Calculation of the load-bearing capacity of a brick wall. Strength calculation of brickwork

The need to calculate brickwork when building 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 houses. It has a large specific gravity, attractive appearance, high strength. Limited use due to the high cost of the material.

The most popular and in demand material is red brick. It has sufficient strength with relatively little specific gravity, easy to process, little exposed to the environment. Disadvantages - sloppy surfaces with great roughness, the ability to absorb water when high humidity. Under normal operating conditions, this ability does not manifest itself.

There are two methods for laying bricks:

• tychkovy;
• spoon

When laying using the butt 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 lengthwise. Outside it turns out side surface. Using this method, you can lay out half-brick walls - 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 obtained will be approximate and overestimated. For a more accurate calculation, the following factors must be taken into account:

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

Manufacturers quite often various reasons can't stand it standard sizes products. According to GOST, red masonry bricks must have dimensions of 250x120x65 mm. To avoid unnecessary mistakes material costs It is advisable to check with suppliers about the sizes of available bricks.

Optimal thickness external walls for most regions is 500 mm, or 2 bricks. This size provides high strength building, good thermal insulation. The disadvantage is the large weight of the structure and, as a result, pressure on the foundation and lower layers of 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 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 seams and the method of laying the brick, you can save some of it.

For example, let's take a seam thickness of 6 mm and the spoon method of laying brick walls. If the wall thickness is 0.5 m, you need to lay 4 bricks 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 area of ​​the masonry will be approximately 1.25 m2. If the bricks are laid closely, without gaps, 240 pieces fit in 1 m2. Taking into account the gaps, the material consumption will be approximately 236 pieces.

Return to contents

Calculation method for 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 carry out calculations, then carry them out in reality. When planning the construction of 2 floors, you should calculate the amount of material in stages for each floor.

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

• length = 15 m;
• width = 10 m;
• height = 3 m;
• The thickness of the walls is 2 bricks.

Using these dimensions you need to determine the perimeter of the building:

(15 + 10) x 2 = 50

3 x 50 = 150 m2

By calculating the total area, you can determine maximum amount bricks for building a wall. To do this, you need to multiply the previously determined number of bricks for 1 m2 by the total area:

236 x 150 = 35,400

The result is inconclusive, the walls must have openings for installing doors and windows. The number of 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 sizes and location are determined by the internal layout of the building.

As an example, you can take 3 window openings per 10-meter wall, 4 per 15-meter walls. It is advisable to make one of the walls blank, without openings. Volume doorways can be determined by standard sizes. If the sizes differ from standard ones, the volume can be calculated using overall dimensions, adding to them the width of the installation gap. 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 standard values, we get:

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

Volume 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 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.

Purchase required amount should be done with a small margin, because errors and other unforeseen situations are possible during operation.

Brick is a fairly durable building material, especially solid ones, and when building houses with 2-3 floors, the walls are made of ordinary ceramic bricks As a rule, additional calculations are not needed. Nevertheless, situations are different, for example, it is planned two-storey house with a terrace on the second floor. The metal crossbars, on which the metal beams of the terrace will also rest, are planned to be supported by brick columns from the front hollow brick 3 meters high, higher there will be columns 3 m high on which the roof will rest:

A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea of ​​laying columns of clay bricks, and even more so the walls of a house, is far from new and all possible aspects of the calculations of brick walls, piers, pillars, which are the essence of the column, are described in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures." This is exactly what normative document and should be used as a guide when making calculations. The calculation below is nothing more than an example of using the specified SNiP.

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

with central compression

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

With this design scheme, the maximum load will be on the middle lower column. This is exactly what you should count on for strength. The load on the column depends on many factors, in particular the construction area. For example, snow load for roofing in St. Petersburg is 180 kg/m², and in Rostov-on-Don - 80 kg/m². Taking into account the weight of the roof itself, 50-75 kg/m², 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 current loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab It’s not exactly planned, but it is assumed that the ceiling will be wooden, from separate edged boards, then to calculate the load from the terrace, you can take a uniformly distributed load of 600 kg/m², then the concentrated force from the terrace acting on the central column will be:

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

The dead weight of columns 3 m long will be:

N from 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 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 temporary load from snow, maximum in winter time, and the temporary load on the floor, maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability coefficient 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 M75 brick grade means that the brick must withstand a load of 75 kgf/cm2, 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. Design compressive strengths for brickwork

But that's not all. The same SNiP II-22-81 (1995) clause 3.11 a) recommends that for the area of ​​pillars and piers less than 0.3 m², multiply the value of the design resistance by the operating conditions coefficient γ s =0.8. And since the cross-sectional area of ​​our column is 0.25x0.25 = 0.0625 m², we will have to use this recommendation. As you can see, for M75 grade brick, even when using M100 masonry mortar, the strength of the masonry will not exceed 15 kgf/cm2. As a result, the calculated resistance for our column will be 15·0.8 = 12 kg/cm², then the maximum compressive stress will be:

10300/625 = 16.48 kg/cm² > R = 12 kgf/cm²

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

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 still 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)

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

φ - longitudinal bending coefficient, 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 a structure are not outlined here, we only note that according to SNiP II-22-81 (1995) clause 4.3: “Calculation heights of walls and pillars l o when determining buckling coefficients φ depending on the conditions of supporting them on horizontal supports, the following should be taken:

a) with fixed hinged supports l o = N;

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 pinched supporting sections - taking into account the actual degree of pinching, but not less l o = 0.8N, Where N- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the clear distance between them."

At first glance, our calculation scheme can be considered as satisfying the conditions of point b). i.e. you can take it l o = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only in the case when the lower support is really rigid. If a brick column is laid on a layer of roofing felt waterproofing laid on the foundation, then such a support should rather be considered as hinged rather than rigidly clamped. And in this case, our design 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 the specified plane. There are 4 possible 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 beams will be welded; then, for aesthetic reasons, the metal columns can be covered with facing brick of any brand, since the entire load will be carried by the metal. In this case, it is true that the metal columns need to be calculated, but the calculated length can be taken l o = 1.25H.

2. Make another overlap, for example from sheet materials, which will allow us to consider both the upper and lower supports of the column as hinged, in this case l o = H.

3. Make a stiffening 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 supports of the column as hinged, 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 (though not made of brick) without any knowledge of the resistance of materials, without the use of metal anchors, and even so carefully written building codes and there were no rules in those days, however, some columns still stand to this day.

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

λ h = l o /h (1.2) or

λ i = l o (1.3)

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

Determining the radius of gyration is not difficult in principle; you need to divide the moment of inertia of the section by the cross-sectional area, and then extract from the result Square root, however, in this case there is no great need for this. Thus λ h = 2 300/25 = 24.

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

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

In this case, the elastic characteristics of the masonry α determined by the table:

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

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

N р = m g φγ with RF = 1 0.6 0.8 22 625 = 6600 kg< N с об = 9400 кг

This means that the adopted cross-section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, it is most optimal to increase the cross-section of the column. For example, if you lay out a column with a void inside of one and a half bricks, measuring 0.38 x 0.38 m, then not only will the cross-sectional area of ​​the column increase to 0.13 m or 1300 cm, but the radius of inertia of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the coefficient value φ = 0.8. In this case, the maximum load on the central column will be:

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

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

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

That seems to be all, but it is advisable to take into account one more detail. In this case, it is better to make the foundation strip (united for all three columns) rather than 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 most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of ​​such columns will be 2601 cm2.

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

The outer 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 still, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the outer columns not in the center of the column section. Where exactly 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 e o. 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 transferred as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be subject to a bending moment equal to M = Ne o, and this point must be taken into account when calculating. In general, stability testing can be performed using the following formula:

N = φRF - MF/W (2.1)

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

N р = 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 load application, we have a more than double 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.

To perform a wall stability calculation, you first need to understand their classification (see SNiP II -22-81 “Stone and reinforced masonry structures”, as well as a manual for SNiP) and understand what types of walls there are:

1. Load-bearing walls- these are the walls on which floor slabs, roof structures, etc. rest. The thickness of these walls must be at least 250 mm (for brickwork). These are the most important walls in the house. They need to be designed for strength and stability.

2. Self-supporting walls- these are walls on which nothing rests, but they are subject to the load from all the floors above. In fact, in a three-story house, for example, such a wall will be three floors high; the load on it only from the own weight of the masonry is significant, but at the same time the question of the stability of such a wall is also very important - the higher the wall, the greater the risk of its deformation.

