It is required to determine the calculated carrying capacity of the building wall area with a rigid structural scheme *
Calculation of the bearing capacity of a building of the bearing wall of a building with a rigid structural circuit.
Calculated longitudinal force is applied to the section of the wall of the rectangular section. N.\u003d 165 kN (16.5 TC), from long loads N. g. \u003d 150 kN (15 TC), short-term N. st. \u003d 15 kN (1.5Ts). The size of the section is 0.40x1.00 m, the height of the floor is 3 m, the lower and the upper supports of the wall are hinged, fixed. The wall is designed from the four-layer blocks of the design grade for the strength of the M50, using the construction solution of the design grade M50.
It is required to check the carrier ability of the wall element in the middle of the height of the floor when building a building in summer conditions.
In accordance with paragraph for bearing walls, a thickness of 0.40 M, random eccentricity should not be considered. We carry out the calculation by the formula
N. ≤ m. g. RA. ,
where N. - Calculated longitudinal force.
An example of the calculation given in this annex is made by formulas, tables and clauses SNIP P-22-81 * (listed in square brackets) and these recommendations.
Section area of \u200b\u200belement
BUT \u003d 0.40 ∙ 1.0 \u003d 0.40m.
Estimated resistance to masonry compression R.table 1 of these recommendations, taking into account the coefficient of working conditions with \u003d 0.8, see p., Equal
R. \u003d 9.2-0.8 \u003d 7.36 kgf / cm 2 (0.736MPA).
An example of the calculation given in this annex is made by formulas, tables and clauses SNIP P-22-81 * (listed in square brackets) and these recommendations.
The calculated length of the element according to the features., P. Equal
l. 0 = Η \u003d Z m.
The flexibility of the element is equal
.
Elastic characteristic of masonry adopted according to "Recommendations" is equal
The coefficient of longitudinal bend determine the table.
The coefficient, taking into account the effect of long-term load with a thickness of 40 cm wall, accept m. g. = 1.
Coefficient For masonry from four-layer blocks is accepted by table. equal to 1.0.
Estimated bearing capacity of the wall N. cC. equal
N. cC. \u003d Mg. m. g. ∙ ∙R.∙A.∙ \u003d 1.0 ∙ 0.9125 ∙ 0.736 ∙ 10 3 ∙ 0.40 ∙ 1.0 \u003d 268.6 kN (26.86 TC).
Estimated longitudinal force N.less N. cC. :
N. \u003d 165 kN.< N. cC. \u003d 268.6 kN.
Consequently, the wall satisfies the requirements for the bearing capacity.
II An example of calculating the resistance of heat transfer walls of buildings from four-layer heat efficient blocks
Example. Determine the heat transfer resistance of the wall with a thickness of 400 mm of four-layer heat efficient blocks. The inner surface of the wall on the side of the room is facing with plasterboard sheets.
The wall is designed for premises with normal humidity and moderate outdoor climate, construction area - Moscow and Moscow region.
When calculating, we take masonry from four-layer blocks with layers having characteristics:
The inner layer is a ceramzite concrete thickness of 150 mm, a density of 1800 kg / m 3 - \u003d 0.92 W / m ∙ 0 s;
The outer layer is a picked ceramzite concrete with a thickness of 80 mm, a density of 1800 kg / m 3 - \u003d 0.92 W / m ∙ 0 s;
The heat insulating layer - polystyrene with a thickness of 170 mm, - 0.05 W / m ∙ 0 s;
Dry plastering of gypsum sheat sheets with a thickness of 12 mm - \u003d 0.21 W / m ∙ 0 S.
The reduced resistance of the heat transfer of the outer wall is calculated by the main constructive 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 resistance of the heat transfer of the wall is determined by SNiP 23-02-2003 "Thermal protection of buildings", based on the power saving conditions according to Table 1B * for residential buildings.
For the conditions of Moscow and the Moscow region, the required resistance of heat transfer walls of buildings (II stage)
HSOP \u003d (20 + 3.6) ∙ 213 \u003d 5027 degrees. SUT.
General resistance heat transfer R. o. The adopted wall design is determined by the formula
,(1)
where 
and 
- the heat transfer coefficients of the inner and outer surface of the wall,
accepted on SNIP 23-2-2003- 8.7 W / m 2 ∙ 0 s and 23 W / m 2 ∙ 0 s
respectively;
R. 1 ,R. 2 ...R. n. - Thermal resistance of individual layers of block designs
n. - layer thickness (m);
n. - coefficient of thermal conductivity of the layer (W / m 2 ∙ 0 s)
\u003d 3.16 m 2 ∙ 0 C / W.
Determine the resistance to the heat transfer wall R. o. Without plastering inner layer.
R. o.
=
\u003d 0.115 + 0.163 + 3.4 + 0.087 + 0.043 \u003d 3.808 m 2 ∙ 0 C / W.
If necessary, from the side of the indoor plastering layer, from drywall sheets of heat transfer resistance, the walls increases on
R. pC.
=
\u003d 0.571 m 2 ∙ 0 C / W.
Thermal resistance of the wall will be
R. o. \u003d 3.808 + 0.571 \u003d 4.379 m 2 ∙ 0 C / W.
Thus, the design of the outer wall of four-layer heat efficient blocks with a thickness of 400 mm with an inland plastering layer of plasterboard sheets with a thickness of 12 mm 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 the outer Fencing structures of buildings in the climatic conditions of Moscow and the Moscow region.
