The bearing capacity of the inner wall is one brick. Brick wall thickness
To calculate the stability of a wall, you first need to understand their classification (see SNiP II -22-81 "Stone and reinforced-stone structures", as well as a manual to SNiP) and understand what types of walls there are:
1. Load-bearing walls- these are walls on which floor slabs, roof structures, etc. are supported. The thickness of these walls must be at least 250 mm (for brickwork). These are the most important walls in the house. They must be counted on for strength and stability.
2. Self-supporting walls - these are walls on which nothing rests, but the load from all the overlying floors acts on them. In fact, in a three-story house, for example, such a wall would be three stories high; the load on it only from the own weight of the masonry is significant, but the question of the stability of such a wall is also very important - the higher the wall, the greater the risk of its deformations.
3. Curtain walls- these are external walls that rest on the floor (or on other structural elements) and the load on them falls from the height of the floor only from the own weight of the wall. The height of the curtain walls must be no more than 6 meters, otherwise they become self-supporting.
4. Partitions are internal walls with a height of less than 6 meters, taking only the load from their own weight.
Let's deal with the issue of wall stability.
The first question that arises from the "uninitiated" person: well, where can the wall go? Let's find the answer using an analogy. Take a hardcover book and place it on its edge. The larger the format of the book, the less its sustainability will be; on the other hand, the thicker the book, the better it will stand on the edge. The situation is the same with the walls. The stability of the wall depends on the height and thickness.
Now let's take the worst option: put a thin large-format notebook on its edge - it will not only lose stability, but also bend. So the wall, if the conditions for the ratio of thickness and height are not met, will begin to bend out of the plane, and over time - crack and collapse.
What is needed to avoid such a phenomenon? It is necessary to study paragraphs. 6.16 ... 6.20 SNiP II -22-81.
Consider the issues of determining the stability of walls using examples.
Example 1. Given is a partition made of M25 aerated concrete on a solution of M4 brand 3.5 m high, 200 mm thick, 6 m wide, not associated with overlap. In the partition, the doorway is 1x2.1 m. It is necessary to determine the stability of the partition.
From table 26 (p. 2) we determine the masonry group - III. From table 28 we find? = 14. Since the partition is not fixed in the upper section, it is necessary to reduce the β value 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 its 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 a partition with openings;
means k = k 1 k 3 = 1.4 * 0.9 = 1.26.
Finally, β = 1.26 * 9.8 = 12.3.
Let us 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 a thickness with a given geometry cannot be made.
How can you solve this problem? Let's try to increase the grade of the solution 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 constructive 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. Given is an external curtain wall made of lightweight brickwork of M50 grade on M25 grade mortar. The height of the wall is 3 m, the thickness is 0.38 m, the length of the wall is 6 m. A wall with two windows measuring 1.2 x 1.2 m. It is necessary to determine the stability of the wall.
From table 26 (p. 7) we determine the masonry group - I. From table 28 we find β = 22. the wall is not fixed in the upper section, the β value must be reduced by 30% (according to clause 6.20), i.e. β = 15.4.
We find the coefficients k from table s 29:
k 1 = 1.2 - for a wall, not a bearing load with a thickness of 38 cm;
k 2 = √А n / A b = √1.37 / 2.28 = 0.78 - for a wall with openings, where A b = 0.38 * 6 = 2.28 m 2 is the area of the horizontal section of the wall, taking into account windows, And n = 0.38 * (6-1.2 * 2) = 1.37 m 2;
means k = k 1 k 2 = 1.2 * 0.78 = 0.94.
Finally, β = 0.94 * 15.4 = 14.5.
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.
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0 # 212 Alexey 02/21/2018 07:08 AM
I quote Irina:
profiles will not replace fittings
I quote Irina:
about 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 bearing capacity. That is, there should be a thin layer below reinforced concrete.
And what kind of foundation - tape or slab? What are the soils?
Soils are not yet known, most likely there will be an open field of all kinds of loam, initially I thought it would be a slab, but it will come out low, I want it higher, and I also have to remove the upper fertile layer, so I tend to a ribbed or even box-shaped foundation. I don't need a lot of the bearing capacity of the soil - the house was still decided on the 1st floor, and the expanded clay concrete is not very heavy, freezing there is no more than 20 cm (although according to the old Soviet standards, 80).
I think to take off 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 in it, although I'm not sure), on top of the waterproofing with a primer
and then there is already a dilemma - even if you tie the reinforcement frames with a width of 150-200mm x 400-600mm in height and lay them in one meter increments, then you still need to form some voids 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 with a thin layer under a 60-100mm screed on top) - I think the PPS slabs should be monolithic as voids - theoretically it will be possible to pour this in 1 run with vibration.