3. Curtain walls- these are external walls that rest on the ceiling (or other structural elements) and the load on them comes from the height of the floor only from the wall’s own weight. The height of non-load-bearing walls should be no more than 6 meters, otherwise they become self-supporting.

4. Partitions are internal walls less than 6 meters high that only support the load from their own weight.

Let's look at the issue of wall stability.

The first question that arises for an “uninitiated” person is: where can the wall go? Let's find the answer using an analogy. Let's take a hardcover book and place it on its edge. The larger the book format, the less stable it will be; on the other hand, the thicker the book, the better it will stand on its edge. The situation is the same with walls. The stability of the wall depends on the height and thickness.

Now let's take the worst case scenario: a thin, large-format notebook and place it on its edge - it will not only lose stability, but will also bend. Likewise, the wall, if the conditions for the ratio of thickness and height are not met, will begin to bend out of plane, and over time, crack and collapse.

What is needed to avoid this phenomenon? You need to study pp. 6.16...6.20 SNiP II -22-81.

Let's consider the issues of determining the stability of walls using examples.

Example 1. Given a partition made of aerated concrete grade M25 on mortar grade M4, 3.5 m high, 200 mm thick, 6 m wide, not connected to the ceiling. The partition has a doorway of 1x2.1 m. It is necessary to determine the stability of the partition.

From Table 26 (item 2) we determine the masonry group - III. From the tables do we find 28? = 14. Because the partition is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 9.8.

k 1 = 1.8 - for a partition that does not carry a load with a thickness of 10 cm, and k 1 = 1.2 - for a partition 25 cm thick. By interpolation, we find for our partition 20 cm thick k 1 = 1.4;

k 3 = 0.9 - for partitions with openings;

that means k = k 1 k 3 = 1.4*0.9 = 1.26.

Finally β = 1.26*9.8 = 12.3.

Let's find the ratio of the height of the partition to the thickness: H /h = 3.5/0.2 = 17.5 > 12.3 - the condition is not met, a partition of such thickness cannot be made with the given geometry.

How can this problem be solved? Let's try to increase the grade of mortar to M10, then the masonry group will become II, respectively β = 17, and taking into account the coefficients β = 1.26*17*70% = 15< 17,5 - этого оказалось недостаточно. Увеличим марку газобетона до М50, тогда группа кладки станет I , соответственно β = 20, а с учетом коэффициентов β = 1,26*20*70% = 17.6 >17.5 - the condition is met. It was also possible, without increasing the grade of aerated concrete, to lay structural reinforcement in the partition in accordance with clause 6.19. Then β increases by 20% and the stability of the wall is ensured.

Example 2. An external non-load-bearing wall is made of lightweight masonry made of M50 grade brick with M25 grade mortar. Wall height 3 m, thickness 0.38 m, wall length 6 m. Wall with two windows measuring 1.2x1.2 m. It is necessary to determine the stability of the wall.

From Table 26 (clause 7) we determine the masonry group - I. From Table 28 we find β = 22. Because the wall is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 15.4.

We find the coefficients k from tables 29:

k 1 = 1.2 - for a wall that does not bear a load with a thickness of 38 cm;

k 2 = √A n /A b = √1.37/2.28 = 0.78 - for a wall with openings, where A b = 0.38*6 = 2.28 m 2 - horizontal sectional area of ​​the wall, taking into account windows, A n = 0.38*(6-1.2*2) = 1.37 m2;

that means k = k 1 k 2 = 1.2*0.78 = 0.94.

Finally β = 0.94*15.4 = 14.5.

Let's find the ratio of the height of the partition to the thickness: H /h = 3/0.38 = 7.89< 14,5 - условие выполняется.

It is also necessary to check the condition stated in clause 6.19:

H + L = 3 + 6 = 9 m< 3kβh = 3*0,94*14,5*0,38 = 15.5 м - условие выполняется, устойчивость стены обеспечена.

Attention! For the convenience of answering your questions, a new section “FREE CONSULTATION” has been created.

class="eliadunit">

Comments

« 3 4 5 6 7 8

0 #212 Alexey 02/21/2018 07:08

I quote Irina:

profiles will not replace reinforcement

I quote Irina:

Regarding the foundation: voids in the concrete body are permissible, but not from below, so as not to reduce the bearing area, which is responsible for the load-bearing capacity. That is, there should be a thin layer below reinforced concrete.
What kind of foundation - strip or slab? What soils?

