Picture 1. Design scheme for brick columns the building being designed.
In this case, a natural question arises: what is the minimum section of the columns that will provide the required strength and stability? Of course, the idea of laying out columns from clay brick, and even more so the walls of the 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 masonry structures". It is this normative document that should be followed in the calculations. The calculation below is nothing more than an example of using the specified SNiP.
To determine the strength and stability of the columns, you need to have a lot of initial data, such as: the brand of brick for strength, the area of support of the crossbars on the columns, the load on the columns, the sectional area of the column, and if none of this is known at the design stage, then you can do in the following way:
An example of calculating a brick column for stability under central compression
Designed:
Terrace measuring 5x8 m. Three columns (one in the middle and two along the edges) from the front hollow brick with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The brick grade for strength is M75.
Design assumptions:
.With such a design scheme, the maximum load will be on the middle lower column. It is she who should be counted on strength. The load on the column depends on many factors, in particular on the area of construction. For example, in St. Petersburg it is 180 kg / m 2, and in Rostov-on-Don - 80 kg / m 2. Taking into account the weight of the roof itself 50-75 kg / m 2, the load on the column from the roof for Pushkin, Leningrad Region, can be:
N from the roof = (180 1.25 + 75) 5 8/4 = 3000 kg or 3 tons
Since the actual loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculations of the load from the terrace can be taken uniformly distributed load 600 kg / m 2, then the concentrated force from the terrace, acting on the central column, will be:
N from the terrace = 600 5 8/4 = 6000 kg or 6 tons
The own weight of columns 3 m long will be:
N per column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons
Thus, the total load on the middle lower column in the section of the column near the foundation will be:
N with about \u003d 3000 + 6000 + 2 650 \u003d 10300 kg or 10.3 tons
However, in this case it can be taken into account that there is not a very high probability that the temporary load from snow, maximum in winter time, and temporary load on the ceiling, maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:
N with about \u003d (3000 + 6000) 0.9 + 2 650 \u003d 9400 kg or 9.4 tons
The calculated load on the outer columns will be almost two times less:
N cr = 1500 + 3000 + 1300 = 5800 kg or 5.8 tons
2. Determination of the strength of brickwork.
The brand of brick M75 means that the brick must withstand a load of 75 kgf / cm 2, however, the strength of the brick and the strength brickwork- Different things. The following table will help you understand this:
Table 1. Calculated compressive strength for brickwork (according to SNiP II-22-81 (1995))
But that's not all. All the same SNiP II-22-81 (1995) p.3.11 a) recommends that if the area of pillars and piers is less than 0.3 m 2, multiply the value of the design resistance by working conditions coefficient γ s =0.8. And since the cross-sectional area of our column is 0.25x0.25 \u003d 0.0625 m 2, we will have to use this recommendation. As you can see, for a brick of the M75 brand, even when using masonry mortar M100 masonry strength will not exceed 15 kgf / cm 2. As a result, the calculated resistance for our column will be 15 0.8 = 12 kg / cm 2, then the maximum compressive stress will be:
10300/625 \u003d 16.48 kg / cm 2\u003e R \u003d 12 kgf / cm 2
Thus, to ensure the necessary strength of the column, it is necessary either to use a brick of greater strength, for example, M150 (the calculated compressive strength with a brand of mortar M100 will be 22 0.8 = 17.6 kg / cm 2) or increase the section of the column or use transverse reinforcement of the masonry. For now, let's focus on using a more durable face brick.
3. Determination of the stability of a brick column.
The strength of brickwork and the stability of a brick column are also different things and all the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:
N ≤ m g φRF (1.1)
Where m g- coefficient taking into account the influence of long-term load. In this case, relatively speaking, we are lucky, since at the height of the section h≈ 30 cm, the value of this coefficient can be taken equal to 1.
Note: Actually, with the coefficient m g, everything is not so simple, the details can be found in the comments to the article.
φ - coefficient of buckling, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l 0 , but it does not always coincide with the height of the column. The subtleties of determining the estimated length of the structure are set out separately, here we just note that according to SNiP II-22-81 (1995) p. 4.3: "The estimated heights of walls and pillars l 0 when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, one should take:
a) with fixed hinged supports l 0 = H;
b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l 0=1.5H, for multi-span buildings l 0=1.25H;
c) for free standing structures l 0 = 2N;
d) for structures with partially pinched support sections - taking into account the actual degree of pinching, but not less than l 0 = 0.8N, Where H- the distance between ceilings or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light.
At first glance, our calculation scheme can be considered as satisfying the conditions of paragraph b). i.e. you can take l 0 = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only if the lower support is really rigid. If the brick column will be laid out on a roofing material waterproofing layer laid on the foundation, then such a support should rather be considered as hinged, and not rigidly clamped. And in this case, our construction in a plane parallel to the plane of the wall is geometrically variable, since the construction of the ceiling (separately lying boards) does not provide sufficient rigidity in this plane. There are 4 ways out of this situation:
1. Apply a fundamentally different design scheme
For example - metal columns, rigidly embedded in the foundation, to which the crossbars of the floor will be welded, then, for aesthetic reasons, metal columns can be overlaid with face bricks of any brand, since the metal will carry the entire load. In this case, it is true that metal columns need to be calculated, but the estimated length can be taken l 0=1.25H.
2. Make another cover,
for example from sheet materials, which will allow us to consider both the upper and lower column supports as hinged, in this case l 0=H.
3. Make a hardness diaphragm
in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower column supports as hinged ones, but in this case it is necessary to additionally calculate the stiffness diaphragm.
