Exterior load-bearing walls should, as a minimum, be dimensioned for strength, stability, localized crushing and heat transfer resistance. To find out how thick should the 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, then we will consider all calculations for this category.

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

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

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

Reinforcement of brickwork is allowed only if an increase in the grade of brick and mortar does not allow providing the required bearing capacity.

An example of calculating a brick wall.

The bearing capacity of brickwork depends on many factors - on the brand of brick, the 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 on hinged fixed supports. When calculating walls for horizontal (wind) loads, the wall is considered rigidly restrained. It is important not to confuse these diagrams as the moment diagrams will be different.

The choice of the design section.

In blank walls, the design section is I-I at the level of the bottom of the floor with a longitudinal force N and a maximum bending moment M. 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 take a look at section I-I.

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

N = G + P 1 = 3.7t + 1.8t = 5.5t

The floor slab rests on the wall at a distance of a = 150mm. The longitudinal force P 1 from the overlap will be at a distance of 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 G floors is considered to be applied at the center.

Since the load from the floor slab (P 1) is applied not 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 = M / N = 13.5 / 5.5 = 2.5 cm

Because bearing wall 25cm thick, then the calculation should take into account the value of the random eccentricity e ν = 2cm, then the total eccentricity is equal to:

e 0 = 2.5 + 2 = 4.5 cm

y = h / 2 = 12.5cm

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

The strength of the cage 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 considered section 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 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 homes. He has a large proportion, attractive appearance, high strength. Limited use due to the high cost of the material.

The most popular and demanded material is red brick. It possesses sufficient strength with a relatively low specific gravity, is easy to process, is little affected environment... Disadvantages - sloppy surfaces with great roughness, the ability to absorb water when high humidity... Under normal operating conditions, this ability is not manifested.

There are two methods for laying bricks:

• bonded;
• 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 be composed of the end faces of the material.

With the spoon method, the brick is laid along. The side surface is outside. 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 overestimated. For a more accurate calculation, it is necessary to take into account the following factors:

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

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

The optimum external wall thickness 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 you use coarse-grained sand to prepare the mixture, the width of the seam will increase, with fine-grained sand, the seam can be made thinner. The optimum thickness of the 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 the way the brick is laid, you can save some amount of it.

For example, let's take a joint thickness of 6 mm and a spoon method for laying brick walls. With a wall thickness of 0.5 m, you need to lay 4 bricks wide.

The total width of the gaps is 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. In this case, the total masonry area will be approximately 1.25 m 2. If the bricks are stacked close, without gaps, 240 pieces are placed in 1 m 2. Taking into account the gaps, the material consumption will be approximately 236 pieces.

Back to the table of contents

Method for calculating load-bearing walls

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

First, the external walls on the ground floor are calculated. For example, you can 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 structure:

(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, you need to multiply the previously determined number of bricks for 1 m 2 by the total area:

236 x 150 = 35 400

The result is inconclusive, the walls must have openings for installing doors and windows. Quantity entrance doors may vary. Small private houses usually have one door. For buildings large sizes it is advisable to plan two entrances. The number of windows, their size and location are determined internal layout building.

As an example, you can take 3 window openings on a 10-meter wall, 4 each on a 15-meter wall. It is advisable to make one of the walls blind, without openings. The volume of doorways can be determined from the standard dimensions. 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 in the same way. With window sizes 1.4 x 2.05 m, the volume will be 7450 pieces. Determining the number of bricks per temperature gap is simple: you need to multiply the length of the perimeter by 4. The result is 200 pieces.

35400 — (200 + 7450 + 849) = 26 901.

To acquire required amount follows with a small margin, because errors and other unforeseen situations are possible during operation.

Brick is a fairly durable building material, especially corpulent, and when building houses with 2-3 floors, the walls are made of ordinary ceramic brick additional calculations are usually not needed. Nevertheless, situations are different, for example, it is planned two-storey house with a terrace on the second floor. Metal girders, which will also be supported metal beams the ceiling of the terrace, it is planned to lean on brick columns made of facing hollow bricks 3 meters high, there will be more columns 3 meters high, on which the roof will rest:

This raises a natural question: what is the minimum column cross-section that will provide the required strength and stability? Of course, the idea of ​​laying out columns of clay bricks, and even more so the walls of a house, is far from new and all possible aspects of calculating brick walls, piers, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures". It is by this regulatory document and should be guided in the calculations. The calculation given 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 strength grade, the bearing area of ​​the crossbars on the columns, the load on the columns, the column cross-sectional area, and if at the design stage none of this is known, then you can do in the following way:

with central compression

Designed by: Terrace measuring 5x8 m. Three columns (one in the middle and two at the edges) of facing hollow bricks with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. Brick strength is M75.

