	Design value kN/m
1. Coating design
2. Attic floor
3. Interfloor overlap
4. Outer wall t=680 mm

The calculation of eccentrically compressed unreinforced elements of stone structures should be made according to the formula 13

Exterior load-bearing walls should, at a minimum, be designed for strength, stability, local collapse and resistance to heat transfer. To find out how thick should a brick wall be , you need to calculate it. In this article we will consider the calculation of the bearing capacity of brickwork, and in the following articles - the rest of the calculations. In order not to miss the release of a new article, subscribe to the newsletter and you will find out what the thickness of the wall should be after all the calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, we will consider all calculations for this category.

carriers walls are called that perceive the load from floor slabs, coatings, beams, etc. resting on them.

You should also take into account the brand of brick for frost resistance. Since everyone builds a house for himself, at least for a hundred years, then with a dry and normal humidity regime of the premises, a grade (M rz) of 25 and above is accepted.

When building a house, cottage, garage, outbuildings and other structures with dry and normal humidity conditions, it is recommended to use hollow bricks for external walls, since its thermal conductivity is lower than that of solid bricks. Accordingly, with a thermal engineering calculation, the thickness of the insulation will turn out to be less, which will save money when buying it. Solid brick for external walls should be used only if it is necessary to ensure the strength of the masonry.

Reinforcement of masonry allowed only in the case when the increase in the grade of brick and mortar does not allow to provide the required bearing capacity.

An example of the calculation of a brick wall.

The bearing capacity of brickwork depends on many factors - on the brand of brick, brand of mortar, on the presence of openings and their sizes, on the flexibility of the walls, etc. The calculation of the bearing capacity begins with the definition of the design scheme. When calculating walls for vertical loads, the wall is considered to be supported by hinged-fixed supports. When calculating walls for horizontal loads (wind), the wall is considered to be rigidly clamped. It is important not to confuse these diagrams, since the moment diagrams will be different.

Choice of design section.

In blank walls, the section I-I at the level of the bottom of the floor with the longitudinal force N and the maximum bending moment M is taken as the calculated one. It is often dangerous section II-II, since the bending moment is slightly less than the maximum and is equal to 2/3M, and the coefficients m g and φ are minimal.

In walls with openings, the section is taken at the level of the bottom of the lintels.

Let's look at the section I-I.

From a previous article Collection of loads on the wall of the first floor we take the obtained value of the total load, which includes the loads from the floor of the first floor P 1 \u003d 1.8t and the overlying floors G \u003d G P + P 2 +G 2 = 3.7t:

N \u003d G + P 1 \u003d 3.7t + 1.8t \u003d 5.5t

The floor slab rests on the wall at a distance a=150mm. The longitudinal force P 1 from the overlap will be at a distance a / 3 = 150 / 3 = 50 mm. Why 1/3? Because the stress diagram under the support section will be in the form of a triangle, and the center of gravity of the triangle is just 1/3 of the support length.

The load from the overlying floors G is considered to be applied in the center.

Since the load from the floor slab (P 1) is not applied in the center of the section, but at a distance from it equal to:

e = h / 2 - a / 3 = 250mm / 2 - 150mm / 3 = 75 mm = 7.5 cm,

then it will create a bending moment (M) in section I-I. Moment is the product of force on the shoulder.

M = P 1 * e = 1.8t * 7.5cm = 13.5t * cm

Then the eccentricity of the longitudinal force N will be:

e 0 \u003d M / N \u003d 13.5 / 5.5 \u003d 2.5 cm

Since the load-bearing wall is 25cm thick, the calculation should take into account the random eccentricity e ν = 2cm, then the total eccentricity is:

e 0 \u003d 2.5 + 2 \u003d 4.5 cm

y=h/2=12.5cm

When e 0 \u003d 4.5 cm< 0,7y=8,75 расчет по раскрытию трещин в швах кладки можно не производить.

The strength of the masonry of an eccentrically compressed element is determined by the formula:

N ≤ m g φ 1 R A c ω

Odds m g And φ 1 in the section under consideration, I-I are equal to 1.

The need to calculate brickwork during the construction of a private house is obvious to any developer. In the construction of residential buildings, clinker and red bricks are used; finishing bricks are used to create an attractive appearance of the outer surface of the walls. Each brand of brick has its own specific parameters and properties, but the difference in size between different brands is minimal.

The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick.

Clinker bricks are used for the construction of luxury houses. It has a large specific gravity, attractive appearance, high strength. The limited use is caused by the high cost of the material.