4. Ignore the above options and count the columns as free-standing with a rigid bottom support, i.e. l 0 = 2N
In the end, the ancient Greeks set up their columns (although not of brick) without any knowledge of the resistance of materials, without the use of metal anchors, and even so carefully written out building codes and there were no rules in those days, nevertheless, some columns stand to this day.
Now, knowing the estimated length of the column, you can determine the coefficient of flexibility:
λ h =l 0 /h (1.2) or
λ i =l 0 /i (1.3)
Where h- the height or width of the section of the column, and i- radius of inertia.
In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the area of the section, and then extract from the result Square root, but in this case it is not really necessary. Thus λh = 2 300/25 = 24.
Now, knowing the value of the coefficient of flexibility, we can finally determine the coefficient of buckling from the table:
table 2. Buckling coefficients for masonry and reinforced masonry structures (according to SNiP II-22-81 (1995))
At the same time, the elastic characteristic of the masonry α determined by the table:
Table 3. Elastic characteristic of masonry α (according to SNiP II-22-81 (1995))
As a result, the value of the buckling coefficient will be about 0.6 (with the value of the elastic characteristic α = 1200, according to item 6). Then the maximum load on the central column will be:
N p \u003d m g φγ with RF \u003d 1x0.6x0.8x22x625 \u003d 6600 kg< N с об = 9400 кг
This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then in this way not only the cross-sectional area of \u200b\u200bthe column will increase to 0.13 m 2 or 1300 cm 2, but the radius of gyration of the column will also increase to i= 11.45 cm. Then λ i = 600/11.45 = 52.4, and the value of the coefficient φ = 0.8. In this case, the maximum load on the central column will be:
N p \u003d m g φγ with RF \u003d 1x0.8x0.8x22x1300 \u003d 18304 kg\u003e N with about \u003d 9400 kg
This means that a section of 38x38 cm is enough to ensure the stability of the lower central centrally compressed column with a margin, and even the brand of brick can be reduced. For example, with the originally adopted brand M75, the ultimate load will be:
N p \u003d m g φγ with RF \u003d 1x0.8x0.8x12x1300 \u003d 9984 kg\u003e N with about \u003d 9400 kg
It seems to be everything, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the section of columns 0.51x0.51 m will be the most optimal, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of such columns will be 2601 cm 2.
An example of calculating a brick column for stability under eccentric compression
The extreme columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then all the same, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the extreme columns not in the center of the column section. In which place the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the elastic moduli of the crossbars and columns, and a number of other factors, which are discussed in detail in the article "Calculation of the support section of the beam for collapse". This displacement is called the load application eccentricity e o. In this case, we are interested in the most unfavorable combination of factors, in which the floor load on the columns will be transferred as close as possible to the edge of the column. This means that, in addition to the load itself, the bending moment will also act on the columns, equal to M = Ne o, and this moment must be taken into account in the calculations. In general, stability testing can be performed using the following formula:
N = φRF - MF/W (2.1)
Where W- section modulus. In this case, the load for the lower extreme columns from the roof can be conditionally considered to be centrally applied, and the eccentricity will be created only by the load from the ceiling. With an eccentricity of 20 cm
N p \u003d φRF - MF / W \u003d1x0.8x0.8x12x2601- 3000 20 2601· 6/51 3 = 19975, 68 - 7058.82 = 12916.9 kg >N cr = 5800 kg
Thus, even with a very large load application eccentricity, we have more than a double margin of safety.
Note: SNiP II-22-81 (1995) "Stone and reinforced masonry structures" recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore I do not give the calculation method recommended by SNiP here.
Greetings to all readers! What should be the thickness of the brick exterior walls - 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 bricks solves the issues of building buildings and structures of almost any architectural form.
Starting to carry out the project, the design company calculates all structural elements - including the thickness of the brick outer walls.
The walls in the building perform various functions:
- If the walls are only a building envelope- in this case, they must comply with thermal insulation requirements in order to ensure a constant temperature and humidity microclimate, as well as have soundproofing qualities.
- load-bearing walls should 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, fire resistance.
Features of calculating the thickness of the walls
- 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 harsher 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 the “nonsense” turns out - the thickness of the masonry, determined thermotechnical calculation, often, according to the requirements of strength, it turns out to be excessive.
- Therefore, from the point of view of material costs and subject to 100% use of its strength, it is necessary to lay solid masonry of solid brick walls only 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 apply lightweight masonry.
- This does not apply to external walls in buildings where there is an increased percentage of humidity (for example, in laundries, baths). They are usually built with protective layer from a vapor barrier material from the inside and from a 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 \u003d 130 * n -10, where
B - wall thickness in millimeters
130 - the size of half a brick, taking into account the seam (vertical = 10mm)
n - integer half of the brick (= 120mm)
The value of continuous masonry obtained by calculation is rounded up to the nearest whole number of half-bricks.
Based on this, the following values (in mm) of brick walls are obtained:
- 120 (to the floor of a brick, but this is considered a partition);
- 250 (into one);
- 380 (one and a half);
- 510 (at two);
- 640 (in two and a half);
- 770 (at three o'clok).
In order to save material resources (brick, mortar, fittings, etc.), the number of machine hours of mechanisms, the calculation of wall thickness is tied to the bearing capacity of the building. And the thermotechnical component is obtained by insulating the facades of buildings.
How can you insulate the exterior walls of a brick building? In the article warming the house with polystyrene foam from the outside, I indicated the reasons why it is impossible to insulate brick walls 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 to the outside. That is why it is advisable to insulate a brick wall heat-insulating plaster or mineral wool boards, the nature of which is vapor-permeable. Expanded polystyrene is suitable for warming the base of concrete or reinforced concrete. "The nature of the insulation must correspond to the nature of the load-bearing wall."