With this design scheme, the maximum load will be on the middle bottom column. It is her that should be counted on for strength. The column load depends on many factors, in particular the area of ​​construction. For example, snow load on the roof in St. Petersburg is 180 kg / m & sup2, and in Rostov-on-Don - 80 kg / m & sup2. Taking into account the weight of the roof itself 50-75 kg / m & sup2, 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 acting loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab it 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 & sup2, 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 dead weight of the columns with a length of 3 m will be:

N from the 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 column section near the foundation will be:

N with rev = 3000 + 6000 + 2 · 650 = 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 live load from snow, the maximum in winter time, and the live load on the floor, maximum in summer time will be attached at the same time. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:

N with rev = (3000 + 6000) 0.9 + 2 650 = 9400 kg or 9.4 tons

The design 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 brick grade M75 means that the brick must withstand a load of 75 kgf / cm & sup2, however, the strength of the brick and the strength of the brickwork are different things. The following table will help you understand this:

Table 1... Calculated compressive strengths for masonry

But that's not all. All the same SNiP II-22-81 (1995) clause 3.11 a) recommends that, with the area of ​​pillars and walls less than 0.3 m & sup2, multiply the value of the design resistance by the coefficient of working conditions γ c = 0.8... And since the cross-sectional area of ​​our column is 0.25x0.25 = 0.0625 m & sup2, you will have to use this recommendation. As you can see, for brick grade M75, even when using masonry mortar M100 masonry strength will not exceed 15 kgf / cm & sup2. Eventually design resistance for our column will be 15 · 0.8 = 12 kg / cm & sup2, then the maximum compressive stress will be:

10300/625 = 16.48 kg / cm & sup2> R = 12 kgf / cm & sup2

Thus, to ensure the required strength of the column, either use a brick of greater strength, for example, M150 (the calculated compressive strength for the M100 solution grade will be 22 0.8 = 17.6 kg / cm2) or increase the column section or use transverse reinforcement of the masonry. For now, let's focus on using a more durable facing brick.

3. Determination of the stability of a brick column.

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

N ≤ m g φRF (1.1)

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

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 o, and it does not always coincide with the height of the column. The subtleties of determining the design length of the structure are not set out here, we just note that according to SNiP II-22-81 (1995) clause 4.3: "The design heights of walls and pillars l o when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, the following should be taken:

a) with fixed hinged bearings l o = H;

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

c) for free-standing structures l o = 2H;

d) for structures with partially restrained support sections - taking into account the actual degree of restraint, but not less l o = 0.8H, where N- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light. "

At first glance, our design scheme can be considered as satisfying the conditions of item b). that is, you can take l o = 1.25H = 1.25 3 = 3.75 meters or 375 cm... However, we can confidently use this value only when the lower support is really rigid. If a brick column will be laid out on a layer of waterproofing made of roofing material 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 structure of the floor (separately lying boards) does not provide sufficient rigidity in the indicated plane. There are 4 ways out of this situation:

1. Apply a fundamentally different constructive scheme , for example - metal columns, rigidly embedded in the foundation, to which the floor girders will be welded, then, for aesthetic reasons, the metal columns can be overlaid with facing bricks of any brand, since the metal will bear the entire load. In this case, it is true that you need to calculate the metal columns, but the estimated length can be taken l o = 1.25H.

2. Make another overlap, for example from sheet materials, which will allow considering both the upper and bottom support columns, as articulated, in this case l o = H.

3. Make diaphragm stiffness in a plane parallel to the plane of the wall. For example, lay not columns at the edges, but rather piers. This will also make it possible to consider both the upper and lower support of the column as articulated, but in this case it is necessary to additionally calculate the stiffness diaphragm.

4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l o = 2H... In the end, the ancient Greeks erected their columns (albeit not made of bricks) without any knowledge of the resistance of materials, without the use of metal anchors, and there were no such carefully written building codes at that time, nevertheless, some columns stand and to this day.

Now, knowing the calculated length of the column, you can determine the slenderness factor:

λ h = l o / h (1.2) or

λ i = l o (1.3)

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

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 sectional area, and then extract from the result Square root, however, in this case, this is not very necessary. Thus λ h = 2 300/25 = 24.