The most popular and demanded material is red brick. It has sufficient strength with a relatively low specific weight, is easy to process, and is little affected by the environment. Disadvantages - sloppy surfaces with high roughness, the ability to absorb water at high humidity. Under normal operating conditions, this ability does not manifest itself.

There are two methods for laying bricks:

• bonder;
• spoon.

When laying with the bonding method, the brick is laid across the wall. The wall thickness must be at least 250 mm. The outer surface of the wall will consist of the end surfaces of the material.

With the spoon method, the brick is laid along. Outside is the side surface. In this way, you can lay out the walls in half a brick - 120 mm thick.

What you need to know to calculate

The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick. The result will be approximate and inflated. For a more accurate calculation, the following factors must be taken into account:

• the size of the masonry seam;
• exact dimensions of the material;
• the thickness of all walls.

Manufacturers quite often, for various reasons, do not withstand the standard sizes of products. Red masonry brick according to GOST should have dimensions of 250x120x65 mm. In order to avoid errors, unnecessary material costs, it is advisable to check with suppliers for the dimensions of the bricks available.

The optimal thickness of the outer walls for most regions is 500 mm, or 2 bricks. This size provides high strength of the building, good thermal insulation. The disadvantage is the large weight of the structure and, as a result, the pressure on the foundation and the lower layers of the masonry.

The size of the masonry joint will primarily depend on the quality of the mortar.

If coarse-grained sand is used to prepare the mixture, the width of the seam will increase, with fine-grained sand, the seam can be made thinner. The optimal thickness of masonry joints is 5-6 mm. If necessary, it is allowed to make seams with a thickness of 3 to 10 mm. Depending on the size of the joints and how the bricks are laid, some amount can be saved.

For example, let's take a seam thickness of 6 mm and a spoon method for laying brick walls. With a wall thickness of 0.5 m, 4 bricks must be laid wide.

The total width of the gaps will be 24 mm. Laying 10 rows of 4 bricks will give a total thickness of all gaps of 240 mm, which is almost equal to the length of a standard product. The total masonry area in this case will be approximately 1.25 m 2. If the bricks are laid closely, without gaps, 240 pieces are placed in 1 m 2. Taking into account the gaps, the material consumption will be approximately 236 pieces.

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Method for calculating load-bearing walls

When planning the external dimensions of a building, it is advisable to choose values ​​that are multiples of 5. With such numbers, it is easier to perform the calculation, then perform it in reality. When planning the construction of 2 floors, the amount of material should be calculated in stages, for each floor.

First, the calculation of the outer walls on the first floor is performed. For example, take a building with dimensions:

• length = 15 m;
• width = 10 m;
• height = 3 m;
• wall thickness 2 bricks.

According to these dimensions, you need to determine the perimeter of the building:

(15 + 10) x 2 = 50

3 x 50 = 150 m 2

By calculating the total area, you can determine the maximum number of bricks for building a wall. To do this, multiply the previously determined number of bricks for 1 m 2 by the total area:

236 x 150 = 35,400

The result is not final, the walls should have openings for installing doors and windows. The number of entrance doors may vary. Small private houses usually have one door. For large buildings, it is desirable to plan two entrances. The number of windows, their size and location are determined by the internal layout of the building.

As an example, you can take 3 window openings for a 10-meter wall, 4 for 15-meter walls. It is desirable to perform one of the walls deaf, without openings. The volume of doorways can be determined by standard sizes. If the dimensions differ from the standard ones, the volume can be calculated from the overall dimensions by adding the width of the mounting gap to them. To calculate, use the formula:

2 x (A x B) x 236 = C

where: A is the width of the doorway, B is the height, C is the volume in the number of bricks.

Substituting the standard values, we get:

2 x (2 x 0.9) x 236 = 849 pcs.

The volume of window openings is calculated similarly. With window sizes of 1.4 x 2.05 m, the volume will be 7450 pieces. Determining the number of bricks per expansion gap is simple: you need to multiply the perimeter length by 4. The result will be 200 pieces.

35400 — (200 + 7450 + 849) = 26 901.

The required quantity should be purchased with a small margin, because errors and other unforeseen situations are possible during operation.

Picture 1. Calculation scheme for brick columns of the designed building.

In this case, a natural question arises: what is the minimum section of the columns that will provide the required strength and stability? Of course, the idea of ​​laying clay brick columns, and even more so the walls of the house, is far from new, and all possible aspects of the calculations of brick walls, walls, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced masonry structures". It is this normative document that should be followed in the calculations. The calculation below is nothing more than an example of using the specified SNiP.