The soils are not yet known, most likely it will be an open field of all sorts of loam, initially I thought of a slab, but it will be a little low, I want it higher, and I will also have to remove the top fertile layer, so I am leaning towards a ribbed or even box-shaped foundation. I don’t need a lot of bearing capacity of the soil - after all, the house was built on the 1st floor, and expanded clay concrete is not very heavy, freezing there is no more than 20 cm (although according to old Soviet standards it is 80).

I'm thinking about renting upper layer 20-30 cm, lay out geotextiles, cover with river sand and level with compaction. Then a light preparatory screed - for leveling (it seems like they don’t even make reinforcement into it, although I’m not sure), waterproofing with a primer on top
and then there’s a dilemma - even if you tie reinforcement frames with a width of 150-200mm x 400-600mm in height and lay them in steps of a meter, then you still need to form voids with something between these frames and ideally these voids should be on top of the reinforcement (yes also with some distance from the preparation, but at the same time they will also need to be reinforced on top thin layer under a 60-100mm screed) - I’m thinking of monolithing the PPS slabs as voids - theoretically it would be possible to fill this in one go with vibration.

Those. It looks like a slab of 400-600mm with powerful reinforcement every 1000-1200mm, the volumetric structure is uniform and light in other places, while inside about 50-70% of the volume there will be foam plastic (in unloaded places) - i.e. in terms of consumption of concrete and reinforcement - quite comparable to a 200mm slab, but + a lot of relatively cheap polystyrene foam and more work.

If we somehow replaced the foam plastic with simple soil/sand, it would be even better, but then instead of light preparation, it would be wiser to do something more serious with reinforcement and moving the reinforcement into the beams - in general, I lack both theory and practical experience here.

0 #214 Irina 02.22.2018 16:21

Quote:

It’s a pity, in general they just write that lightweight concrete (expanded clay concrete) has a poor connection with the reinforcement - how to deal with this? I understand what stronger than concrete and the larger the surface area of ​​the reinforcement, the better the connection will be, i.e. you need expanded clay concrete with the addition of sand (and not just expanded clay and cement) and thin reinforcement, but more often

why fight it? you just need to take it into account in the calculations and design. You see, expanded clay concrete is quite good wall material with its own list of advantages and disadvantages. Just like any other materials. Now, if you wanted to use it for monolithic ceiling, I would dissuade you, because
Quote:

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

A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea of ​​laying columns of clay bricks, and even more so the walls of a house, is far from new and all possible aspects of the calculations of brick walls, piers, pillars, which are the essence of the column, are described in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures." It is this regulatory document that should be used as a guide when making calculations. The calculation below is nothing more than an example of using the specified SNiP.

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

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

Designed:

Terrace dimensions 5x8 m. Three columns (one in the middle and two at 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 strength grade of the brick is M75.

Calculation prerequisites:

.

With this design scheme, the maximum load will be on the middle lower column. This is exactly what you should count on for strength. The load on the column depends on many factors, in particular the construction area. For example, in St. Petersburg it is 180 kg/m2, and in Rostov-on-Don - 80 kg/m2. Taking into account the weight of the roof itself 50-75 kg/m2, 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 current loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but a reinforced concrete slab is definitely not planned, and it is assumed that the floor will be wooden, from separately lying edged boards, then to calculate the load from the terrace you can accept a uniformly distributed load of 600 kg/m2, then the concentrated force from the terrace acting on the central column will be:

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

The dead weight of columns 3 m long will be:

N from 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 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 temporary load from snow, maximum in winter, and the temporary load on the floor, maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability coefficient 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 M75 brick grade means that the brick must withstand a load of 75 kgf/cm2, 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. Design compressive strengths for brickwork (according to SNiP II-22-81 (1995))

But that's not all. All the same SNiP II-22-81 (1995) clause 3.11 a) recommends that for the area of ​​pillars and piers less than 0.3 m 2, multiply the value of the design resistance by working conditions factor γ s =0.8. And since the cross-sectional area of ​​our column is 0.25x0.25 = 0.0625 m2, we will have to use this recommendation. As you can see, for M75 grade brick, even when using M100 masonry mortar, the strength of the masonry will not exceed 15 kgf/cm2. As a result, the calculated resistance for our column will be 15·0.8 = 12 kg/cm2, then the maximum compressive stress will be:

10300/625 = 16.48 kg/cm 2 > R = 12 kgf/cm 2

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

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 still 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, we were, relatively speaking, 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 m g coefficient, everything is not so simple; details can be found in the comments to the article.