In the case of self-designing a brick house there is an urgent need to calculate whether the loads that are laid in the project can withstand the brickwork. A particularly serious situation consists of laying sites, weakened by window and doorways. In the event of a large load, these areas may not withstand and be destroyed.
The accurate calculation of the stability of the seal to compression by the overlying floors is quite complex and is determined by the formulas embedded in the SNIP-2-22-81 regulatory document (hereinafter referred to<1>). In engineering calculations of the strength of the wall to compression, many factors are taken into account, including the configuration of the wall, resistance to compression, the strength of this type of materials and much more. However, approximately, "on the eyes", you can estimate the resistance of the wall to compression, using the estimated tables in which the strength (in tons) is linked dependent on the wall width, as well as brick and mortar brands. The table is drawn up for the height of the wall of 2.8 m.
Table Strength of the Brick Wall, Tons (example)
| Brands | The width of the site, see | |||||||||||
| 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 |
In case the value of the width of the simpleness is in the interval between the specified, it is necessary to focus on the minimum number. At the same time, it should be remembered that there are not all factors that can adjust the stability, the strength of the design and the resistance of the brick wall to the compression in a fairly wide range.
By load time there are temporary and constant.
Permanent:
- weight of elements of structures (weight of fences, carrier and other structures);
- pressure of soils and rocks;
- 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, holly, wind, etc.);
- and many others.
When analyzing the loads of structures, the total effects should be taken into account. Below is an example of calculating the main loads on the simplest floor of the building.
Loading brick masonry
For accounting to the projected section of the wall of force, you need to summarize the load:
In the case of low-rise construction, the task is greatly simplified, and many temporary load factors can be neglected by asking a certain margin of safety at the design stage.
However, in the case of construction of 3 or more floors, careful analysis is needed on special formulas, taking into account the addition of loads from each floor, the angle of the application of force and much more. In some cases, the strength of the simpleness is achieved by reinforcement.
Example of load calculation
This example shows an analysis of the existing loads on the simplicity of the 1st floor. Here are only permanent loads from the various structural elements of the building, taking into account the uneven weight of the structure and an angle of the application of forces.
Source analysis data:
- number of floors - 4 floors;
- the thickness of the walls of bricks T \u003d 64cm (0.64 m);
- the proportion of masonry (brick, solution, plaster) m \u003d 18 kN / m3 (the indicator is taken from reference data, Table 19<1>);
- the width of the window openings is: sh1 \u003d 1.5 m;
- the height of the window openings - B1 \u003d 3 m;
- sexy section of 0.64 * 1.42 m (loaded area, where the weight of the overlying structural elements is applied);
- height of the floor vet \u003d 4.2 m (4200 mm):
- pressure is distributed at an angle of 45 degrees.
- An example of a load of the wall (layer of plaster 2 cm)
NST \u003d (3-4sh1V1) (H + 0.02) MyF \u003d (* 3-4 * 3 * 1.5) * (0.02 + 0.64) * 1.1 * 18 \u003d 0, 447MN.
The width of the loaded area P \u003d WET * B1 / 2-w / 2 \u003d 3 * 4.2 / 2.0-0.64 / 2.0 \u003d 6 m
NP \u003d (30 + 3 * 215) * 6 \u003d 4,072mn
ND \u003d (30 + 1.26 + 215 * 3) * 6 \u003d 4,094MN
H2 \u003d 215 * 6 \u003d 1,290mn,
including H2L \u003d (1.26 + 215 * 3) * 6 \u003d 3,878mn
- Own weight of simpleness
NPR \u003d (0.02 + 0.64) * (1.42 + 0.08) * 3 * 1.1 * 18 \u003d 0.0588 mn
The overall load will be the result of the combination of these loads on the simpleness of the building, for its calculation, the loads of loads from the wall are performed, from the overlaps of the 2 meter floor and the weight of the projected area).
Scheme of the load and strength analysis scheme
To calculate the simpleness of the brick wall, you will need:
- the length of the floor (it is the height of the site) (VET);
- number of floors (CET);
- wall thickness (T);
- brick wall width (w);
- masonry parameters (brick type, brick brand, solution brand);
- Square simplest (P)
- Table 15.<1> It is necessary to determine the coefficient A (characteristic of elasticity). The coefficient depends on the type, brick and mortar brand.
- Flexibility indicator (g)
- Depending on the indicators A and G, according to Table 18<1> You need to see the bend coefficient f.
- Finding the height of a compressed part
where E0 is an EXCRENSISITE indicator.
- Finding the area of \u200b\u200ba compressed part of the section
PSG \u003d P * (1-2 E0 / t)
- Determination of the flexibility of the compressed part of the simple
GSG \u003d WET / hard
- Table definition. 18<1> FSG coefficient based on GSG and coefficient a.
- Calculation of the averaged FSR coefficient
FSR \u003d (F + FSG) / 2
- Definition of the coefficient Ω (Table 19<1>)
ω \u003d 1 + e / t<1,45
- Calculation of force affecting the section
- Determination of stability
Y \u003d kdv * FSR * R * PSG * Ω
KDV - long-term coefficient
R - masonry resistance to compression, can be defined on table 2<1>, in MPa
- Recharge
Example of calculating masonry strength
- WET - 3.3 m
- CET - 2
- T - 640 mm
- W - 1300 mm
- Masonry parameters (clay brick made by plastic pressing, cement-sandy solution, brick brand - 100, Mark solution - 50)
- Area (P)
N \u003d 0,64 * 1,3 \u003d 0,832
- Table 15.<1> Determine the coefficient a.