Now, knowing the value of the slenderness factor, we can finally determine the buckling factor from the table:

table 2... Buckling coefficients for stone 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 ultimate load on the central column will be:

N p = m g φγ with RF = 1 0.6 0.8 22 625 = 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 not only will the sectional area of ​​the column increase to 0.13 m & sup2 or 1300 cm & sup2, but the radius of inertia 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 ultimate load on the central column will be:

N p = m g φγ with RF = 1 0.8 0.8 22 1300 = 18304 kg> N with rev = 9400 kg

This means that the sections of 38x38 cm are enough to ensure the stability of the lower centrally compressed column with a margin, and it is even possible to reduce the grade of brick. For example, with the originally adopted M75 grade, the maximum load will be:

N p = m g φγ with RF = 1 0.8 0.8 12 1300 = 9984 kg> N with rev = 9400 kg

It seems to be all, 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), in 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 most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, this section is optimal. The cross-sectional area of ​​such columns will be 2601 cm & sup2.

An example of calculating a brick column for stability
eccentric compression

The outer columns in the projected house will not be centrally compressed, since the girders will rest on them only on one side. And even if the girders are laid on the entire column, still, due to the deflection of the girders, the load from the floor and the 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. This displacement is called the eccentricity of the load application eo. In this case, we are interested in the most unfavorable combination of factors, in which the load from the floor to the columns will be transmitted as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be affected by a bending moment equal to M = Ne o, and this point 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)

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

N p = φRF - MF / W =1 0.8 0.8 12 2601- 3000 20 2601· 6/51 3 = 19975.68 - 7058.82 = 12916.9 kg>N cr = 5800 kg

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

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, the calculation method recommended by SNiP is not given here.

Greetings to all readers! What should be the thickness of the brick exterior walls - the topic of today's article. The most commonly used small stone walls are brick walls. This is due to the fact that the use of bricks solves the issues of creating buildings and structures of almost any architectural form.

Starting to carry out the project, the design firm calculates all structural elements - including the calculation of the thickness of the brick exterior walls.

The walls in the building have different functions:

• If the walls are only a building envelope- in this case, they must meet the thermal insulation requirements in order to ensure a constant temperature and humidity microclimate, as well as have sound insulating qualities.
• Load-bearing walls must be distinguished by the necessary strength and stability, but also as enclosing, have heat-shielding properties. In addition, based on the purpose of the building, its class, the thickness of the bearing walls must correspond to the technical indicators of its durability and fire resistance.

Features of calculating wall thickness

• The thickness of the walls according to the heat engineering calculation does not always coincide with the calculation of the value according to the strength characteristics. Naturally, the more severe the climate, the thicker the wall should be in terms of thermal performance.
• But according to the conditions of strength, for example, it is enough to lay out the outer walls in one brick or one and a half. This is where it turns out "nonsense" - the thickness of the masonry, determined heat engineering calculation, often, due to the strength requirements, it turns out to be excessive.
• Therefore, laying a solid brickwork of solid brick walls in terms of material costs and provided that its strength is 100% use should be done only in the lower floors of high-rise buildings.
• In low-rise buildings, as well as in the upper floors of high-rise buildings, it should be used for outdoor masonry hollow or lightweight brick, lightweight masonry can be used.
• This does not apply to external walls in buildings where there is a high percentage of humidity (for example, in laundries, baths). They are usually erected with a protective layer of vapor barrier material inside and out of full-bodied clay material.

Now I will tell you about the calculation of the thickness of the outer walls.

It is determined by the formula:

B = 130 * n -10, where

B - wall thickness in millimeters

130 - the size of half of the brick, taking into account the seam (vertical = 10mm)

n - an integer of a half brick (= 120mm)

The size of the solid masonry obtained by calculation is rounded up to an integer number of half-bricks.

Based on this, the following values ​​(in mm) of brick walls are obtained:

• 120 (brick floor, but this is considered a partition);
• 250 (into one);
• 380 (one and a half);
• 510 (at two);
• 640 (two and a half);
• 770 (at three o'clok).

In order to save material resources (brick, mortar, fittings, etc.), the number of machine - clock mechanisms, the calculation of the wall thickness is tied to the bearing capacity of the building. And the heat engineering component is obtained by insulating the facades of buildings.

How can you insulate the outer walls of a brick building? In the article on insulating a house with expanded polystyrene outside, I indicated the reasons why brick walls should not be insulated with this material. Check out the article.

The point is that brick is a porous and permeable material. And the absorbency of expanded polystyrene is zero, which prevents moisture migration outward. That is why it is advisable to insulate a brick wall thermal insulation plaster or mineral wool slabs, the nature of which is vapor-permeable. Expanded polystyrene is suitable for insulating a base made of concrete or reinforced concrete. "The nature of the insulation must match the nature of the load-bearing wall."