φ - longitudinal bending coefficient, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l 0 , and it does not always coincide with the height of the column. The subtleties of determining the design length of a structure are set out separately; here we only note that according to SNiP II-22-81 (1995) clause 4.3: “Calculated heights of walls and pillars l 0 when determining buckling coefficients φ depending on the conditions of supporting them on horizontal supports, the following should be taken:

a) with fixed hinged supports l 0 = N;

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 = 2H;

d) for structures with partially pinched supporting sections - taking into account the actual degree of pinching, but not less l 0 = 0.8N, Where N- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the clear distance between them."

At first glance, our calculation scheme can be considered as satisfying the conditions of point b). i.e. you can take it l 0 = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only in the case when the lower support is really rigid. If a brick column is laid on a layer of roofing felt waterproofing laid on the foundation, then such a support should rather be considered as hinged rather than rigidly clamped. And in this case, our design 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 specified plane. There are 4 possible 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 beams will be welded; then, for aesthetic reasons, the metal columns can be covered with facing bricks of any brand, since the entire load will be carried by the metal. In this case, it is true that the metal columns need to be calculated, but the calculated length can be taken l 0 = 1.25H.

2. Make another overlap,

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

3. Make a stiffening 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 supports of the column as hinged, 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 0 = 2H

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

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

λ h = l 0 /h (1.2) or

λ i = l 0 /i (1.3)

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

Determining the radius of inertia is, in principle, not difficult; you need to divide the moment of inertia of the section by the cross-sectional area, and then take the square root of the result, but in this case there is no great need for this. Thus λ h = 2 300/25 = 24.

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

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

In this case, the elastic characteristics of the masonry α determined by the table:

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

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

N р = m g φγ with RF = 1х0.6х0.8х22х625 = 6600 kg< N с об = 9400 кг

This means that the adopted cross-section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, it is most optimal to increase the cross-section of the column. For example, if you lay out a column with a void inside of one and a half bricks, measuring 0.38x0.38 m, then not only will the cross-sectional area of ​​the column increase to 0.13 m2 or 1300 cm2, but the radius of inertia of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the coefficient value φ = 0.8. In this case, the maximum load on the central column will be:

N r = m g φγ with RF = 1x0.8x0.8x22x1300 = 18304 kg > N with rev = 9400 kg

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

N r = m g φγ with RF = 1x0.8x0.8x12x1300 = 9984 kg > N with rev = 9400 kg

That seems to be all, but it is advisable to take into account one more detail. In this case, it is better to make the foundation strip (united for all three columns) rather than 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 most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of ​​such columns will be 2601 cm2.

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

The outer 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 still, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the outer columns not in the center of the column section. Where exactly the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the modulus of elasticity 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 a beam for bearing". This displacement is called the eccentricity of the load application e o. 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 transferred as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be subject to a bending moment equal to M = Ne o, and this point must be taken into account when calculating. In general, stability testing can be performed using the following formula:

N = φRF - MF/W (2.1)

Where W- section moment of resistance. In this case, the load for the lower outermost columns from the roof can be conditionally considered centrally applied, and eccentricity will only be created by the load from the floor. At eccentricity 20 cm

N р = φRF - MF/W =1x0.8x0.8x12x2601- 3000 20 2601· 6/51 3 = 19975, 68 - 7058.82 = 12916.9 kg >N cr = 5800 kg

Thus, even with a very large eccentricity of load application, we have a more than double 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 I do not present the calculation method recommended by SNiP here.