- Flexibility (g)
R \u003d 3.3 / 0.64 \u003d 5,156
- Bending coefficient (Table 18<1>).
- Height of compressed part
Hard \u003d 0,64-2 * 0,045 \u003d 0.55 m
- Area of \u200b\u200bcompressed part of the section
PSG \u003d 0.832 * (1-2 * 0.045 / 0.64) \u003d 0.715
- Flexibility of compressed part
GSG \u003d 3.3 / 0.55 \u003d 6
- fSG \u003d 0.96
- FSR calculation
FSR \u003d (0.98 + 0.96) / 2 \u003d 0.97
- Table. 19<1>
ω \u003d 1 + 0.045 / 0.64 \u003d 1.07<1,45
To determine the current load, we need to calculate the weight of all structural elements that have an impact on the designed section of the building.
- Determination of stability
Y \u003d 1 * 0.97 * 1.5 * 0,715 * 1.07 \u003d 1,113 mn
- Recharge
The condition is fulfilled, the strength of the masonry and the strength of its elements is sufficient
Insufficient resistance of simpleness
What to do if the calculated resistance of the pressure of pressure is not enough? In this case, it is necessary to strengthen the wall with the help of reinforcement. Below is an example of the analysis of the desired modernization of the structure with insufficient compression resistance.
For convenience, you can use the tabular data.
The bottom line presents indicators for the wall, reinforced with a wire mesh with a diameter of 3 mm, with a cell 3 cm, class B1. Reinforcement of each third row.
The increase in strength is about 40%. Typically, this resistance to compression is sufficient. It is better to make a detailed analysis by calculating the change in the strength characteristics in accordance with the designed structural enhancement method.
Below is an example of such a calculation.
An example of calculating the amplification of common
Source data - see the previous example.
- the height of the floor is 3.3 m;
- wall thickness - 0.640 m;
- masonry width 1,300 m;
- typical Characteristics of Masonry (type of bricks - clay bricks made by pressing, type of solution - cement with sand, brand of bricks - 100, mortar - 50)
In this case, the condition in\u003e \u003d n is not performed (1,113<1,5).
It is required to increase the compression resistance and the strength of the structure.
Gain
k \u003d u1 / y \u003d 1.5 / 1,113 \u003d 1.348,
those. It is necessary to increase the strength of the structure by 34.8%.
Strengthening reinforced concrete clip
The reinforcement is made by a cable of 0.060 m befended by a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 in increments of 0.150 m.
Dimensions of the section of the enhanced design:
Sh_1 \u003d 1300 + 2 * 60 \u003d 1.42
T_1 \u003d 640 + 2 * 60 \u003d 0.76
With such indicators, the condition in\u003e \u003d n is performed. Resistance to compression and strength of the structure is sufficient.
The external bearing walls should be at least calculated for strength, stability, local crumpled and heat transfer resistance. To find out which thickness should be a brick wall , It is necessary to make its calculation. In this article, we will consider the calculation of the carrier ability of brickwork, and in the following articles - the remaining calculations. In order not to miss the output of the new article, subscribe to the newsletter and you will ounce which there should be a wall thickness after all calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, then we will consider all calculations for this category.
Carriers It is called walls that perceive the load from the slabs of overlapping, coatings, beams, etc.
You should also consider the brick stamp on frost resistance. Since everyone builds a house for itself, at least one hundred years, then with a dry and normal humidity mode of the premises, a brand (M RZ) is taken from 25 and higher.
During the construction of a house, cottage, garage, a host. Bruks and other systems with a dry and normal humidity regime, it is recommended to use hollow bricks for outer walls, as its thermal conductivity is lower than that of full-time. Accordingly, with a heat engineering calculation, the thickness of the insulation will be less, which will save money when purchasing it. Full-year bricks for external walls must be applied only if necessary to ensure the strength of the masonry.
Reinforcement of brick masonry It is allowed only if the increase in the brick and solution brand does not allow to provide the required carrying ability.
An example of calculating the brick wall.
The carrying ability of brickwork depends on many factors - from the brand of bricks, the grade of the solution, from the presence of openings and their size, from the flexibility of walls, etc. The calculation of the bearing capacity begins with the definition of the calculation scheme. When calculating walls on vertical loads, the wall is considered an operated on hinged-fixed supports. When calculating the walls on horizontal loads (wind), the wall is considered rigidly pinched. It is important not to confuse these schemes, since the moments will be different.
Selection of estimated section.
In the deaf walls for the calculated, the cross section of I-I at the level of overlapping with the longitudinal force N and the maximum bending moment M. is often dangerous section II-IISince the bending moment is slightly smaller than the maximum and equal to 2 / 3m, and the coefficients M G and φ are minimal.
In the walls with openings, the section is accepted at the bottom level of the jumpers.
Let's consider the cross section I-i.
From the past article Harvesting load on the wall of the first floor Take the resulting value of the full load, which includes loads from the overlap of the first floor P 1 \u003d 1.8T and the above floors G \u003d G P + P. 2 + G. 2 = 3.7T:
N \u003d G + P 1 \u003d 3.7T + 1.8T \u003d 5.5T
The slab overlap relies on the wall at a distance A \u003d 150mm. The longitudinal force P 1 from the overlap will be at a distance of a / 3 \u003d 150/3 \u003d 50 mm. Why 1/3? Because the stress plot under the support area will be in the form of a triangle, and the center of gravity of the triangle is just 1/3 of the length of the support.