To determine the strength and stability of the columns, you need to have a lot of initial data, such as: the brick brand for strength, the area of ​​support of the crossbars on the columns, the load on the columns, the sectional area of ​​​​the column, and if none of this is known at the design stage, then you can do in the following way:

An example of calculating a brick column for stability under central compression

Designed:

Terrace with dimensions of 5x8 m. Three columns (one in the middle and two along the edges) made of facing hollow brick with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The brick strength grade is M75.

Design assumptions:

With such a design scheme, the maximum load will be on the middle lower column. It is she who should be counted on strength. The load on the column depends on many factors, in particular on the area of ​​construction. For example, in St. Petersburg it is 180 kg / m 2, and in Rostov-on-Don - 80 kg / m 2. Taking into account the weight of the roof itself 50-75 kg / m 2, the load on the column from the roof for Pushkin, Leningrad Region, can be:

N from the roof = (180 1.25 + 75) 5 8/4 = 3000 kg or 3 tons

Since the actual loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but the reinforced concrete slab is not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculating the load from the terrace you can take a uniformly distributed load of 600 kg / m 2, then the concentrated force from the terrace, acting on the central column, will be:

N from the terrace = 600 5 8/4 = 6000 kg or 6 tons

The own weight of columns 3 m long will be:

N per column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons

Thus, the total load on the middle lower column in the section of the column near the foundation will be:

N with about \u003d 3000 + 6000 + 2 650 \u003d 10300 kg or 10.3 tons

However, in this case, it can be taken into account that there is not a very high probability that the temporary load from snow, which is maximum in winter, and the temporary load on the floor, which is maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:

N with about \u003d (3000 + 6000) 0.9 + 2 650 \u003d 9400 kg or 9.4 tons

The calculated load on the outer columns will be almost two times less:

N cr = 1500 + 3000 + 1300 = 5800 kg or 5.8 tons

2. Determination of the strength of brickwork.

The brand of brick M75 means that the brick must withstand a load of 75 kgf / cm 2, however, the strength of the brick and the strength of the brickwork are two different things. The following table will help you understand this:

Table 1. Calculated compressive strength for brickwork (according to SNiP II-22-81 (1995))

But that's not all. All the same SNiP II-22-81 (1995) p.3.11 a) recommends that if the area of ​​​​pillars and piers is less than 0.3 m 2, multiply the value of the design resistance by working conditions coefficient γ s =0.8. And since the cross-sectional area of ​​​​our column is 0.25x0.25 \u003d 0.0625 m 2, we will have to use this recommendation. As you can see, for a brick of the M75 brand, even when using the M100 masonry mortar, the masonry strength will not exceed 15 kgf / cm 2. As a result, the calculated resistance for our column will be 15 0.8 = 12 kg / cm 2, then the maximum compressive stress will be:

10300/625 \u003d 16.48 kg / cm 2\u003e R \u003d 12 kgf / cm 2

Thus, to ensure the necessary strength of the column, it is necessary either to use a brick of greater strength, for example, M150 (the calculated compressive strength with a brand of mortar M100 will be 22 0.8 = 17.6 kg / cm 2) or increase the section of the column or use transverse reinforcement of the masonry. For now, let's focus on using a more durable face brick.

3. Determination of the stability of a brick column.

The strength of brickwork and the stability of a brick column are also different things and all the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:

N ≤ m g φRF (1.1)

where m g- coefficient taking into account the influence of long-term load. In this case, relatively speaking, we are lucky, since at the height of the section h≈ 30 cm, the value of this coefficient can be taken equal to 1.

Note: Actually, with the coefficient m g, everything is not so simple, the details can be found in the comments to the article.

φ - coefficient of buckling, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l 0 , but it does not always coincide with the height of the column. The subtleties of determining the estimated length of the structure are set out separately, here we just note that according to SNiP II-22-81 (1995) p. 4.3: "The estimated heights of walls and pillars l 0 when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, one should take:

a) with fixed hinged supports l 0 = H;

b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l 0=1.5H, for multi-span buildings l 0=1.25H;

c) for free-standing structures l 0 = 2N;

d) for structures with partially pinched support sections - taking into account the actual degree of pinching, but not less than l 0 = 0.8N, where H- the distance between ceilings or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light.