Lots of heat insulating plasters- the difference lies in the components. But the principle of application is the same. It is done in layers and overall thickness can reach up to 150mm (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 an ordinary clay brick, depending on the area and the climatic conditions of the area, at the average winter ambient temperature, looks like this in millimeters:
- - 5 degrees - thickness = 250;
- - 10 degrees = 380;
- - 20 degrees = 510;
- - 30 degrees = 640.
I would like to summarize the above. The thickness of the outer walls of brick 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 exterior walls without the use of insulation. If the house is uncomfortable 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 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 performing the functions of 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 sight Brick wall very simple design, it has height, width and thickness. The heaviness 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 about the thickness of the brick wall? - you ask. Despite the fact that in construction, much 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, the thicker it must be, especially 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 he, in itself, is a 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 abandoned the idea of building houses resembling a bunker. In order to have strong bearing surfaces and at the same time good thermal insulation began to use a multilayer scheme. Where one layer acts silicate masonry, of sufficient weight to withstand all the loads to which it is subject, the second layer is an insulating material, and the third is a lining, 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 greater 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 ability for thermal insulation 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 under consideration 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 one we are all used to. Its dimensions are as follows: 250X120X65 mm.
One and a half or thickened - has a large weight, and its dimensions look like this: 250X120X88 mm. Double - respectively, has a cross section of two single blocks 250X120X138 mm.
The half one is a baby among its brethren, it has, as you probably already guessed, half the thickness of a single one - 250X120 X12 mm.
As you can see, the only differences in the size 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 building massive surfaces, for example, these are often load-bearing surfaces and smaller blocks for partitions.
wall thickness
We have already considered the parameters on which the thickness of the outer walls of brick depends. As we remember, this is stability, strength, thermal insulation properties. In addition, different types of surfaces should 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 affected external factors, such as winds, precipitation, in addition, their own weight presses on them. Therefore, their heaviness, compared with non-bearing surfaces and internal partitions, should be the highest.
In modern realities, for most two and three-story houses, 25 cm of thickness or one block is enough, less often one and a half or 38 cm. Such masonry will have enough strength 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 mortar 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 the description from paragraph 7 is also true for it. 26. After that, we look in table 28 and find the value of β, which means the allowable ratio of the weight 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 1.2 (k1=1.2).
- k2=√Аn/Аb where:
An - cross-sectional area of the bearing surface horizontally, the calculation is simple 0.25 * 5 \u003d 1.25 square meters. m
Ab is the horizontal cross-sectional area of the wall, taking into account the window openings, we do not have any, therefore k2 = 1.25
- The value of k4 is given, and for a height of 2.5 m it is equal to 0.9.
Now knowing 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 considered surface is 1.35*22=29.7, and the allowable ratio of height and thickness is 2.5:0.25=10, which is much 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 required by the norms of SNiP.
Well, we figured out the bearing surfaces, but 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 loads, the stability formula, which we discussed above, is also valid. But since from above, such a wall will not be fixed, the coefficient β must be reduced by a third, and the calculations should be continued with a different value.
Laying in half a brick, brick, one and a half, two bricks
In conclusion, let's look at how bricklaying is carried out depending on the heaviness of the surface. Laying in half a brick, the simplest of all, since there is no need to make complex dressings of rows. It is enough to put the first row of material on a perfectly even base and make sure that the solution lays down 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 dressing of vertical seams, which should not coincide. For this masonry option, it is important to follow the chosen system from beginning to end, of which there are at least two, single-row and multi-row. They differ in the way of dressing and laying blocks.
Before proceeding with the consideration of issues related to the calculation of the thickness of the brick wall of the house, it is necessary to understand what this is for. For example, why not build an outer wall half a brick thick, because the brick is so hard and durable?
Many non-specialists do not even have basic ideas about the characteristics of enclosing structures, however, they undertake independent construction.
In this article, we will consider 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 terms.
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 protective function, and also serve as supports for slabs or beams of a ceiling or roof structure. 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, where walls thicker than 1.5 bricks would be required, logically 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 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 (in one brick of the strength grade M50, M75, M100) will correspond to the calculations of bearing loads. It’s not worth it to play it safe, because the calculations already take into account snow, wind loads and many coefficients that provide a 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 are stacked on top of each other, the less stable the column of them becomes. The elementary laws of physics acting on cubes act in the same way on a brick wall, because the principle of laying is the same. Obviously, there is some relationship between the thickness of the wall and its height, which ensures the stability of the structure. That's what we'll talk about in the first half of this article.
Wall stability, as well as building standards for bearing and other loads, is described in detail in SNiP II-22-81 "Stone and reinforced masonry structures". These standards are a guide for designers, and for the "uninitiated" may seem rather difficult to understand. So it is, 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 possibilities of the information web, today almost everyone, if desired, can understand the most complex issues.
To begin with, let's try to understand the issue of 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 that's a lot of money today. To avoid the destruction of the wall or unnecessary financial expenses, let's turn to a mathematical calculation.
All the necessary data for calculating the stability of the wall are available in the relevant tables of SNiP II-22-81. On specific example consider how to determine whether the stability of an external load-bearing brick (M50) wall on a mortar M25 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 I-th masonry group 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 masonry mortar. The required parameter β is the ratio of the wall height 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 paragraph 6.20, the value of β should be reduced by 30%. Thus, β is no longer equal to 22, but 15.4.