The load from the overlying floors G is considered applied in the center.
Since the load from the ceiling slab (P 1) is applied not in the center of the section, but at a distance from it equal:
e \u003d H / 2 - A / 3 \u003d 250mm / 2 - 150mm / 3 \u003d 75 mm \u003d 7.5 cm,
it will create a bending moment (M) in cross section I - I. The moment is the work of strength on the shoulder.
M \u003d p 1 * e \u003d 1,800 * 7.5 cm \u003d 13.5 t * cm
Then the eccentricity of the longitudinal force N will be:
e 0 \u003d m / n \u003d 13.5 / 5.5 \u003d 2.5 cm
Since the carrier wall with a thickness of 25 cm, then calculated the value of the random eccentricity E ν \u003d 2 cm, then the total eccentricity is:
e 0 \u003d 2.5 + 2 \u003d 4.5 cm
y \u003d H / 2 \u003d 12.5 cm
At e 0 \u003d 4.5 cm< 0,7y=8,75 расчет по раскрытию трещин в швах кладки можно не производить.
The strength of the adki high-centranenially compressed element is determined by the formula:
N ≤ m g φ 1 r a c ω
Factors m G. and Φ 1. In the considered section of the I - i are equal to 1.
The article presents an example of calculating the carrying capacity of the brick wall of a three-story frameless building, taking into account the defects detected during its inspection. Such calculations refer to the category "verification" and are usually carried out within the framework of a detailed visual-instrumental examination of buildings.
The carrying ability of the central and high-centrian-compressed stone pillars is determined on the basis of data on the actual strength of brickwork materials (bricks, mortar) in accordance with section 4.
To take into account the defects identified during the survey, an additional lowering coefficient is introduced into formula, taking into account the reduction of the bearing ability of stone structures (CTR), depending on the nature and degree of damage to the challenges. four .
Example of calculation
We check the carrying capacity of the inner bearing stone wall of the 1st floor along the "8" axis of the M / O "B" - "B" on the operation of operational loads, taking into account the defects and damage identified during its survey.
Initial data:
- Wall thickness: dET \u003d 0.38 m
- Width of the Eastern: b \u003d 1.64 m
- The height of the sealer to the bottom of the slabs of overlap 1st floor: H \u003d 3.0 m
- The height of the clinch of the masonry: h \u003d 6.5 m
- Load collection area from overlapping and coating: SGR \u003d 9.32 m2
- Estimated masonry resistance to the compression: R \u003d 11.05 kg / cm2
During the inspection of the wall along the axis "8", the following defects and damage were recorded (see photo below): Mass loss of masonry seams at a depth of more than 4 cm; displacement (curvature) of horizontal rows of masonry vertically up to 3 cm; Multiple vertically oriented cracks with a disclosure of 2-4 mm (incl. for dissolved seams), crossing from 2 to 4 horizontal rows of masonry (up to 2 cracks per 1 m wall).
 |
 |
|
| Pustose | Cracking bricks | Curvature of rows of masonry |
By the aggregate of identified defects (taking into account their nature, degree of development and distribution area), in accordance with, the carrying ability of the simpleness under consideration should be reduced by at least 30%. Those. The coefficient of reducing the carrier ability of the simpleness is taken equal to - KTR \u003d 0.7. The scheme for collecting loads on the simpleness is shown below in Fig.1.
Fig.1. Scheme for gathering loads on simple
I. Calculation of settlement loads on common
II. Calculation of the carrier ability of common
(p. 4.1 SNiP II-22-81)
Quantitative assessment of the actual bearing capacity of a brick centrally compressed simpleness (taking into account the influence of detected defects) on the action of the estimated longitudinal force n applied without eccentricity, is reduced to checking the following condition (Formula 10):
Ns \u003d mg × φ × r × a × ktr ≥ n(1)
According to the results of the strength tests, the calculated resistance of the wall masonry along the "8" axis compression is R \u003d 11.05 kg / cm2.
The elastic characteristic of the masonry according to P.9 Table 15 (K) is equal to: α \u003d 500.
Estimated pillar height: l0 \u003d 0.8 × H \u003d 0.8 × 300 \u003d 240 cm.
Flexibility of the element of the rectangular solid section: λh \u003d L0 / DST \u003d 240/38 \u003d 6.31.
The coefficient of longitudinal bend φ
for α \u003d 500. and λh \u003d 6,31 (Table 18): φ \u003d 0.90.
Cross cross section area (simple): A \u003d b × dst \u003d 164 × 38 \u003d 6232 cm2.
Because The thickness of the calculated wall is more than 30 cm (dst \u003d 38 cm), coefficient mG. It is accepted equal to unity: mg \u003d 1.
Substituting the obtained values \u200b\u200binto the left part of formula (1), we define the actual carrier ability of the central compressed unarmed brick simpleness NS.:
NC \u003d 1 × 0.9 × 11.05 × 6232 × 0.7 \u003d 43 384 kgf
III. Checking the performance of strength (1)
[NC \u003d 43384 kgf]\u003e [n \u003d 36340,5 kgf]
The condition of strength is fulfilled: Brick pillar bearing NS. Taking into account the influence of the identified defects, the total load value was N..