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

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

The walls in a building perform various functions:

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

Features of calculating wall thickness

• The thickness of the walls according to thermal engineering calculations does not always coincide with the calculation of the value based on strength characteristics. Naturally, the more severe the climate, the thicker the wall should be in terms of thermal performance indicators.
• But in terms of strength, for example, it is enough to lay out the outer walls in one or one and a half bricks. This is where it turns out to be “nonsense” - the thickness of the masonry, a certain thermotechnical calculation, often, due to strength requirements, it turns out to be excessive.
• Therefore, laying solid brick walls from the point of view of material costs and subject to 100% use of its strength should only be done in the lower floors of high-rise buildings.
• In low-rise buildings, as well as in the upper floors of high-rise buildings, hollow or light brick, you can use lightweight masonry.
• 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 built with protective layer from vapor barrier material from the inside and from a solid clay material.

Now I’ll tell you about the calculation used to determine the thickness of external walls.

It is determined by the formula:

B = 130*n -10, where

B – wall thickness in millimeters

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

n – integer half of a brick (= 120mm)

The calculated value of the solid masonry is rounded up to the whole number of half-bricks.

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

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

To save money material resources(bricks, mortar, fittings, etc.), the number of machine hours of mechanisms, the calculation of wall thickness is tied to the load-bearing capacity of the building. And the thermal component is obtained by insulating the facades of buildings.

How can you insulate the external walls of a brick building? In the article insulating a house with polystyrene foam from the outside, I indicated the reasons why brick walls cannot 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 the migration of moisture outward. That is why it is advisable to insulate a brick wall with heat-insulating plaster or mineral wool slabs, the nature of which is vapor permeable. Expanded polystyrene is suitable for insulating concrete or reinforced concrete bases. “The nature of the insulation must correspond to the nature of the load-bearing wall.”

There are many heat-insulating plasters– the difference lies in the components. But the principle of application is the same. It is performed in layers and the total thickness can reach up to 150mm (for large values, reinforcement is required). In most cases, this value is 50 - 80 mm. It depends on the climate zone, thickness of the base walls, other factors. I will not go into detail, since this is the topic of another article. Let's return to our bricks.

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

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

I would like to summarize the above. We calculate the thickness of external brick walls based on the strength characteristics, and solve the heat-technical side of the issue using the method of wall insulation. As a rule, a design firm designs external walls without the use of insulation. If the house is uncomfortably cold and the need for insulation arises, then carefully consider the selection of insulation.

When building your house, one of the main points is the construction of walls. The laying of load-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 design, it has height, width and thickness. The weight of the wall that interests us 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 depends on 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 surface load is the thermal conductivity of the material. An ordinary solid block has quite high thermal conductivity. This means that it, in itself, is a poor thermal insulator. Therefore, in order to achieve 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 preserve common sense, people abandoned the idea of ​​​​building houses that resemble a bunker. In order to have strong load-bearing surfaces and at the same time good thermal insulation, they began to use a multilayer scheme. Where one layer is silicate masonry, heavy enough to withstand all the loads to which it is subject, 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 solid and perforated. Solid materials have greater strength, cost, and thermal conductivity.

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

The dimensions of any type of material in question may also vary. He can be:

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

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

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

The half is the baby among its brothers, it has, as you probably already guessed, half the thickness of the single - 250X120X12 mm.

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

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

Wall thickness

We have already examined the parameters on which the thickness of 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.

Load-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 external factors, such as winds, precipitation, in addition, their own weight presses on them. Therefore, their weight, compared to non-load-bearing surfaces and internal partitions, should be the highest.

In modern realities, for most two- and three-story houses, 25 cm in thickness or one block is enough, less often one and a half or 38 cm. The strength of such masonry will be sufficient 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 SNiP II-22-8 standards. 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 M50 material, on M25 mortar, the calculation will be carried out for one load-bearing surface, without windows. So let's get started.

Table No. 26

According to the data from the table above, we know that the characteristics of our masonry belong to the first group, and the description from point 7 is also valid for it. Table. 26. After this, we look at table 28 and find the value β, which means the permissible ratio of the load of the wall to its height, taking into account the type of mortar used. For our example, this value is 22.

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

Аn – horizontal cross-sectional area of ​​the load-bearing surface, the calculation is simple: 0.25*5=1.25 sq. m

Ab is the horizontal cross-sectional area of ​​the wall, taking into account window openings we do not have, so k2 = 1.25

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

Now that you know all the variables, you can find the overall coefficient “k” by multiplying all the values. K=1.2*1.25*0.9=1.35 Next, we find out the total 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 significantly less than the obtained indicator 29.7. This means that 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 that required by SNiP standards.