There are a lot of heat-insulating plasters- the difference lies in the components. But the principle of application is the same. It is carried out in layers and the total thickness can reach 150 mm (with a large value, reinforcement is required). In most cases, this value is 50 - 80 mm. It depends on the climatic zone, the thickness of the walls of the base, and other factors. I will not dwell in detail, since this is a topic for another article. We return to our bricks.

The average wall thickness for ordinary clay brick, depending on the area and climatic conditions of the area at the average winter ambient temperature, looks like this in millimeters:

1. - 5 degrees - thickness = 250;
2. - 10 degrees = 380;
3. - 20 degrees = 510;
4. - 30 degrees = 640.

I would like to summarize the above. The thickness of the outer brick walls is calculated based on the strength characteristics, and the heat engineering side of the issue is solved by the method of wall insulation. As a rule, the design firm calculates the external walls without the use of insulation. If the house is uncomfortably cold and there is a need for insulation, then carefully consider the selection of insulation.

When building your home, one of the main points is the construction of walls. The laying of bearing surfaces is most often carried out using bricks, but what should be the thickness of the brick wall in this case? In addition, the walls in the house are not only load-bearing, but also serve as partitions and cladding - what should be the thickness of the brick wall in these cases? I will talk about this in today's article.

This question is very relevant for all people who are building their own brick house and are just learning the basics of construction. At first glance, the brick wall is very simple construction, it has height, width and thickness. The weight of the wall we are interested in depends primarily on its final total area. That is, the wider and higher the wall, the thicker it should be.

But what does the thickness of the brick wall have to do with it? - you ask. Despite the fact that in construction, a lot is tied to the strength of the material. Brick, like other building materials, has its own GOST, which takes into account its strength. Also, the weight of the masonry depends on its stability. The narrower and higher the bearing surface is, the thicker it must be, especially for the base.

Another parameter that affects the overall weight of the surface is the thermal conductivity of the material. An ordinary solid block has a rather high thermal conductivity. This means that it is, in itself, poor thermal insulation. Therefore, in order to reach standardized thermal conductivity indicators, building a house exclusively from silicate or any other blocks, the walls must be very thick.

But, in order to save money and save common sense, people gave up the idea of ​​building houses resembling a bunker. In order to have strong bearing surfaces and at the same time good thermal insulation, they began to use a multi-layer scheme. Where one layer is silicate masonry, of sufficient weight to withstand all the loads to which it is exposed, the second layer is an insulating material, and the third is a cladding, which can also be a brick.

Brick selection

Depending on what it should be, you need to choose a certain type of material that has different sizes and even structure. So, according to their structure, they can be divided into full-bodied and perforated. Solid materials have great strength, cost, and thermal conductivity.

Building material with cavities inside in the form through holes not so strong, has a lower cost, but at the same time the perforated block has a higher capacity for thermal insulation. This is achieved due to the presence of air pockets in it.

The sizes of any type of material in question may also vary. He might be:

• Single;
• One and a half;
• Double;
• Half.

A single block is a building material of standard sizes, such to which we are all accustomed to. Its dimensions are as follows: 250X120X65 mm.

One and a half or thickened - has a great weight, and its dimensions look like this: 250X120X88 mm. Double - respectively, has a cross-section of two single blocks 250X120X138 mm.

Half is a kid among his fellows, he has, as you probably already guessed, half the thickness of a single - 250X120X12 mm.

As you can see, the only differences in the dimensions of this building material are in its thickness, and the length and width are the same.

Depending on the thickness of the brick wall, it is economically feasible to choose larger ones when erecting massive surfaces, for example, such are often load-bearing surfaces and smaller blocks for partitions.

Wall thickness

We have already considered the parameters on which the thickness of the external brick walls depends. As we remember, this is stability, strength, thermal insulation properties... In addition, different types of surfaces must have completely different dimensions.

Bearing surfaces are, in fact, the support of the entire building, they take on the main load from the entire structure, including the weight of the roof, they are also influenced by external factors such as wind, precipitation, and besides, their own weight presses on them. Therefore, their weight, in comparison with non-bearing surfaces and internal partitions, should be the highest.

In modern realities, most two and three-storey houses need 25 cm of thickness or one block, less often one and a half or 38 cm. The strength of such masonry will be enough for a building of this size, but what about stability. Everything is much more complicated here.

In order to calculate whether the stability will be sufficient, you need to refer to the norms of SNiP II-22-8. Let's calculate whether our brick house, with walls 250 mm thick, 5 meters long and 2.5 meters high. For masonry, we will use material M50, on a solution of M25, we will carry out the calculation for one bearing surface, without windows. So let's get started.