We proceed to the definition of correction factors from table 29, which will help to find the cumulative factor k:
- for a wall with a thickness of 38 cm, not load-bearing, k1=1.2;
- k2=√Аn/Аb, where An is the area of the horizontal section of the wall, taking into account window openings, Аb - the area of the horizontal section, excluding windows. In our case, An= 0.38×6=2.28 m², and Ab=0.38×(6-1.2×2)=1.37 m². We perform the calculation: k2=√1.37/2.28=0.78;
- k4 for a 3 m high wall is 0.9.
By multiplying all the correction factors, we find the total 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 allowable ratio of the wall to the required parameters in our case is 12.98. Available ratio h/h= 3:0.38 = 7.89. It's less allowable relation 12.98, and this means that our wall will be quite stable, because. the condition H/h
According to paragraph 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 vast majority brick houses have a multilayer wall structure, consisting of lightweight brickwork, insulation and facade decoration. According to SNiP II-3-79 (Construction heating engineering), the outer walls of residential buildings with a need 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 particular 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 get the total thermal resistance of a multilayer building envelope, it is necessary to add thermal resistance all layers of the wall structure. Consider the following with a specific example.
The task is to determine how thick the wall of silicate brick so that its thermal conductivity resistance corresponds to SNiP II-3-79 for the lowest standard 1.2 m².°C/W. The thermal conductivity coefficient of silicate 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. Substitute the values and solve: δ \u003d 1.2 × 0.7 \u003d 0.84 m.
Now let's calculate with what layer of expanded polystyrene you need to insulate a wall of silicate brick with a thickness of 25 cm 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 brick 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) determine the thickness of expanded polystyrene 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 the wall in one brick to the standard thermal resistance (1.2 m². ° С / W), insulation with a layer of polystyrene foam 3.3 cm thick will be required.
Using this technique, you will be able to independently calculate the thermal resistance of the walls, taking into account the region of construction.
Modern residential construction claims high requirements to parameters such as strength, reliability and thermal protection. Exterior walls built of bricks have excellent bearing capacity, but have little heat-shielding properties. If you comply with 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 wall thickness
A building material such as brick has been used for construction for several hundred years. The material has standard sizes 250x12x65, regardless of the type. Determining what should be the thickness of a brick wall, they proceed from these classical parameters.
Bearing walls are a rigid frame of a structure that cannot be destroyed and re-planned, as the reliability and strength of the building are violated. Load-bearing walls can withstand enormous loads - this is the roof, ceilings, own weight and partitions. The most suitable and time-tested material for the construction of load-bearing walls is 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 strength.
Properly built bearing wall brick has a service life of more than one hundred years. For low-rise buildings apply solid brick with insulation or perforated.
Brick wall thickness parameters
Both external and internal walls. Inside the structure, the thickness of the wall must be at least 12 cm, that is, the floor of a brick. The cross section of the 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 edge". The thickness of a brick wall made by this method must be reinforced metal frame every 2 rows. Reinforcement will allow the walls to acquire additional strength and withstand more substantial loads.
The combined masonry method is very popular when the walls are made up of several layers. This decision allows to achieve greater reliability, strength and heat resistance. This wall includes:
- Brickwork consisting of porous or slotted material;
- Insulation - mineral wool or polystyrene;
- Cladding - panels, plaster, facing bricks.
outer thickness combined wall determined climatic conditions region and the type of insulation used. In fact, the wall may standard thickness, and thanks to the correctly selected insulation, all norms for the thermal protection of the building are achieved.
One brick wall
The most common masonry wall in one brick, makes it possible to obtain a wall thickness of 250 mm. Bricks in this masonry do not fit next to each other, since the wall will not have the desired 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 the correct dressing of vertical seams connecting materials. The brick from the top row must certainly overlap the bottom vertical seam. Such dressing significantly increases the strength of the structure and evenly distributes the load on the wall.
Types of dressings:
- Vertical seam;
- A transverse seam that does not allow materials to be shifted along the length;
- A longitudinal seam that prevents the bricks from moving horizontally.
The laying of a wall in one brick should be carried out according to a strictly chosen scheme - it is single-row or multi-row. In a single-row system, the first row of bricks is laid on the spoon side, the second on the bond side. The transverse seams are shifted by half the brick.
A multi-row system involves alternating through a row, and through several spoon rows. If a thickened brick is used, then the spoon rows are no more than five. This method ensures maximum structural strength.
The next row is laid in the opposite order, thus forming a mirror image of the first row. Such masonry has a special strength, since the vertical seams do not match anywhere and are overlapped by upper bricks.
If it is planned to create a masonry in two bricks, then, accordingly, the thickness of the wall 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 is one of the main building materials in low-rise construction. The main advantages of brickwork are strength, fire resistance, moisture resistance. Below we will give data on the consumption of bricks per 1 sq.m with different thicknesses of brickwork.
Currently, there are several ways to perform brickwork (standard brickwork, Lipetsk masonry, Moscow, etc.). But when calculating the consumption of bricks, the method of making brickwork is not important, the thickness of the brickwork and the size of the brick are important. Brick is produced various sizes, characteristics and purpose. The main typical brick sizes are 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 masonry, as a rule, the thickness of the vertical mortar joint is on average about 10 mm, the thickness of the horizontal joint is 12 mm. Brickwork It happens different thickness: 0.5 bricks, 1 brick, 1.5 bricks, 2 bricks, 2.5 bricks, etc. As an exception, there is brickwork in a quarter of a brick.
Quarter-brick masonry is used for small partitions that do not carry loads (for example, brick wall between bathroom and toilet). Half-brick brickwork is often used for one-story outbuildings(barn, toilet, etc.), gables of residential buildings. With one brick laying, you can build a garage. 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 storeys, type of floors, individual features buildings).