Well, we figured out the load-bearing surfaces, but what about the partitions and those that do not bear the load. It is advisable to make partitions half the thickness - 12 cm. For surfaces that do not bear a load, the stability formula that we discussed above is also valid. But since such a wall will not be secured from above, the β coefficient must be reduced by a third, and calculations must be continued with a different value.

Laying half a brick, brick, one and a half, two bricks

In conclusion, let's look at how bricklaying is carried out depending on the load of the surface. Half-brick masonry is the simplest of all, since there is no need to make complex row dressings. It is enough to place the first row of material on a perfectly flat base and make sure that the solution lies evenly and does not exceed 10 mm in thickness.

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

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

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

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

The walls of a house, depending on their place in the project diagram, can be load-bearing, self-supporting, non-load-bearing and partitions. Load-bearing walls perform an enclosing function and also serve as supports for slabs or floor beams or roof structures. The thickness of load-bearing brick walls cannot be less than one brick (250 mm). Most modern houses are built with walls of one or 1.5 bricks. Projects of private houses, which would require walls thicker than 1.5 bricks, logically should not exist. Therefore, choosing the thickness of the outer brick wall according to by and large- the matter is decided. If you choose between a thickness of one brick or one and a half, then from a purely technical point of view, for a cottage with a height of 1-2 floors, a brick wall with a thickness of 250 mm (one brick of strength grade M50, M75, M100) will correspond to the calculations bearing loads. There is no need to play it safe, since the calculations already take into account snow, wind loads and many coefficients that provide the brick wall with a sufficient margin of safety. However, there is a very important point that really affects the thickness of a brick wall - stability.

Everyone once played with cubes in childhood and noticed that the more cubes you stack on top of each other, the less stable the column of them becomes. The elementary laws of physics that act on cubes act in exactly 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, ensuring the stability of the structure. We will talk about this dependence in the first half of this article.

Wall stability, as well as construction standards for load-bearing and other loads, are described in detail in SNiP II-22-81 “Stone and reinforced masonry structures”. These standards are a guide for designers, and for the “uninitiated” they may seem quite difficult to understand. This is true, because to become an engineer you need to study for at least four years. Here we could refer to “contact specialists for calculations” and call it a day. However, thanks to the capabilities of the information web, today almost everyone can understand the most complex issues if they wish.

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, excess reinsurance can increase the cost of the box by 1.5-2 times. And this is a lot of money today. To avoid wall destruction or unnecessary financial expenses, let's turn to mathematical calculations.

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 Let's consider how to determine whether the stability of an external load-bearing brick (M50) wall on M25 mortar with a thickness of 1.5 bricks (0.38 m), a height of 3 m and a length of 6 m with two window openings of 1.2 × 1.2 m is sufficient .

Turning to table 26 (table above), we find that our wall belongs to the first group of masonry and fits the description of point 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 masonry mortar. The required parameter β is the ratio of the height of the wall to its thickness (β=Н/h). In accordance with the data in 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 to 15.4.

Let's move on to determining the correction factors from Table 29, which will help find the total coefficient k:

• for a wall 38 cm thick, not load-bearing, k1=1.2;
• k2=√Аn/Аb, where An is the horizontal sectional area of ​​the wall taking into account window openings, Аb is the horizontal sectional area excluding windows. In our case, An= 0.38×6=2.28 m², and Аb=0.38×(6-1.2×2)=1.37 m². We perform the calculation: k2=√1.37/2.28=0.78;
• k4 for a wall 3 m high is 0.9.

By multiplying all correction factors, we find the overall coefficient k = 1.2 × 0.78 × 0.9 = 0.84. After taking into account the set of 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. Available 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 condition H/h is satisfied

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

Brick wall thickness and heat transfer resistance standards

Today, the overwhelming number of brick houses have a multi-layer wall structure consisting of lightweight brickwork, insulation and facade finishing. According to SNiP II-3-79 (Building heating engineering) external walls of residential buildings with a requirement 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. 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 construction and climatic zones in accordance with SNiP II-3-79 and SP-41-99.

Heat transfer resistance R(thermal resistance, m².°C/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.°C).