Those. a slab of 400-600mm in appearance 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 (in unloaded places) - i.e. in terms of concrete and reinforcement consumption - it is quite comparable to a 200mm slab, but + a bunch of relatively cheap foam plastic and more work.
If you somehow still replace the foam with simple soil / sand, it will be even better, but then instead of light preparation it is wiser to do something more serious with reinforcement and the removal of reinforcement into the beams - in general, I lack both theory and practical experience.
0 # 214 Irina 02/22/2018 16:21
Quote:
why fight it? you just need to take into account in the calculation and when designing. You see, expanded clay concrete is good enough wall material with its own list of advantages and disadvantages. Like any other material. Now, if you wanted to use it for monolithic floor, I would discourage you, becauseIt is a pity, in general, they simply write that in lightweight concrete (expanded clay concrete) there is a poor connection with the reinforcement - how to deal with this? I understand what stronger concrete and what larger area reinforcement surfaces - the better the connection will be, i.e. you need expanded clay concrete with the addition of sand (and not only expanded clay and cement) and thin reinforcement, but more often
Quote:
Brick is strong enough construction material, especially full-bodied, and in the construction of houses in 2-3 floors, walls from an ordinary ceramic brick as a rule, they do not need additional calculations. Nevertheless, situations are different, for example, it is planned two-storey house with a terrace on the second floor. Metal crossbars, which will also be supported metal beams overlapping of the terrace, it is planned to lean on brick columns made of facing hollow bricks 3 meters high, there will be more columns 3 meters high, on which the roof will rest:
This raises a natural question: what is the minimum column cross-section that will provide the required strength and stability? Of course, the idea of laying out columns of clay bricks, and even more so the walls of a house, is far from new and all possible aspects of calculating brick walls, piers, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures". It is by this regulatory document and should be guided in the calculations. The calculation given below is nothing more than an example of using the specified SNiP.
To determine the strength and stability of columns, you need to have a lot of initial data, such as: a brick strength grade, the area of support of the crossbars on the columns, the load on the columns, the cross-sectional area of the column, and if at the design stage none of this is known, then you can do in the following way:
with central compression
Designed by: Terrace measuring 5x8 m. Three columns (one in the middle and two at the edges) of facing hollow bricks with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. Brick strength is M75.
With this design scheme, the maximum load will be on the middle bottom column. It is her that should be counted on for strength. The column load depends on many factors, in particular the area of construction. For example, the snow load on the roof in St. Petersburg is 180 kg / m & sup2, and in Rostov-on-Don - 80 kg / m & sup2. Taking into account the weight of the roof itself 50-75 kg / m & sup2, the load on the column from the roof for Pushkin Leningrad region can be:
N from the roof = (180 1.25 +75) 5 8/4 = 3000 kg or 3 tons
Since the acting loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab it is not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculating the load from the terrace, we can take a uniformly distributed load of 600 kg / m & sup2, then the concentrated force from the terrace acting on the central column will be:
N from the terrace = 600 5 8/4 = 6000 kg or 6 tons
The dead weight of the columns with a length of 3 m will be:
N from the column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons
Thus, the total load on the middle lower column in the column section near the foundation will be:
N with rev = 3000 + 6000 + 2 · 650 = 10300 kg or 10.3 tons
However, in this case, it can be taken into account that there is not a very high probability that the live load from snow, the maximum in winter time, and the live floor load, maximum in summer time, will be applied at the same time. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:
N with rev = (3000 + 6000) 0.9 + 2 650 = 9400 kg or 9.4 tons
The design load on the outer columns will be almost two times less:
N cr = 1500 + 3000 + 1300 = 5800 kg or 5.8 tons
2. Determination of the strength of brickwork.
The brick grade M75 means that the brick must withstand a load of 75 kgf / cm & sup2, however, the strength of the brick and the strength of the brickwork are different things. The following table will help you understand this:
Table 1... Calculated compressive strengths for masonry
But that's not all. All the same SNiP II-22-81 (1995) clause 3.11 a) recommends that, with the area of pillars and walls less than 0.3 m & sup2, multiply the value of the design resistance by the coefficient of working conditions γ c = 0.8... And since the cross-sectional area of our column is 0.25x0.25 = 0.0625 m & sup2, you will have to use this recommendation. As you can see, for brick grade M75, even when using masonry mortar M100 masonry strength will not exceed 15 kgf / cm & sup2. As a result, the calculated resistance for our column will be 15 0.8 = 12 kg / cm & sup2, then the maximum compressive stress will be:
10300/625 = 16.48 kg / cm & sup2> R = 12 kgf / cm & sup2
Thus, to ensure the required strength of the column, either use a brick of greater strength, for example, M150 (the calculated compressive resistance for the M100 solution grade will be 22 0.8 = 17.6 kg / cm2) or increase the column cross-section or use transverse reinforcement of the masonry. For now, let's focus on using a more durable facing brick.