Based on the given data on the dimensions 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 a wall made of brickwork of various thicknesses.
Wall thickness and consumption of bricks with 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.
| 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 |
When self-design brick house there is an urgent need to calculate whether the brickwork can withstand the loads that are laid down in the project. A particularly serious situation develops in masonry areas weakened by window and doorways. In the event of a heavy load, these areas may not withstand and be destroyed.
The exact calculation of the resistance of the wall to compression by the overlying floors is quite complicated and is determined by the formulas laid down in normative document SNiP-2-22-81 (hereinafter reference -<1>). In engineering calculations of the compressive strength of a wall, many factors are taken into account, including the configuration of the wall, the compressive strength, the strength of a given type of material, and more. However, approximately, "by eye", you can estimate the resistance of the wall to compression, using the indicative tables, in which the strength (in tons) is linked depending on the width of the wall, as well as grades of brick and mortar. The table is compiled for a wall height of 2.8 m.
Brick wall strength table, tons (example)
| Stamps | Plot width, cm | |||||||||||
| brick | solution | 25 | 51 | 77 | 100 | 116 | 168 | 194 | 220 | 246 | 272 | 298 |
| 50 | 25 | 4 | 7 | 11 | 14 | 17 | 31 | 36 | 41 | 45 | 50 | 55 |
| 100 | 50 | 6 | 13 | 19 | 25 | 29 | 52 | 60 | 68 | 76 | 84 | 92 |
If the value of the width of the wall is in the range between the indicated ones, it is necessary to focus on minimum number. At the same time, it should be remembered that the tables do not take into account all the factors that can correct the stability, structural strength and resistance of the brick wall to compression in a fairly wide range.
In terms of time, loads are temporary and permanent.
Permanent:
- weight of elements of structures (weight of fences, load-bearing and other structures);
- soil and rock pressure;
- hydrostatic pressure.
Temporary:
- weight of temporary structures;
- loads from stationary systems and equipment;
- pressure in pipelines;
- loads from stored products and materials;
- climatic loads (snow, ice, wind, etc.);
- and many others.
When analyzing the loading of structures, it is necessary to take into account the total effects. Below is an example of calculating the main loads on the walls of the first floor of a building.
Loading brickwork
To take into account the force acting on the designed section of the wall, it is necessary to sum up the loads:
In the case of low-rise construction, the task is greatly simplified, and many live load factors can be neglected by setting a certain margin of safety at the design stage.
However, in the case of the construction of 3 or more storey structures, a thorough analysis is required using special formulas that take into account the addition of loads from each floor, the angle of application of force, and much more. In some cases, the strength of the pier is achieved by reinforcement.
Load Calculation Example
This example shows the analysis of the existing loads on the walls of the 1st floor. Here only constant acting load from various structural elements building, taking into account the uneven weight of the structure and the angle of application of forces.
Initial data for analysis:
- number of floors - 4 floors;
- brick wall thickness T = 64 cm (0.64 m);
- specific weight of masonry (brick, mortar, plaster) M = 18 kN / m3 (the indicator is taken from the reference data, Table 19<1>);
- the width of the window openings is: W1=1.5 m;
- height of window openings - B1 = 3 m;
- section of the wall 0.64 * 1.42 m (loaded area, where the weight of the overlying structural elements);
- floor height Vet=4.2 m (4200 mm):
- pressure is distributed at an angle of 45 degrees.
- Example of determining the load from the wall (plaster layer 2 cm)
Hst \u003d (3-4SH1V1) (h + 0.02) Myf \u003d (* 3-4 * 3 * 1.5) * (0.02 + 0.64) * 1.1 * 18 \u003d 0, 447 MN.
Width of the loaded area П=Вет*В1/2-Ш/2=3*4.2/2.0-0.64/2.0=6 m
Np \u003d (30 + 3 * 215) * 6 \u003d 4.072 MN
Nd \u003d (30 + 1.26 + 215 * 3) * 6 \u003d 4.094 MN
H2 \u003d 215 * 6 \u003d 1.290 MN,
including H2l=(1.26+215*3)*6= 3.878MN
- Self weight of piers
Npr \u003d (0.02 + 0.64) * (1.42 + 0.08) * 3 * 1.1 * 18 \u003d 0.0588 MN
The total load will be the result of a combination of the specified loads on the walls of the building, to calculate it, the summation of the loads from the wall, from the floors of the 2nd floor and the weight of the projected area is performed).
Scheme of analysis of load and structural strength
To calculate the pier of a brick wall, you will need:
- the length of the floor (it is also the height of the site) (Wat);
- number of floors (Chat);
- wall thickness (T);
- brick wall width (W);
- masonry parameters (type of brick, brand of brick, brand of mortar);
- Wall area (P)
- According to table 15<1>it is necessary to determine the coefficient a (elasticity characteristic). The coefficient depends on the type, brand of brick and mortar.
- Flexibility index (G)
- Depending on indicators a and D, according to table 18<1>you need to look at the bending factor f.
- Finding the height of the compressed part
where е0 is the extensibility index.
- Finding the area of the compressed part of the section
Pszh \u003d P * (1-2 e0 / T)
- Determination of the flexibility of the compressed part of the wall
Gszh=Vet/Vszh
- Definition according to the table. 18<1>coefficient fszh, based on Gszh and coefficient a.