List of sources:
1. SNIP II-22-81 * "Stone and Armocama Structures".
2. Recommendations for strengthening the stone structures of buildings and structures. TsNII. Kurchenko, Gosstroy.
Greetings all readers! What should be the thickness of the brick outer 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 brick solves the creation of buildings and structures of almost any architectural form.
Starting the project, the project firm produces the calculation of all structural elements - including the thickness of brick outer walls is calculated.
Walls in the building perform various functions:
- If the walls are only enclosing design - In this case, they must comply with thermal insulation requirements in order to provide a permanent temperature and humidity microclimate, and also have soundproofing qualities.
- Bearing walls Must be different with the necessary strength and stability, but also as enclosing, have heat shielding properties. In addition, based on the appointment of the construction, its class, the thickness of the bearing walls must correspond to the technical indicators of its durability, fire resistance.
Features of the calculation of wall thickness
- The thickness of the walls on the heat engineering count does not always coincide with the calculation of the values \u200b\u200bof the strength characteristics. Naturally, that the harsh climate, the thicker should be a wall at heat-technical indicators.
- But according to strength conditions, for example, it is enough to lay out the outer walls in one brick or one and a half. Here, it turns out "nonsense" - the thickness of the masonry, determined by the heat engineering, often, according to the requirements of strength, it turns out unnecessary.
- Therefore, put the solid masonry of the walls of full-scale brick from the point of view of material costs and, provided that 100% of its strength is used only in the lower floors of high-rise buildings.
- In low-rise buildings, as well as in the upper floors, high-rise buildings should be used for outer masonry hollow or light brick, you can apply a lightweight masonry.
- It does not apply to the outer walls in buildings, where there is an increased percentage of humidity (for example, in laundry, baths). They erected, usually, with a protective layer of vapor barrier material from the inside and from a full-skinned clay material.
Now I will tell you about what count is the thickness of the outer walls.
It is determined by the formula:
B \u003d 130 * N -10, where
B - Wall thickness in millimeters
130 - Size of half bricks taking into account seam (vertical \u003d 10mm)
n - integer halves of bricks (\u003d 120mm)
The amount obtained by calculating the size of the solid masonry is rounded to an integer number of half-winged.
Based on this, the following values \u200b\u200b(in mm) of brick walls are obtained:
- 120 (in the floor of the brick, but it is considered a partition);
- 250 (into one);
- 380 (in one and a half);
- 510 (at two);
- 640 (in two and a half);
- 770 (at three o'clok).
In order to save material resources (bricks, mortar, reinforcement and other things), the number of machines - hours of mechanisms, the counting of wall thickness is attached to the bearing capacity of the building. And the thermal component is obtained due to the insulation of the facades of buildings.
How can I insulate the outer walls of the brick building? In the article insulation of the house expanded by polystyrene foam outside, I pointed out the reasons why it is impossible to insulate brick walls with this material. Check out the article.
The meaning is that the brick is porous and the water permeable material. And the absorption capacity of the polystyrene is zero, which prevents moisture migration to the outside. That is why the wall of the brick is advisable to warm the thermal insulating plaster or mineral wool slabs, the nature of which is varying. The polystyrene foam is suitable for insulation of the base of concrete or reinforced concrete. "The nature of the insulation must correspond to the nature of the bearing wall."
Many heat insulating plasters - The difference is the components. But the principle of applying one. It is performed by layers and the total thickness can reach 150 mm (with a large amount of reinforcement). In most cases, this value is 50 - 80 mm. It depends on the climatic belt, the thickness of the walls of the base, other factors. I will not stop in detail, as this is the topic of another article. We return to your bricks.
The average wall thickness for ordinary clay brick, depending on the area and climatic conditions of the area, with the winter average of the current ambient temperature looks in millimeters like this:
- - 5Gradusov - thickness \u003d 250;
- - 10Gradusov \u003d 380;
- - 20graduts \u003d 510;
- - 30 degrees \u003d 640.
I want to summarize the above. The thickness of the outer walls of the brick is calculated based on the strength characteristics, and the heat engineering side of the issue solve the method of insulation of the walls. As a rule, the project firm calculates the outer walls without the use of insulation. If the house will be uncomfortable and the need for insulation will arise, then carefully take care of the sector's selection.
During the construction of his house, one of the main points is the construction of walls. The laying of carrier 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 carriers, but still performing partitions and facing - what should be the thickness of the brick wall in these cases? About this, I will tell in today's article.
This question is very relevant for all people who build their own brick house and only comprehend the aces of construction. At first glance, the brick wall is a very simple design, it has a height, width and thickness. The wall load of interest to us is primarily depends on its final total area. That is, the wider and above the wall, the thicker it should be.
But, where is 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 GOST, which takes into account its strength. Also, the loading of masonry depends on its stability. The already the above will be the carrier surface, the though it is obliged to be, especially the foundation.
Another parameter that affects the overall loadability of the surface is the thermal conductivity of the material. The ordinary full-scale block thermal conductivity is quite high. This means that he, in itself, bad thermal insulation. Therefore, to exit 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 maintained common sense, people refused to build houses reminiscent of the bunker. In order to have durable carrying surfaces and at the same time good thermal insulation, the multilayer scheme began to apply. Where a single layer is a silicate masonry, sufficient loading, to withstand all the loads that it is subject to, the second layer is a warming material, and the third is a cladding that brick can also act.
Choosing bricks
Depending on which it should be, you need to choose a certain type of material having different dimensions and even the structure. So, according to the structure, they can be divided into full-scale and holes. Full-time materials have greater strength, cost, and thermal conductivity.