3. Determination of the stability of a brick column.
The strength of the brickwork and the stability of the brick column are also different things and still the same SNiP II-22-81 (1995) recommends determining the stability of a brick column by the following formula:
N ≤ m g φRF (1.1)
m g- coefficient taking into account the effect of long-term load. In this case, we, relatively speaking, were lucky, since at a section height h≤ 30 cm, the value of this coefficient can be taken equal to 1.
φ - coefficient of buckling, depending on the flexibility of the column λ ... To determine this coefficient, you need to know the estimated length of the column l o, and it does not always coincide with the height of the column. The subtleties of determining the design length of the structure are not set out here, we just note that according to SNiP II-22-81 (1995) clause 4.3: "Design heights of walls and pillars l o when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, the following should be taken:
a) with fixed hinge 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 restrained support sections - taking into account the actual degree of restraint, but not less l o = 0.8H, where H- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light. "
At first glance, our design scheme can be considered as satisfying the conditions of item b). that is, you can take l o = 1.25H = 1.25 3 = 3.75 meters or 375 cm... However, we can confidently use this value only when the lower support is really rigid. If a brick column will be laid out on a layer of waterproofing made of roofing felt laid on the foundation, then such a support should rather be considered as hinged, and not rigidly pinched. And in this case, our structure in a plane parallel to the plane of the wall is geometrically variable, since the structure of the floor (separately lying boards) does not provide sufficient rigidity in the indicated plane. There are 4 ways out of this situation:
1. Apply a fundamentally different design scheme, for example - metal columns, rigidly embedded in the foundation, to which the floor girders will be welded, then, for aesthetic reasons, the metal columns can be overlaid with facing bricks of any brand, since the metal will bear the entire load. In this case, however, you need to calculate the metal columns, but the estimated length can be taken l o = 1.25H.
2. Make another overlap, for example from sheet materials, which will allow considering both the upper and lower support of the column as hinged, in this case l o = H.
3. Make diaphragm stiffness in a plane parallel to the plane of the wall. For example, lay not columns at the edges, but rather piers. This will also make it possible to consider both the upper and lower support of the column as articulated, but in this case it is necessary to additionally calculate the stiffness diaphragm.
4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l o = 2H... In the end, the ancient Greeks erected their columns (albeit not made of bricks) without any knowledge of the resistance of materials, without the use of metal anchors, and even so carefully written building codes and there were no rules in those days, nevertheless, some of the columns stand to this day.
Now, knowing the calculated length of the column, you can determine the slenderness factor:
λ h = l o / h (1.2) or
λ i = l o (1.3)
h- the height or width of the column section, and i- radius of gyration.
In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the sectional area, and then extract from the result Square root, however, in this case, this is not very necessary. Thus λ h = 2 300/25 = 24.
Now, knowing the value of the slenderness factor, we can finally determine the buckling factor from the table:
table 2... Buckling coefficients for stone and reinforced masonry structures
(according to SNiP II-22-81 (1995))
At the same time, the elastic characteristic of the masonry α determined by the table:
Table 3... Elastic characteristic of masonry α (according to SNiP II-22-81 (1995))
As a result, the value of the buckling coefficient will be about 0.6 (with the value of the elastic characteristic α = 1200, according to item 6). Then the ultimate load on the central column will be:
N p = m g φγ with RF = 1 0.6 0.8 22 625 = 6600 kg< N с об = 9400 кг
This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then this will not only increase the cross-sectional area of the column to 0.13 m & sup2 or 1300 cm & sup2, but the radius of inertia of the column will also increase to i= 11.45 cm... Then λ i = 600 / 11.45 = 52.4, and the value of the coefficient φ = 0.8... In this case, the ultimate load on the central column will be:
N p = m g φγ with RF = 1 0.8 0.8 22 1300 = 18304 kg> N with rev = 9400 kg
This means that the cross-sections of 38x38 cm are enough to ensure the stability of the lower central centrally compressed column with a margin, and it is even possible to reduce the brick grade. For example, with the originally adopted M75 grade, the maximum load will be:
N p = m g φγ with RF = 1 0.8 0.8 12 1300 = 9984 kg> N with rev = 9400 kg
It seems to be all, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, this section is optimal. The cross-sectional area of such columns will be 2601 cm & sup2.