- Calculation of the average coefficient fsr
Fsr=(f+fszh)/2
- Determination of the coefficient ω (table 19<1>)
ω =1+e/T<1,45
- Calculation of the force acting on the section
- Definition of sustainability
Y \u003d Kdv * fsr * R * Pszh * ω
Kdv - long-term exposure coefficient
R - resistance of masonry to compression, can be determined from table 2<1>, in MPa
- Reconciliation
Masonry Strength Calculation Example
- Wet - 3.3 m
- Chet - 2
- T - 640 mm
– W – 1300 mm
- masonry parameters (clay brick made by plastic pressing, cement-sand mortar, brick grade - 100, mortar grade - 50)
- Area (P)
P=0.64*1.3=0.832
- According to table 15<1>determine the coefficient a.
- Flexibility (G)
G \u003d 3.3 / 0.64 \u003d 5.156
- Bending factor (table 18<1>).
- Height of the compressed part
Vszh=0.64-2*0.045=0.55 m
- The area of the compressed part of the section
Pszh \u003d 0.832 * (1-2 * 0.045 / 0.64) \u003d 0.715
- Flexibility of the compressed part
Gf=3.3/0.55=6
- fsf=0.96
- Calculation of fsr
Fav=(0.98+0.96)/2=0.97
- According to the table 19<1>
ω=1+0.045/0.64=1.07<1,45
To determine the actual load, it is necessary to calculate the weight of all structural elements that affect the designed section of the building.
- Definition of sustainability
Y \u003d 1 * 0.97 * 1.5 * 0.715 * 1.07 \u003d 1.113 MN
- Reconciliation
The condition is met, the strength of the masonry and the strength of its elements is sufficient
Insufficient wall resistance
What to do if the calculated pressure resistance of the walls is not enough? In this case, it is necessary to strengthen the wall with reinforcement. Below is an example of an analysis of the necessary structural modifications in case of insufficient compressive strength.
For convenience, you can use tabular data.
The bottom line shows the values for a wall reinforced with wire mesh 3 mm in diameter, with a 3 cm cell, class B1. Reinforcement of every third row.
The increase in strength is about 40%. Usually this compression resistance is sufficient. It is better to make a detailed analysis by calculating the change in strength characteristics in accordance with the applied method of strengthening the structure.
Below is an example of such a calculation.
An example of calculating the reinforcement of piers
Initial data - see the previous example.
- floor height - 3.3 m;
- wall thickness - 0.640 m;
- masonry width 1,300 m;
- typical masonry characteristics (type of bricks - clay bricks made by pressing, type of mortar - cement with sand, brand of bricks - 100, mortar - 50)
In this case, the condition Y>=H is not satisfied (1.113<1,5).
It is required to increase the compressive strength and structural strength.
Gain
k=Y1/Y=1.5/1.113=1.348,
those. it is necessary to increase the strength of the structure by 34.8%.
Reinforcement of reinforced concrete clip
Reinforcement is made with a clip of concrete B15 with a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 with a step of 0.150 m.
Cross-sectional dimensions of the reinforced structure:
Ш_1=1300+2*60=1.42
Т_1=640+2*60=0.76
With such indicators, the condition Y>=H is fulfilled. Compressive strength and structural strength are sufficient.
It is required to determine the design bearing capacity of a building wall section with a rigid structural scheme *
Calculation of the bearing capacity of a section of the bearing wall of a building with a rigid structural scheme.
An estimated longitudinal force is applied to a section of a rectangular wall N= 165 kN (16.5 tf), from continuous loads N g= 150 kN (15 tf), short-term N st= 15 kN (1.5 tf). Section size - 0.40x1.00 m, floor height - 3 m, lower and upper wall supports - articulated, fixed. The wall was designed from four-layer blocks of design grade M50 strength, using mortar of design grade M50.
It is required to check the bearing capacity of the wall element in the middle of the floor height during the construction of the building in summer conditions.
In accordance with clause for load-bearing walls with a thickness of 0.40 m, random eccentricity should not be taken into account. We calculate according to the formula
N ≤ m g RA ,
Where N- calculated longitudinal force.
The calculation example given in this Appendix is made according to the formulas, tables and paragraphs of SNiP P-22-81 * (given in square brackets) and these Recommendations.
Sectional area of the element
A= 0.40 ∙ 1.0 = 0.40m.
Design compressive strength of masonry R according to Table 1 of these Recommendations, taking into account the coefficient of working conditions With\u003d 0.8, see paragraph , is equal to
R\u003d 9.2-0.8 \u003d 7.36 kgf / cm 2 (0.736 MPa).
The calculation example given in this Appendix is made according to the formulas, tables and paragraphs of SNiP P-22-81 * (given in square brackets) and these Recommendations.
The estimated length of the element according to the drawing, p. is equal to
l 0 = Η = 3 m.
The flexibility of the element is
.
Elastic characteristic of masonry , taken according to these "Recommendations", is equal to
Buckling ratio determined according to the table.
The coefficient taking into account the effect of long-term load with a wall thickness of 40 cm is taken m g = 1.
Coefficient for masonry of four-layer blocks is taken according to table. equal to 1.0.
Estimated bearing capacity of the wall section N cc is equal to
N cc= mg m g ∙ ∙R∙A∙ \u003d 1.0 ∙ 0.9125 ∙ 0.736 ∙ 10 3 ∙ 0.40 ∙ 1.0 \u003d 268.6 kN (26.86 tf).
Estimated longitudinal force N less N cc :
N= 165 kN< N cc= 268.6 kN.
Therefore, the wall satisfies the requirements for bearing capacity.
II example of calculating the resistance to heat transfer of the walls of buildings from four-layer heat-efficient blocks
Example. Determine the heat transfer resistance of a 400 mm thick wall of four-layer heat-efficient blocks. The inner surface of the wall from the side of the room is lined with plasterboard sheets.