Building material with cavities inside in the form of through holes is not as durable, has a smaller cost, but the ability to thermal insulation in the hole block is higher. This is achieved by the presence of air pockets in it.
The dimensions of any species of the material under consideration may also be varied. He might be:
- Single;
- Half;
- Double;
- Half.
Single block, it is building material, standard sizes, such to whom we are all used to. Its sizes are as follows: 250x120x65 mm.
One-hour or thickened - has a greater loading, and its size looks like this: 250x120x88 mm. Double - respectively, has a section of two single blocks 250x120x138 mm.
A half is a baby among his fellow, he has, as you probably have already guessed, half the thickness of the single - 250x120 x12 mm.
As can be seen, the only differences in the sizes of this building material in its thickness, and the length and width are the same.
Depending on what will be the thickness of the brick wall, economically appropriate, choose larger when erecting massive surfaces, for example, there are often carrier surfaces and smaller blocks for partitions.
Wall thickness
We have already considered the parameters on which the thickness of the outer walls of the brick depends. As we remember, it is stability, strength, thermal insulation properties. In addition, different types of surfaces should have a completely different dimension.
The carrier surfaces are, in fact, the support of the whole building, they take on the main burden, from the whole structure, including the weight of the roof, the external factors, such as wind, precipitation are also influenced by their own weight. Therefore, their loadability, compared with the surfaces of an undesuctive nature and internal partitions, should be highest.
In modern realities, most of the two and three-storey houses are sufficient to 25 cm thick or one block, less often a half or 38 cm. These masonry will be enough for the building of such sizes, but how to be resistant. Everything is much more complicated here.
In order to calculate whether sustainability will be sufficient to refer to the standards of SNIP II-22-8. Let's calculate whether our brick house will be resistant, with walls with a thickness of 250 mm, 5 meters long and 2.5 meters high. For masonry we will use the M50 material, on the M25 solution, we will calculate for one carrier surface, without windows. So, proceed.
Table number 26.
According to the data from the table above, we know that the characteristic of our masonry refers to the first group, as well as a description of it from clause 7. Tab. 26. After that, we look at Table 28 and find the value of β, which means the allowable ratio of the load of the wall to its height, given, the type of solution used. For our example, this value is 22.
- k1 for the cross section of our masonry is 1.2 (k1 \u003d 1.2).
- k2 \u003d √An / AB where:
An is the cross-sectional area of \u200b\u200bthe carrier surface horizontally, the calculation is simple 0.25 * 5 \u003d 1.25 square meters. M.
AB - the area of \u200b\u200bthe sections of the wall horizontally, given the window openings, we are missing, therefore, K2 \u003d 1.25
- The value of K4 is specified, and for a height of 2.5 m is 0.9.
Now learning, all variables can be found a general coefficient "k", by multiplying all values. K \u003d 1.2 * 1.25 * 0.9 \u003d 1.35 Next, we learn the cumulative value of the correction coefficients and in fact we learn how much the resistant surface is 1.35 * 22 \u003d 29.7, and the allowable ratio of height and thickness is 2.5: 0.25 \u003d 10, which is significantly less than the resulting indicator 29.7. This means that the laying of a thickness of 25 cm with a width of 5 m and a 2.5 meter high height has a stability almost three times higher than it is necessary on the standp standards.
Well with the supporting surfaces figured out, and what with partitions and with those that do not bear the load. Partitions, it is advisable to take half the thickness - 12 cm. For surfaces that do not carry loads, the stability formula that we have considered above is also fair. But since from above, such a wall will not be fixed, the ratio of the β coefficient must be reduced by a third, and continue the calculations with the other value.
Masonry in Pollipich, brick, one and a half, two bricks
In conclusion, let's look at how the brickwork is performed depending on the loadability of the surface. Masonry in Polkirpich, the most simple of all, as there is no need to make complex dressing of the ranks. Enough, put the first series of material, on the perfectly level base and ensure that the solution is evenly laid, and did not exceed 10 mm in the thickness.
The main criterion of high-quality masonry with a cross section of 25 cm is the implementation of high-quality dressing of vertical seams that should not coincide. For this option, masonry is important from beginning to end the selected system, which have at least two, single-row and multi-row. They differ in the way of dressing and laying blocks.
Before proceeding to consider issues related to the calculation of the brick wall thickness of the house, it is necessary to understand what it is necessary. For example, why it is impossible to build an outer wall thick in Polkirpich, because the brick is so solid and durable?
Very many nonspecialists do not even have basic ideas about the characteristics of the enclosing structures, nevertheless, are taken for independent construction.
In this article, we will look at two main criteria for calculating the thickness of brick walls - carrier loads and heat transfer resistance. But before immersed in boring figures and formulas, let me explain some moments in simple language.
The walls of the house, depending on their place in the project scheme, can be carriers, self-supporting, nonsense and partitions. Bearing walls are performed by a protective function, and also serve as supports with plates or beams of overlapping or roof design. The thickness of the bearing brick walls can not be less than one brick (250 mm). Most modern homes are built with walls in one or 1.5 bricks. Projects of private houses, where the walls need the thickness of 1.5 bricks, according to the logic of things should not exist. Therefore, the choice of the thickness of the outer brick wall by and large is solved. If you choose between a thickness of one brick or one and a half, then with a purely technical point of view for a cottage height of 1-2 floors a brick wall with a thickness of 250 mm (in one brick of the strength of the M50, M75, M100) will correspond to the calculations of carrier loads. It is not worth the reinsure, because the calculations already take into account snow, wind loads and many coefficients that provide a brick wall sufficient margin of safety. However, there is a very important point truly affecting the thickness of the brick wall - stability.