To obtain the total thermal resistance of a multilayer enclosing structure, it is necessary to add up the thermal resistances of all layers of the wall structure. Let's consider the following using 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 1.2 m².°C/W. The thermal conductivity coefficient of sand-lime brick 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. We substitute the values ​​and solve: δ =1.2×0.7=0.84 m.

Now let’s calculate what layer of expanded polystyrene needs to be used to insulate a 25 cm thick sand-lime brick wall in order to reach a figure of 1.2 m².°C/W. The thermal conductivity coefficient of expanded polystyrene (PSB 25) is no more than 0.039 W/(m°C), and that of sand-lime brick is 0.7 W/(m°C).

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

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

3) determine the thickness of polystyrene foam required to obtain a thermal resistance equal to 0.85 m².°C/W: 0.85×0.039=0.033 m.

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

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

Modern residential construction places high demands on such parameters as strength, reliability and thermal protection. External walls built of brick have excellent load-bearing capabilities, but have poor heat-insulating 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.

Thickness of load-bearing brick wall

Building materials such as brick have been used for construction for several hundred years. The material has standard dimensions 250x12x65, regardless of type. When determining what the thickness of a brick wall should be, we proceed from these classical parameters.

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

A properly constructed load-bearing brick wall has a service life of hundreds of years. For low-rise buildings use solid brick with insulation or perforated brick.

Brick wall thickness parameters

Both external and internal walls are made of brick. Inside the structure, the wall thickness should be at least 12 cm, that is, half a brick. The cross-section of the pillars and partitions is at least 25x38 cm. The partitions inside the building can be 6.5 cm thick. This method of masonry is called “on edge”. The thickness of a brick wall made using this method must be reinforced with a metal frame every 2 rows. Reinforcement will allow the walls to acquire additional strength and withstand more significant loads.

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

• Brickwork consisting of porous or slotted material;
• Insulation – mineral wool or polystyrene foam;
• Facing – panels, plaster, facing bricks.

The thickness of the external combined wall is determined by the climatic conditions of the region and the type of insulation used. In fact, the wall may have standard thickness, and thanks to the correctly selected insulation, all standards for thermal protection of the building are achieved.

Laying a wall in one brick

The most common wall laying in one brick makes it possible to obtain a wall thickness of 250 mm. The bricks in this masonry are not laid next to each other, since the wall will not have the required strength. Depending on the expected loads, the thickness of a 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 connecting the materials. The brick from the top row must certainly overlap the bottom vertical seam. This bandaging significantly increases the strength of the structure and distributes the load evenly on the wall.

Types of dressings:

• Vertical seam;
• A transverse seam that does not allow materials to shift along their length;
• A longitudinal seam that prevents the bricks from moving horizontally.

The laying of a single brick wall must be carried out according to a strictly chosen pattern - single-row or multi-row. In a single-row system, the first row of bricks is laid with the tongue side, the second with the butt side. Transverse seams shift by half a brick.

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

The next row is laid in the opposite order, thereby forming mirror reflection first row. This type of masonry is especially strong, since the vertical seams do not coincide anywhere and are overlapped by the top bricks.

If you plan to create a masonry of two bricks, then the thickness of the wall will be 51 cm. Such construction is only necessary in regions with severe frosts or in construction where insulation is not intended 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, and moisture resistance. Below we will provide data on brick consumption per 1 sq.m. for different thicknesses of brickwork.

Currently, there are several ways to make brickwork (standard brickwork, Lipetsk brickwork, Moscow, etc.). But when calculating brick consumption, the method of making brickwork is not important, what is important is the thickness of the brickwork and the size of the brick. Brick is produced various sizes, characteristics and purpose. Main standard sizes bricks are considered to be the so-called “single” and “one and a half” bricks:

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

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

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

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

Based on the given data on the size of the brick and the thickness of the connecting mortar joints, you can easily calculate the number of bricks required to build 1 sq.m of wall made of brickwork of various thicknesses.

Wall thickness and brick consumption for different brickwork

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

Type of brickwork	Wall thickness, mm	Number of bricks per 1 sq.m of wall
0.25 bricks	65	31
0.5 bricks	120	52
1 brick	250	104
1.5 bricks	380	156
2 bricks	510	208
2.5 bricks	640	260
3 bricks	770	312