The wall is designed for rooms with normal humidity and a moderate outdoor climate, the construction area is Moscow and the Moscow region.
When calculating, we accept masonry from four-layer blocks with layers having the following characteristics:
Inner layer - expanded clay concrete 150 mm thick, density 1800 kg / m 3 - \u003d 0.92 W / m ∙ 0 C;
The outer layer is porous expanded clay concrete 80 mm thick, with a density of 1800 kg / m 3 - \u003d 0.92 W / m ∙ 0 C;
Heat-insulating layer - polystyrene 170 mm thick, - 0.05 W/m ∙ 0 С;
Dry plaster from gypsum sheathing sheets 12 mm thick - \u003d 0.21 W / m ∙ 0 C.
The reduced resistance to heat transfer of the outer wall is calculated according to the main structural element, the most repeated in the building. The design of the building wall with the main structural element is shown in Fig. 2, 3. The required reduced resistance to heat transfer of the wall is determined according to SNiP 23-02-2003 "Thermal protection of buildings", based on the conditions of energy saving according to table 1b* for residential buildings.
For the conditions of Moscow and the Moscow region, the required resistance to heat transfer of the walls of buildings (stage II)
GSOP \u003d (20 + 3.6) ∙ 213 \u003d 5027 deg. day
Total resistance to heat transfer R o of the accepted wall design is determined by the formula
,(1)
Where 
And 
- heat transfer coefficients of the inner and outer surface of the wall,
accepted according to SNiP 23-2-2003 - 8.7 W / m 2 ∙ 0 С and 23 W / m 2 ∙ 0 С
respectively;
R 1 ,R 2 ...R n- thermal resistance of individual layers of block structures
n- layer thickness (m);
n- coefficient of thermal conductivity of the layer (W / m 2 ∙ 0 С)
\u003d 3.16 m 2 ∙ 0 C / W.
Determine the reduced heat transfer resistance of the wall R o without plaster inner layer.
R o
=
\u003d 0.115 + 0.163 + 3.4 + 0.087 + 0.043 \u003d 3.808 m 2 ∙ 0 C / W.
If it is necessary to apply an internal plaster layer of plasterboard sheets from the side of the room, the resistance to heat transfer of the wall increases by
R PC.
=
\u003d 0.571 m 2 ∙ 0 C / W.
The thermal resistance of the wall will be
R o\u003d 3.808 + 0.571 \u003d 4.379 m 2 ∙ 0 C / W.
Thus, the construction of the outer wall of four-layer heat-efficient blocks 400 mm thick with an internal plaster layer of gypsum boards 12 mm thick with a total thickness of 412 mm has a reduced heat transfer resistance equal to 4.38 m 2 ∙ 0 C / W meets the requirements for the heat-shielding qualities of outdoor enclosing structures of buildings in the climatic conditions of Moscow and the Moscow region.
Brick is a fairly strong building material, especially full-bodied, and when building houses of 2-3 floors, walls made of ordinary ceramic bricks usually do not need additional calculations. Nevertheless, the situations are different, for example, a two-story house with a terrace on the second floor is planned. The metal crossbars, on which the metal beams of the terrace floor will also rest, are planned to be supported on brick columns made of face hollow brick 3 meters high, there will be more columns 3 meters high, on which the roof will rest:
In this case, a natural question arises: what is the minimum section of the columns that will provide the required strength and stability? Of course, the idea of laying clay brick columns, and even more so the walls of the house, is far from new, and all possible aspects of the calculations of brick walls, walls, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced masonry structures". It is this normative document that should be followed in the calculations. The calculation below is nothing more than an example of using the specified SNiP.
To determine the strength and stability of the columns, you need to have a lot of initial data, such as: the brand of brick for strength, the area of support of the crossbars on the columns, the load on the columns, the sectional area of the column, and if none of this is known at the design stage, then you can do in the following way:
with central compression
Designed: Terrace with dimensions of 5x8 m. Three columns (one in the middle and two along the edges) made of facing hollow brick with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The brick strength grade is M75.
With such a design scheme, the maximum load will be on the middle lower column. It is she who should be counted on strength. The load on the column depends on many factors, in particular on the area of construction. For example, the snow load on the roof 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 actual loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but the reinforced concrete slab is not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculating the load from the terrace it is possible to 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 the terrace = 600 5 8/4 = 6000 kg or 6 tons
The own weight of columns 3 m long will be:
N from the column \u003d 1500 3 0.38 0.38 \u003d 649.8 kg or 0.65 tons
Thus, the total load on the middle lower column in the section of the column near the foundation will be:
N with about \u003d 3000 + 6000 + 2 650 \u003d 10300 kg or 10.3 tons
However, in this case, it can be taken into account that there is not a very high probability that the temporary load from snow, which is maximum in winter, and the temporary load on the floor, which is maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:
N with about \u003d (3000 + 6000) 0.9 + 2 650 \u003d 9400 kg or 9.4 tons
The calculated load on the outer columns will be almost two times less:
N kr \u003d 1500 + 3000 + 1300 \u003d 5800 kg or 5.8 tons
2. Determination of the strength of brickwork.
The brand of brick M75 means that the brick must withstand a load of 75 kgf / cm & sup2, however, the strength of the brick and the strength of the brickwork are two different things. The following table will help you understand this:
Table 1. Calculated compressive strengths for masonry
But that's not all. All the same SNiP II-22-81 (1995) p. 3.11 a) recommends that if the area of pillars and piers is less than 0.3 m2, multiply the value of the design resistance by the coefficient of working conditions γ c \u003d 0.8. And since the cross-sectional area of our column is 0.25x0.25 \u003d 0.0625 m & sup2, we will have to use this recommendation. As you can see, for a brick of the M75 brand, even when using the M100 masonry mortar, the strength of the masonry will not exceed 15 kgf / cm². As a result, the 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/cm² > R = 12 kgf/cm²
Thus, to ensure the necessary strength of the column, it is necessary either to use a brick of greater strength, for example, M150 (the calculated compressive strength with a brand of mortar M100 will be 22 0.8 = 17.6 kg / cm & sup2) or increase the section of the column or use transverse reinforcement of the masonry. For now, let's focus on using a more durable face brick.