Everyone once played cubes in childhood, and noticed that the more putting the cubes on each other, the less stable the column of them becomes. The elementary laws of physics acting on the cubes are also acting on the brick wall, for the principle of masonry is the same. Obviously, there is some dependence between the wall thickness and its height, which ensures the stability of the structure. Here we will talk about this dependence in the first half of this article.
Stability of wallsAs well as building standards of carriers and other loads, it is described in detail in SNIP II-22-81 "Stone and Armocatament Designs". These standards are a manual for designers, and for the "uninitiated" may seem rather difficult to understand. So it is, because to become an engineer, it is necessary to learn at least four years. Here it would be possible to refer to "contact the calculations to the specialists" and put the point. However, due to the possibilities of the information cobweb, today almost everyone may understand the most difficult issues.
To begin with, let's try to figure out the stability of the brick wall. If the wall is high and long, then the thickness in one brick will be little. At the same time, excessive reinsurance can increase the cost of the box 1.5-2 times. And this is considerable money today. To avoid the destruction of the wall or unnecessary financial spending, we turn to the mathematical calculation.
All necessary data for calculating the stability of the wall are available in the relevant SNIP II-22-81 tables. In a specific example, we consider how to determine whether the stability of the outer carrier brick (M50) walls on the M25 solution of 1.5 bricks (0.38 m) is sufficient (0.38 m), 3 m height and 6 m long with two window openings 1.2 × 1 2 m.
Turning to Table 26 (Table at the top), we find that our wall refers to the I-th masonry group and is suitable for a description of paragraph 7 of this table. Further, we need to know the permissible ratio of the height of the wall to its thickness, taking into account the brand of a masonry solution. The desired parameter β is the ratio of the height of the wall to its thickness (β \u003d n / h). In accordance with the data Table. 28 β \u003d 22. However, our wall is not fixed in the upper section (otherwise the calculation was required only by strength), therefore, according to claim 6.20, the value of β should be reduced by 30%. Thus, β is no longer 22, but 15.4.
Go to the definition of correction coefficients from table 29, which will help to find a cumulative coefficient k.:
- for a wall with a thickness of 38 cm, not carrier load, K1 \u003d 1.2;
- k2 \u003d √An / AB, where AN is the area of \u200b\u200bthe horizontal section of the wall, taking into account the window openings, AB is the area of \u200b\u200bhorizontal section without taking into account windows. In our case, An \u003d 0.38 × 6 \u003d 2.28 m², and AB \u003d 0.38 × (6-1.2 × 2) \u003d 1.37 m². Perform calculation: k2 \u003d √1.37 / 2.28 \u003d 0.78;
- k4 for a wall with a height of 3 m is 0.9.
By multiplying all correction coefficients, we find the total coefficient k \u003d 1.2 × 0.78 × 0.9 \u003d 0.84. After taking into account the aggregate of the correction coefficients β \u003d 0.84 × 15,4 \u003d 12.93. This means that the allowable ratio of the wall with the required parameters in our case is 12.98. Available ratio H / H. \u003d 3: 0.38 \u003d 7.89. This is less than the permissible ratio of 12.98, and this means that our wall will be sustainable enough, because H / H condition is performed
According to clause 6.19, another condition must be respected: the sum of height and length ( H.+L.) The walls should be less than the product 3kβh. Substituting the values, we get 3 + 6 \u003d 9
The thickness of the brick wall and heat transfer resistance rate
Today, the overwhelming number of brick houses have a multilayer design of walls consisting of lightweight brickwork, insulation and facade finish. According to SNiP II-3-79 (construction heat engineering) outer walls of residential buildings with the 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 counts, we offer the following table, where the required thermal resistance of the walls is shown for a number of cities of Russia located in different construction and climatic zones according to SNiP II-3-79 and SP-41-99.
Resistance heat transfer R. (Thermal resistance, m². ° C / W) The layer of the enclosing structure is determined by the formula:
R.=δ /λ where
δ - layer thickness (m), λ - The thermal conductivity coefficient of W / (m. ° C).
To obtain the overall thermal resistance of a multi-layer enclosing structure, it is necessary to add thermal resistance of all layers of the wall structure. Consider the following on a specific example.
The task is to determine which thickness should be in a 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. Suppose our material has a thermal conductivity coefficient of 0.7. Thus, we obtain an equation with one unknown Δ \u003d Rλ.. We substitute the values \u200b\u200band decide: δ \u003d 1.2 × 0.7 \u003d 0.84 m.
Now we calculate how the layer of polystyrene is needed to insulate the wall of silicate brick with a thickness of 25 cm to exit the figure of 1.2 m². ° C / W. The coefficient of thermal conductivity of polystyrene foam (PSB 25) is not more than 0.039 W / (m. ° C), and in silicate brick 0.7 W / (m. ° C).
1) Determine R. Brick layer: R.=0,25:0,7=0,35;
2) Calculate the missing thermal resistance: 1.2-0.35 \u003d 0.85;
3) Determine the thickness of the polystyrene foam, necessary to obtain thermal resistance of 0.85 m². ° C / W: 0.85 × 0.039 \u003d 0.033 m.
The way it has been established that to bring the wall into one brick to the regulatory thermal resistance (1.2 m². ° C / W), it is necessary to insulation with a layer of expanded polystyrene with a thickness of 3.3 cm.