An example of calculating a brick column for stability
with eccentric compression
The extreme columns in the projected house will not be centrally compressed, since the girders will rest on them only on one side. And even if the girders are laid on the entire column, still, due to the deflection of the girders, the load from the floor and the roof will be transferred to the extreme columns not in the center of the column section. In which place the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the elastic moduli of the crossbars and columns, and a number of other factors. This displacement is called the eccentricity of the load application eo. In this case, we are interested in the most unfavorable combination of factors, in which the load from the floor to the columns will be transmitted as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be affected by a bending moment equal to M = Ne o, and this point must be taken into account in the calculations. In general, stability testing can be performed using the following formula:
N = φRF - MF / W (2.1)
W- the moment of resistance of the section. In this case, the load for the lower extreme columns from the roof can be conventionally considered to be centrally applied, and the eccentricity will be created only by the load from the floor. With an eccentricity of 20 cm
N p = φRF - MF / W =1 0.8 0.8 12 2601- 3000 20 2601· 6/51 3 = 19975.68 - 7058.82 = 12916.9 kg>N cr = 5800 kg
Thus, even with a very large eccentricity of the load application, we have more than two times the safety margin.
Note: SNiP II-22-81 (1995) "Stone and reinforced masonry structures" recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore, the calculation method recommended by SNiP is not given here.
V.V. Gabrusenko
Design standards (SNiP II-22-81) allow to take minimum thickness carriers stone walls for masonry of the I group within the range from 1/20 to 1/25 of the floor height. With a floor height of up to 5 m, these restrictions fit well Brick wall thickness of only 250 mm (1 brick), which is what designers use - especially often lately.
In terms of formal requirements, designers are legitimate and vigorously resist when someone tries to hinder their intentions.
Meanwhile, thin walls react most strongly to all kinds of deviations from the design characteristics. Moreover, even for those that are officially permissible by the Norms of the rules for the production and acceptance of work (SNiP 3.03.01-87). Among them: deviations of the walls by the displacement of the axes (10 mm), by the thickness (15 mm), by the deviation of one floor from the vertical (10 mm), by the displacement of the floor slab supports in the plan (6 ... 8 mm), etc.
What these deviations lead to, let us consider an example inner wall 3.5 m high and 250 mm thick made of grade 100 brick on grade 75 mortar, bearing the design load from the ceiling of 10 kPa (slabs with a span of 6 m on both sides) and the weight of the overlying walls. The wall is designed for central compression. Its design bearing capacity, determined according to SNiP II-22-81, is 309 kN / m.
Let us assume that bottom wall offset from the axis by 10 mm to the left, and the top wall - 10 mm to the right (figure). In addition, the floor slabs are displaced 6 mm to the right of the axis. That is, the floor load N 1= 60 kN / m applied with an eccentricity of 16 mm, and the load from the overlying wall N 2- with an eccentricity of 20 mm, then the eccentricity of the resultant will be 19 mm. With such an eccentricity, the bearing capacity of the wall will decrease to 264 kN / m, i.e. by 15%. And this - in the presence of only two deviations and provided that the deviations do not exceed permissible by the Norms values.
If we add here the asymmetric loading of the floors by the temporary load (more on the right than on the left) and the "tolerances" that builders allow themselves - thickening of horizontal seams, traditionally poor filling of vertical seams, poor-quality dressing, curvature or inclination of the surface, "rejuvenation" of the mortar, excessive the use of half timber, etc., etc., then the bearing capacity may decrease by at least 20 ... 30%. As a result, the wall overload will exceed 50 ... 60%, after which an irreversible process of destruction begins. This process does not always manifest itself immediately, it happens - years after the completion of construction. Moreover, it should be borne in mind that the smaller the section (thickness) of the elements, the stronger the negative effect of overloads, since with decreasing thickness, the possibility of stress redistribution within the section due to plastic deformations of the masonry decreases.
If we add more uneven deformations of the foundations (due to soaking of soils), fraught with a turn of the base of the foundation, "hanging" of the outer walls on the internal load-bearing walls, the formation of cracks and a decrease in stability, then we will talk not just about overloading, but about a sudden collapse.
Thin-wall proponents may argue that all of this requires too much of a combination of defects and unfavorable deviations. Let us answer them: the overwhelming majority of accidents and disasters in construction occur precisely when several negative factors are gathered in one place and at the same time - in this case, "too many" of them do not happen.
conclusions
The load-bearing walls must be at least 1.5 bricks (380 mm) thick. Walls with a thickness of 1 brick (250 mm) may only be used for one-story or for last floors multi-storey buildings.