3. Determination of the stability of a brick column.
The strength of brickwork and the stability of a brick column are also different things and all the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:
N ≤ m g φRF (1.1)
m g- coefficient taking into account the influence of long-term load. In this case, relatively speaking, we are lucky, since at the height of the section h≤ 30 cm, the value of this coefficient can be taken equal to 1.
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, but it does not always coincide with the height of the column. The subtleties of determining the estimated length of the structure are not set out here, we only note that according to SNiP II-22-81 (1995) p. 4.3: "The estimated heights of walls and pillars l o when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, one should take:
a) with fixed hinged supports l 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 pinched support sections - taking into account the actual degree of pinching, but not less than l o = 0.8N, Where H- the distance between ceilings or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light.
At first glance, our calculation scheme can be considered as satisfying the conditions of paragraph b). i.e. you can take l o = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only if the lower support is really rigid. If the brick column will be laid out on a roofing material waterproofing layer laid on the foundation, then such a support should rather be considered as hinged, and not rigidly clamped. And in this case, our construction in a plane parallel to the plane of the wall is geometrically variable, since the structure of the ceiling (separately lying boards) does not provide sufficient rigidity in this plane. There are 4 ways out of this situation:
1. Apply a fundamentally different design scheme, for example - metal columns rigidly embedded in the foundation, to which the floor crossbars will be welded, then, for aesthetic reasons, the metal columns can be overlaid with any brand of face brick, since the metal will carry the entire load. In this case, it is true that metal columns need to be calculated, but the estimated length can be taken l o = 1.25H.
2. Make another cover, for example, from sheet materials, which will allow us to consider both the upper and lower support of the column as hinged, in this case l o=H.
3. Make a hardness diaphragm in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower column supports as hinged ones, but in this case it is necessary to additionally calculate the stiffness diaphragm.
4. Ignore the above options and count the columns as free-standing with a rigid bottom support, i.e. l o = 2H. In the end, the ancient Greeks put up their columns (though not of brick) without any knowledge of the resistance of materials, without the use of metal anchors, and there were no such carefully written building codes in those days, nevertheless, some columns stand and to this day.
Now, knowing the estimated length of the column, you can determine the coefficient of flexibility:
λ h =l o /h (1.2) or
λ i =l o (1.3)
h- the height or width of the section of the column, and i- radius of inertia.
In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the area of the section, and then extract the square root from the result, but in this case this is not very necessary. Thus λh = 2 300/25 = 24.
Now, knowing the value of the coefficient of flexibility, we can finally determine the coefficient of buckling from the table:
table 2. Buckling coefficients for masonry and reinforced masonry structures
(according to SNiP II-22-81 (1995))
At the same time, the elastic characteristic of the masonry α determined by the table:
Table 3. Elastic characteristic of masonry α (according to SNiP II-22-81 (1995))
As a result, the value of the buckling coefficient will be about 0.6 (with the value of the elastic characteristic α = 1200, according to item 6). Then the maximum load on the central column will be:
N p \u003d m g φγ with RF \u003d 1 0.6 0.8 22 625 \u003d 6600 kg< N с об = 9400 кг
This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then in this way not only the cross-sectional area of \u200b\u200bthe column will increase to 0.13 m2 or 1300 cm2, but the radius of gyration of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the value of the coefficient φ = 0.8. In this case, the maximum load on the central column will be:
N p \u003d m g φγ with RF \u003d 1 0.8 0.8 22 1300 \u003d 18304 kg > N with vol \u003d 9400 kg
This means that a section of 38x38 cm is enough to ensure the stability of the lower central centrally compressed column with a margin, and even the brand of brick can be reduced. For example, with the originally adopted brand M75, the ultimate load will be:
N p \u003d m g φγ with RF \u003d 1 0.8 0.8 12 1300 \u003d 9984 kg\u003e N with about \u003d 9400 kg
It seems to be everything, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the section of columns 0.51x0.51 m will be the most optimal, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of such columns will be 2601 cm².
An example of calculating a brick column for stability
under eccentric compression
The extreme columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then all the same, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the extreme columns not in the center of the column section. Where exactly the resultant of this load will be transferred 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 load application eccentricity e o. In this case, we are interested in the most unfavorable combination of factors, in which the floor load on the columns will be transferred as close as possible to the edge of the column. This means that, in addition to the load itself, the bending moment will also act on the columns, equal to M = Ne o, and this moment must be taken into account in the calculations. In general, stability testing can be performed using the following formula:
N = φRF - MF/W (2.1)
W- section modulus. In this case, the load for the lower extreme columns from the roof can be conditionally considered to be centrally applied, and the eccentricity will be created only by the load from the ceiling. With an eccentricity of 20 cm
N p \u003d φRF - MF / W \u003d1 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 load application eccentricity, we have more than a double margin of safety.
Note: SNiP II-22-81 (1995) "Stone and reinforced masonry structures" recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore the calculation method recommended by SNiP is not given here.
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