Table No. 26

According to the data from the table above, we know that the characteristic of our clutch belongs to the first group, and also the description from paragraph 7 is valid for it. 26. After that, we look at table 28 and find the value of β, which means the permissible ratio of the weight of the wall to its height, taking into account the type of solution used. For our example, this value is 22.

• k1 for the section of our masonry is 1.2 (k1 = 1.2).
• k2 = √Аn / Аb where:

An is the cross-sectional area of ​​the bearing surface horizontally, the calculation is simple 0.25 * 5 = 1.25 sq. m

Ab - the cross-sectional area of ​​the wall horizontally, taking into account the window openings, we do not have such, therefore k2 = 1.25

• The k4 value is given and for a height of 2.5 m it is 0.9.

Now, having found out, all the variables can be found overall ratio"K", by multiplying all values. K = 1.2 * 1.25 * 0.9 = 1.35 Next, we find out the cumulative value of the correction factors and actually find out how stable the surface under consideration is 1.35 * 22 = 29.7, and the permissible ratio of height and thickness is 2.5: 0.25 = 10, which is much less than the obtained indicator of 29.7. This means that a masonry with a thickness of 25 cm, a width of 5 m and a height of 2.5 meters has a stability almost three times higher than it is necessary according to SNiP standards.

We figured out well with the supporting surfaces, and what about the partitions and those that do not bear the load. Partitions, it is advisable to make half the thickness - 12 cm. For surfaces that do not bear a load, the stability formula, which we considered above, is also valid. But since from above, such a wall will not be fixed, the indicator of the β coefficient must be reduced by a third, and the calculations should be continued with a different value.

Masonry in half a brick, brick, one and a half, two bricks

In conclusion, let's look at how brick laying is carried out, depending on the weight of the surface. Laying in half a brick is the simplest of all, since there is no need to make complex bandaging of the rows. It is enough to put the first row of material on a perfectly flat base and make sure that the mortar lays down evenly and does not exceed 10 mm in thickness.

The main criterion for high-quality masonry with a section of 25 cm is the implementation of high-quality dressing of vertical seams, which should not coincide. For this masonry option, it is important to observe the selected system from start to finish, of which there are at least two, single-row and multi-row. They differ in the way of bandaging and laying blocks.

Before proceeding to consider issues related to calculating the thickness of a brick wall at home, you need to understand what it is for. For example, why can't you build an outer wall half a brick thick, because the brick is so hard and durable?

Many non-specialists do not even have a basic understanding of the characteristics of the enclosing structures, nevertheless, they undertake independent construction.

In this article, we will look at two main criteria for calculating the thickness of brick walls - bearing loads and heat transfer resistance. But before diving into boring numbers and formulas, let me clarify some points in simple language.

The walls of the house, depending on their place in the project scheme, can be load-bearing, self-supporting, non-bearing and partitions. Load-bearing walls perform a fencing function, and also serve as supports for slabs or floor beams or roof structures. The thickness of the bearing brick walls cannot be less than one brick (250 mm). Most modern houses are built with one or 1.5 brick walls. Projects of private houses, where walls thicker than 1.5 bricks would be required, by the logic of things should not exist. Therefore, the choice of the thickness of the outer brick wall is, by and large, a settled matter. If you choose between one brick or one and a half thick, then from a purely technical point of view, for a cottage with a height of 1-2 floors, a brick wall 250 mm thick (one brick of strength grades M50, M75, M100) will correspond to the calculations of bearing loads. You should not be reinsured, since the calculations already take into account snow, wind loads and many coefficients that provide a brick wall with an adequate margin of safety. However, there is a very important point that really affects the thickness of a brick wall - stability.

Once in childhood, everyone played with cubes, and noticed that the more cubes are placed on top of each other, the less stable the column of them becomes. The elementary laws of physics that act on cubes act in the same way on a brick wall, because the principle of masonry is the same. Obviously, there is a certain relationship between the thickness of the wall and its height, which ensures the stability of the structure. We will talk about this dependence in the first half of this article.

Wall stability as well as building codes bearing and other loads, is described in detail in SNiP II-22-81 "Stone and reinforced stone structures". These standards are a guide for designers, and for the "uninitiated" may seem rather difficult to understand. This is so, because to become an engineer, you need to study for at least four years. Here one could refer to “contact specialists for calculations” and put an end to it. However, thanks to the capabilities of the information web, today almost everyone, if desired, can understand the most difficult issues.

First, let's try to understand the issue of the stability of a brick wall. If the wall is high and long, then the thickness of one brick will not be enough. At the same time, extra reinsurance can increase the cost of the box by 1.5-2 times. And this is a lot of money today. To avoid the destruction of the wall or unnecessary financial spending, let's turn to the mathematical calculation.