This requirement should be incorporated into future Territorial Design Codes. building structures and buildings, the need for the development of which is long overdue. In the meantime, we can only recommend that designers avoid using load-bearing walls with a thickness of less than 1.5 bricks.
It is required to determine the design bearing capacity of a section of the wall of a building with a rigid structural scheme *
Calculation of the bearing capacity of the site load-bearing wall buildings with a rigid structural scheme.
A calculated longitudinal force is applied to a section of a wall of a rectangular section 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 - hinged, fixed. The wall is designed from four-layer blocks of the design grade M50 strength, using the design grade M50 mortar.
It is required to check the bearing capacity of a wall element in the middle of the floor height when erecting a 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. The calculation is made according to the formula
N ≤ m g RA ,
where N is the calculated longitudinal force.
The calculation example given in this Appendix is made according to the formulas, tables and clauses of SNiP P-22-81 * (shown in square brackets) and these Recommendations.
Element section area
BUT= 0.40 ∙ 1.0 = 0.40m.
Design resistance to compression of masonry R according to table 1 of these Recommendations, taking into account the coefficient of working conditions with= 0.8, see p., Equals
R= 9.2-0.8 = 7.36 kgf / cm 2 (0.736 MPa).
The calculation example given in this Appendix is made according to the formulas, tables and clauses of SNiP P-22-81 * (shown in square brackets) and these Recommendations.
The estimated length of the element according to drawing, item is equal to
l 0 = Η = W m.
The flexibility of the element is
.
Elastic characteristic of masonry , taken according to these "Recommendations", is equal to
Buckling coefficient we determine according to the table.
The coefficient taking into account the effect of long-term load with a wall thickness of 40 cm is m g = 1.
Coefficient for masonry from four-layer blocks is taken according to table. equal to 1.0.
Estimated bearing capacity of a wall section N cc is equal to
N cc= mg m g ∙ ∙R∙A∙ = 1.0 ∙ 0.9125 ∙ 0.736 ∙ 10 3 ∙ 0.40 ∙ 1.0 = 268.6 kN (26.86 tf).
Calculated longitudinal force N less N cc :
N= 165 kN< N cc= 268.6 kN.
Consequently, the wall meets the load bearing requirements.
II example of calculating the resistance to heat transfer of walls of buildings from four-layer heat-efficient blocks
Example. Determine the resistance to heat transfer of a wall with a thickness of 400 mm from four-layer heat-efficient blocks. The inner surface of the wall from the side of the room is faced with plasterboard sheets.
The wall is designed for rooms with normal humidity and a moderate outdoor climate, construction area - Moscow and the Moscow region.
When calculating, we take a masonry of four-layer blocks with layers that have the following characteristics:
Internal layer - expanded clay concrete 150 mm thick, with a density of 1800 kg / m 3 - = 0.92 W / m ∙ 0 С;
The outer layer - porous expanded clay concrete 80 mm thick, with a density of 1800 kg / m 3 - = 0.92 W / m ∙ 0 С;
Thermal insulation layer - polystyrene 170 mm thick, - 0.05 W / m ∙ 0 С;
Dry plaster made of gypsum sheathing sheets 12 mm thick - = 0.21 W / m ∙ 0 С.
The reduced resistance to heat transfer of the outer wall is calculated according to the main structural element, which is the most repeatable in the building. The structure 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 energy saving conditions 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 = (20 + 3.6) ∙ 213 = 5027 deg. days
Total resistance to heat transfer R o the accepted wall structure is determined by the formula
,(1)
where 
and 
- coefficients of heat transfer of the inner and outer surface of the wall,
adopted 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 individual layers of block structures
n- layer thickness (m);
n- coefficient of thermal conductivity of the layer (W / m 2 ∙ 0 С)
= 3.16 m 2 ∙ 0 С / W.
Determine the reduced resistance to heat transfer of the wall R o without plaster inner layer.
R o
=
= 0.115 + 0.163 + 3.4 + 0.087 + 0.043 = 3.808 m 2 ∙ 0 C / W.
If it is necessary to use an internal plaster layer from the side of the room drywall sheets heat transfer resistance of the wall increases by
R PC.
=
= 0.571 m 2 ∙ 0 С / W.
The thermal resistance of the wall will be
R o= 3.808 + 0.571 = 4.379 m 2 ∙ 0 С / W.
Thus, the structure of the outer wall of four-layer heat-efficient blocks 400 mm thick with an inner plaster layer of gypsum plasterboard sheets 12 mm thick with a total thickness of 412 mm has a reduced heat transfer resistance equal to 4.38 m enclosing structures of buildings in the climatic conditions of Moscow and the Moscow region.