Using this technique, you can independently calculate the thermal resistance of the walls with regard to the construction region.
Modern residential construction declares high requirements for such parameters as durability, reliability and heat protection. The outer walls are built of bricks have excellent supporting abilities, but have small heat-shielding properties. If you comply with the regulations on the heat of the brick wall, then its thickness should be at least three meters - and this is simply not real.
Brick bearing wall thickness
Such a building material as a brick is used to build several hundred years. The material has standard dimensions of 250x2x65, regardless of the type. Determining what should be the thickness of the brick wall proceed from these classical parameters.
Bearing walls are a rigid framework frame, which cannot be chopped and re-posted, as the reliability and strength of the building are disturbed. Bearing walls are withstanding colossal loads - this is the roof, overlap, its own weight and partition. The most suitable and time-tested material for the construction of bearing walls is precisely brick. The thickness of the carrier wall should be at least one brick, or in other words - 25 cm. Such a wall has distinctive thermal insulation characteristics and durability.
A properly built carrier brick wall has a service life of not one hundred years. For low-rise houses, a full-length brick with insulation or a holey is used.
Parameters of brick wall thickness
From bricks are laid out both external and inner walls. Inside the structure, the wall thickness should be at least 12 cm, that is, in the floor of the brick. The cross section of the columns and transplessity is at least 25x38 cm. Partitions inside the building can be a thickness of 6.5 cm. Such a masonry method is called "on the edge". The thickness of the brick wall, made in such a method, should be rejected by a metal frame every 2 rows. Reinforcement will allow the walls to acquire additional strength and withstand more solid loads.
The combined masonry method is enormous popularity when the walls are made up of several layers. This solution allows you to achieve greater reliability, strength and heat resistance. Such a wall includes:
- Brickwork consisting of invoked or slotted material;
- Insulation - minvat or foam;
- Facing - panels, plaster, facing brick.
The thickness of the outer combination wall is determined by the climatic conditions of the region and the type of insulation used. In fact, the wall may have a standard thickness, and thanks to the correctly chosen insulation, all the rules on the thermal building are achieved.
Wall masonry in one brick
The most common masonry wall in one brick makes it possible to obtain a wall thickness of 250 mm. Brick in this masonry does not fit next to each other, as the wall will not have the necessary strength. Depending on the alleged loads, the thickness of the brick wall can be 1.5, 2 and 2.5 bricks.
The most important rule in the laying of this type is a high-quality masonry and the correct burning of vertical seams connecting materials. The brick from the top row must certainly cover the lower vertical seam. Such a gleaming significantly increases the strength of the structure and distributes a uniform load on the wall.
Types of dressings:
- Vertical seam;
- Transverse seam, not allowing to shift materials in length;
- Longitudinal seam that prevents bricks to the horizontal shift.
The wall masonry in one brick should be performed on a strictly selected scheme - it is a single-row or multi-row. In the single-row system, the first row of bricks are put in the spoonful side, the second twitch. Transverse seams are shifted at half a brick.
The multi-row system assumes an alternation through a row, and through several spoon rows. If a thickened brick is used, then spoonful rows are no more than five. This method provides maximum strength of the structure.
The following row is stacked in the opposite order, thus forming a mirror reflection of the first row. Such a laying has a special strength, as the vertical seams do not coincide anywhere and overlap with the upper bricks.
If laying on two bricks are planned, then, accordingly, the wall thickness will be 51 cm. Such construction is necessary only in regions with strong frosts or in construction, where the insulation is not supposed to be used.
The brick has been and still remains one of the main building materials in low-rise construction. The main advantages of brick masonry are strength, refractoriness, moisture resistance. Below we present the data on the consumption of brick per 1 sq. M with different thickness of the brickwork.
Currently, there are several ways to perform brickwork (standard brickwork, Lipetsk laying, Moscow, etc.). But when calculating the consumption of bricks, the method of performing brickwork is not important, the thickness of the masonry and the size of the brick is important. Brick produced various sizes, characteristics and destination. The main typical sizes of brick are the so-called "single" and "one-time" brick:
the size " single"Brick: 65 x 120 x 250 mm
the size " overhead"Brick: 88 x 120 x 250 mm
In brickwork, as a rule, the thickness of the vertical solution is an average of about 10 mm, the thickness of the horizontal seam is 12 mm. Brickwork There can be different thickness: 0.5 bricks, 1 brick, 1.5 bricks, 2 bricks, 2.5 bricks, etc. As an exception, the brickwork is found in a quarter of a brick.
A quarters of a quarter of bricks are used for small partitions that do not carry loads (for example, a brick partition between the bathroom and the toilet). Brick masonry in the floor brick is often used for single-storey business buildings (shed, toilet, etc.), frontones of residential buildings. Masonry in one brick can be built a garage. For the construction of houses (residential premises), brick masonry is used in a brick thickness and more (depending on climate, floors, type of overlaps, individual characteristics of the structure).
Based on the data on the size of the brick and the thickness of the connecting dissolves, you can easily calculate the number of bricks required for the construction of 1 sq. M. Wall made of brickwork of various thicknesses.
Wall thickness and bricks consumption with different brickwork
The data are given for the "single" brick (65 x 120 x 250 mm), taking into account the thickness of the dissolved seams.
| Type of brick masonry | Wall thickness, mm | Number of bricks for 1 sq. M. 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 |