All the necessary data for calculating the stability of the wall are available in the corresponding tables of SNiP II-22-81. On specific example consider how to determine whether the stability of the external load-bearing brick (M50) wall is sufficient on M25 mortar 1.5 bricks (0.38 m) thick, 3 m high and 6 m long with two window openings 1.2 × 1.2 m ...

Referring to table 26 (table above), we find that our wall belongs to the I-th group of masonry and fits the description of paragraph 7 of this table. Next, we need to find out the permissible ratio of the height of the wall to its thickness, taking into account the brand of the masonry mortar. The sought parameter β is the ratio of the height of the wall to its thickness (β = N / h). In accordance with the data in the table. 28 β = 22. However, our wall is not fixed in the upper section (otherwise the calculation was required only for strength), therefore, according to clause 6.20, the value of β should be reduced by 30%. Thus, β is no longer equal to 22, but 15.4.

We turn to the determination of the correction factors from table 29, which will help to find the cumulative factor k:

• for a wall 38 cm thick, not a bearing load, k1 = 1.2;
• k2 = √Аn / Аb, where An is the area of ​​the horizontal section of the wall taking into account the window openings, and Аb is the area of ​​the horizontal section without taking into account the windows. In our case, An = 0.38 × 6 = 2.28 m², and Ab = 0.38 × (6-1.2 × 2) = 1.37 m². We carry out the calculation: k2 = √1.37 / 2.28 = 0.78;
• k4 for a wall with a height of 3 m is 0.9.

By multiplying all the correction factors, we find the total factor k = 1.2 × 0.78 × 0.9 = 0.84. After taking into account the totality of the correction factors β = 0.84 × 15.4 = 12.93. This means that the permissible ratio of the wall with the required parameters in our case is 12.98. The existing ratio H / h= 3: 0.38 = 7.89. This is less than the permissible ratio of 12.98, and this means that our wall will be quite stable, because the condition H / h

According to clause 6.19, one more condition must be met: the sum of the height and length ( H+L) the walls must be less than the product 3kβh. Substituting the values, we get 3 + 6 = 9

Brick wall thickness and heat transfer resistance rates

Today the overwhelming number brick houses have a multi-layer wall structure consisting of lightweight brickwork, insulation and facade decoration... According to SNiP II-3-79 (Building heat engineering), the outer walls of residential buildings with a demand of 2000 ° C / day. must have a heat transfer resistance of at least 1.2 m². ° C / W. To determine the calculated thermal resistance for a specific region, it is necessary to take into account several local temperature and humidity parameters at once. To eliminate errors in complex calculations, we offer the following table, which shows the required thermal resistance of walls for a number of Russian cities located in different building and climatic zones according to SNiP II-3-79 and SP-41-99.

Heat transfer resistance R(thermal resistance, m². ° С / W) of the layer of the enclosing structure is determined by the formula:

R=δ /λ , where

δ - layer thickness (m), λ - coefficient of thermal conductivity of the material W / (m. ° С).

To obtain the total thermal resistance of a sandwich envelope, it is necessary to add the thermal resistances of all layers of the wall structure. Consider the following with a specific example.

The task is to determine how thick the wall should be from sand-lime brick so that its thermal conductivity resistance corresponds SNiP II-3-79 for the lowest standard of 1.2 m². ° С / W. The thermal conductivity coefficient of silicate bricks is 0.35-0.7 W / (m. ° C), depending on the density. Let's say our material has a thermal conductivity coefficient of 0.7. Thus, we obtain an equation with one unknown δ = Rλ... Substitute the values ​​and solve: δ = 1.2 × 0.7 = 0.84 m.

Now let's calculate with what layer of expanded polystyrene you need to insulate a wall of silicate bricks 25 cm thick in order to reach an indicator of 1.2 m². ° C / W. The thermal conductivity coefficient of expanded polystyrene (PSB 25) is not more than 0.039 W / (m. ° C), and for silicate bricks, 0.7 W / (m. ° C).

1) define R brick layer: R=0,25:0,7=0,35;

2) calculate the missing thermal resistance: 1.2-0.35 = 0.85;

3) we determine the thickness of expanded polystyrene necessary to obtain a thermal resistance equal to 0.85 m2. ° С / W: 0.85 × 0.039 = 0.033 m.

Thus, it has been established that in order to bring the wall into one brick to the standard thermal resistance (1.2 m². ° C / W), insulation with a layer of expanded polystyrene with a thickness of 3.3 cm is required.

Using this technique, you can independently calculate the thermal resistance of the walls, taking into account the region of construction.