At first glance, our calculation scheme can be considered as satisfying the conditions of paragraph b). i.e. you can take l 0 = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only if the lower support is really rigid. If a brick column will be laid out on a roofing material waterproofing layer laid on a foundation, then such a support should rather be considered as hinged, and not rigidly clamped. And in this case, our structure in a plane parallel to the plane of the wall is geometrically variable, since the floor structure (separately lying boards) does not provide sufficient rigidity in this plane. There are 4 ways out of this situation:

1. Apply a fundamentally different design scheme

for example - metal columns rigidly embedded in the foundation, to which the floor crossbars will be welded, then, for aesthetic reasons, the metal columns can be overlaid with any brand of face brick, since the metal will carry the entire load. In this case, it is true that metal columns need to be calculated, but the estimated length can be taken l 0=1.25H.

2. Make another cover,

for example, from sheet materials, which will allow us to consider both the upper and lower support of the column as hinged, in this case l 0=H.

3. Make a hardness diaphragm

in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower column supports as hinged ones, but in this case it is necessary to additionally calculate the stiffness diaphragm.

4. Ignore the above options and count the columns as free-standing with a rigid bottom support, i.e. l 0 = 2N

In the end, the ancient Greeks put up their columns (though not of brick) without any knowledge of the resistance of materials, without the use of metal anchors, and there were no such carefully written building codes in those days, nevertheless, some columns stand and to this day.

Now, knowing the estimated length of the column, you can determine the coefficient of flexibility:

λ h = l 0 /h (1.2) or

λ i = l 0 /i (1.3)

where h- the height or width of the section of the column, and i- radius of inertia.

In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the area of ​​the section, and then extract from the result Square root, but in this case it is not really necessary. In this way λh = 2 300/25 = 24.

Now, knowing the value of the coefficient of flexibility, we can finally determine the coefficient of buckling from the table:

table 2. Buckling coefficients for masonry and reinforced masonry structures (according to SNiP II-22-81 (1995))

At the same time, the elastic characteristic of the masonry α determined by the table:

Table 3. Elastic characteristic of masonry α (according to SNiP II-22-81 (1995))

As a result, the value of the buckling coefficient will be about 0.6 (with the value of the elastic characteristic α = 1200, according to item 6). Then the maximum load on the central column will be:

N p \u003d m g φγ with RF \u003d 1x0.6x0.8x22x625 \u003d 6600 kg< N с об = 9400 кг

This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then in this way not only the cross-sectional area of ​​\u200b\u200bthe column will increase to 0.13 m 2 or 1300 cm 2, but the radius of gyration of the column will also increase to i= 11.45 cm. Then λ i = 600/11.45 = 52.4, and the value of the coefficient φ = 0.8. In this case, the maximum load on the central column will be:

N p \u003d m g φγ with RF \u003d 1x0.8x0.8x22x1300 \u003d 18304 kg\u003e N with about \u003d 9400 kg

This means that a section of 38x38 cm is enough to ensure the stability of the lower central centrally compressed column with a margin, and even the brand of brick can be reduced. For example, with the originally adopted brand M75, the ultimate load will be:

N p \u003d m g φγ with RF \u003d 1x0.8x0.8x12x1300 \u003d 9984 kg\u003e N with about \u003d 9400 kg

It seems to be everything, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the section of columns 0.51x0.51 m will be the most optimal, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of ​​such columns will be 2601 cm 2.

An example of calculating a brick column for stability under eccentric compression

The extreme columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then all the same, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the extreme columns not in the center of the column section. In which place the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the elastic moduli of the crossbars and columns, and a number of other factors, which are discussed in detail in the article "Calculation of the support section of the beam for collapse". This displacement is called the load application eccentricity e o. In this case, we are interested in the most unfavorable combination of factors, in which the floor load on the columns will be transferred as close as possible to the edge of the column. This means that, in addition to the load itself, the bending moment will also act on the columns, equal to M = Ne o, and this moment must be taken into account in the calculations. In general, stability testing can be performed using the following formula:

N = φRF - MF/W (2.1)

where W- section modulus. In this case, the load for the lower extreme columns from the roof can be conditionally considered to be centrally applied, and the eccentricity will be created only by the load from the ceiling. With an eccentricity of 20 cm

N p \u003d φRF - MF / W \u003d1x0.8x0.8x12x2601- 3000 20 2601· 6/51 3 = 19975, 68 - 7058.82 = 12916.9 kg >N cr = 5800 kg

Thus, even with a very large load application eccentricity, we have more than a double margin of safety.

Note: SNiP II-22-81 (1995) "Stone and reinforced masonry structures" recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore I do not give the calculation method recommended by SNiP here.

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