Modern residential construction makes high demands on such parameters as strength, reliability and thermal protection. External walls built of bricks have excellent load-bearing capacity, but have little heat-shielding properties. If you follow the standards for thermal protection of a brick wall, then its thickness should be at least three meters - and this is simply not realistic.

Brick load-bearing wall thickness

A building material such as brick has been used for construction for several hundred years. The material has standard dimensions 250x12x65, regardless of the type. Determining what should be the thickness of a brick wall, it is from these classic parameters that they proceed.

Load-bearing walls are a rigid frame of the structure that cannot be torn down and replanned, as the reliability and strength of the building is compromised. The load-bearing walls can withstand colossal loads - these are the roof, floors, dead weight and partitions. The most suitable and time-tested material for the construction of load-bearing walls is precisely brick. The thickness of the bearing wall must be at least one brick, or in other words - 25 cm. Such a wall has distinctive thermal insulation characteristics and durability.

A properly constructed bearing wall made of bricks has a service life of more than one hundred years. For low-rise buildings use solid brick with insulation or perforated.

Brick wall thickness parameters

Both external and internal walls are laid out of bricks. Inside the structure, the wall thickness should be at least 12 cm, that is, in the brick floor. The cross-section of pillars and piers is at least 25x38 cm. Partitions inside the building can be 6.5 cm thick. This method of laying is called “on the edge”. The thickness of a brick wall made in this way must be reinforced metal frame every 2 rows. Reinforcement will allow the walls to acquire additional strength and withstand more substantial loads.

The method of combined masonry, when the walls are made up of several layers, is very popular. This solution allows you to achieve greater reliability, strength and thermal resistance. Such a wall includes:

• Brickwork consisting of porous or slotted material;
• Insulation - mineral wool or foam;
• Cladding - panels, plaster, facing bricks.

Outer thickness combined wall determined climatic conditions region and the type of insulation used. In fact, the wall can have a standard thickness, and thanks to the correctly selected insulation, all the norms for the thermal protection of the building are achieved.

Wall masonry in one brick

The most common wall laying in one brick makes it possible to obtain a wall thickness of 250 mm. Brick in this masonry does not fit next to each other, since the wall will not have the required strength. Depending on the expected loads, the thickness of the brick wall can be 1.5, 2 and 2.5 bricks.

The most important rule in this type of masonry is high-quality masonry and correct dressing of the vertical seams that connect the materials. The brick from the top row must necessarily overlap the bottom vertical seam. Such a dressing significantly increases the strength of the structure and distributes the loads evenly on the wall.

Types of dressings:

• Vertical seam;
• Cross seam that does not allow the materials to shift along the length;
• Longitudinal seam preventing horizontal movement of bricks.

Laying a wall in one brick should be carried out according to a strictly selected scheme - it is single-row or multi-row. In a single-row system, the first row of bricks is laid with the spoon side, the second with the butt side. The transverse joints move half the brick.

The multi-row system assumes alternation through a row, and through several spoon rows. If thick brick is used, then the spoon rows are no more than five. This method provides maximum structural strength.

The next row is stacked in the opposite order, thereby forming a mirror image of the first row. Such masonry has special strength, since the vertical seams do not coincide anywhere and are overlapped by the upper bricks.

If it is planned to create a masonry of two bricks, then, accordingly, the wall thickness will be 51 cm. Such construction is necessary only in regions with severe frosts or in construction where insulation is not supposed to be used.

Brick was and still remains one of the main building materials in low-rise construction. The main advantages of brickwork are strength, fire resistance, moisture resistance. Below we give data on the consumption of bricks per 1 sq. M for different thicknesses of brickwork.

Currently, there are several ways to perform brickwork (standard brickwork, Lipetsk, Moscow, etc.). But when calculating the consumption of bricks, the method of performing the brickwork is not important, the thickness of the brickwork and the size of the brick are important. Brick produced different sizes, characteristics and purpose. The main typical brick sizes are the so-called "single" and "one and a half" bricks:

the size " single"brick: 65 x 120 x 250 mm

the size " one and a half bricks: 88 x 120 x 250 mm

In masonry, as a rule, the thickness of the vertical mortar joint averages about 10 mm, the thickness horizontal seam- 12 mm. Brickwork can be of various thicknesses: 0.5 bricks, 1 brick, 1.5 bricks, 2 bricks, 2.5 bricks, etc. As an exception, there is a quarter-brick brickwork.

Quarter-brick masonry is used for small partitions that do not carry loads (for example, brick partition between bathroom and toilet). Half-brick brickwork is often used for one-story outbuildings (barn, toilet, etc.), gables of residential buildings. One brick masonry can be used to build a garage. For the construction of houses (residential premises), brickwork is used with a thickness of one and a half bricks or more (depending on the climate, number of storeys, type of floors, individual characteristics buildings).

Based on the given data on the dimensions of bricks and the thickness of the mortar joints, you can easily calculate the number of bricks required for the construction of 1 square meter of wall made with brickwork of various thicknesses.

Wall thickness and brick consumption for different brickwork

The data are given for a "single" brick (65 x 120 x 250 mm), taking into account the thickness of the mortar joints.

The article presents an example of calculating the bearing capacity of a brick wall of a three-story frameless building, taking into account the defects identified during its inspection. Such calculations belong to the category of "verification" and are usually performed as part of a detailed visual and instrumental survey of buildings.

The bearing capacity of centrally and eccentrically compressed stone pillars is determined on the basis of data on the actual strength of masonry materials (brick, mortar) in accordance with section 4.

To take into account the defects revealed during the survey, an additional reduction factor is introduced into the SNiP formulas, taking into account the decrease in the bearing capacity of stone structures (Ktr), depending on the nature and degree of damage detected according to the tables of Ch. 4 .

EXAMPLE OF CALCULATION

Let's check the load-bearing capacity of the internal load-bearing stone wall Of the 1st floor along the axis "8" m / o "B" - "V" on the action of operational loads, taking into account the defects and damages revealed during its examination.

Initial data:

- Wall thickness: dst = 0.38 m
- Wall width: b = 1.64 m
- The height of the wall to the bottom of the floor slabs on the 1st floor: H = 3.0 m
- Height of the overlying masonry column: h = 6.5 m
- Area of ​​collection of loads from floors and coverings: Sgr = 9.32 m2
- Design resistance of masonry to compression: R = 11.05 kg / cm2

During the inspection of the wall along the "8" axis, the following defects and damages were recorded (see the photo below): mass loss of mortar from the joints of the masonry to a depth of more than 4 cm; displacement (curvature) of horizontal rows of masonry vertically up to 3 cm; multiple vertically oriented cracks with an opening of 2-4 mm (including along mortar joints), crossing from 2 to 4 horizontal rows of masonry (up to 2 cracks per 1 m of the wall).

Pustoshovka	Cracking bricks	Curving rows of masonry

By the totality of the identified defects (taking into account their nature, degree of development and area of ​​distribution), in accordance with, load bearing capacity the wall in question must be reduced by at least 30%. Those. the coefficient of reducing the bearing capacity of the wall is taken equal to - Ktr = 0.7. The diagram for collecting loads on the wall is shown below in Fig. 1.

FIG. 1. Scheme for collecting loads on the wall

I. Collection of design loads on the wall

II. Calculation of the bearing capacity of the wall

(clause 4.1 SNiP II-22-81)

A quantitative assessment of the actual bearing capacity of a centrally compressed brick wall (taking into account the effect of the detected defects) on the effect of the calculated longitudinal force N applied without eccentricity is reduced to checking the fulfillment of the following condition (formula 10):

Nс = mg × φ × R × A × Ktr ≥ N(1)

According to the results of strength tests, the design resistance of the wall masonry along the "8" axis to compression is R = 11.05 kg / cm2.
The elastic characteristic of the masonry according to clause 9 of Table 15 (K) is equal to: α = 500.
Estimated post height: l0 = 0.8 × H = 0.8 × 300 = 240 cm.
Flexibility of a rectangular solid element: λh = l0 / dst = 240/38 = 6.31.
Buckling coefficient φ at α = 500 and λh = 6.31(according to Table 18): φ = 0.90.
Cross-sectional area of ​​the column (wall): A = b × dst = 164 × 38 = 6232 cm2.
Because the thickness of the calculated wall is more than 30 cm (dst = 38 cm), the coefficient mg taken equal to one: mg = 1.

Substituting the obtained values ​​into the left side of formula (1), we determine the actual bearing capacity of the centrally compressed unreinforced brick wall Nc:

Nс = 1 × 0.9 × 11.05 × 6232 × 0.7 = 43 384 kgf

III. Checking the fulfillment of the strength condition (1)

[Nc = 43384 kgf]> [N = 36340.5 kgf]

The strength condition is met: load bearing capacity brick pillar Nc taking into account the influence of the revealed defects, it turned out more value total load N.

List of sources:
1. SNiP II-22-81 * "Stone and reinforced stone structures."
2. Recommendations for strengthening the stone structures of buildings and structures. TsNIISK them. Kurchenko, Gosstroy.