The load-bearing capacity of the brickwork is 250 mm. Strength calculation of the pier taking into account identified defects. Brickwork load
It is required to determine the calculated load-bearing capacity of a wall section of a building with a rigid structural design*
Calculation of the load-bearing capacity of a section of a load-bearing wall of a building with a rigid structural design.
A calculated longitudinal force is applied to a section of a wall with a rectangular cross-section N= 165 kN (16.5 tf), from long-term loads N g= 150 kN (15 tf), short-term N st= 15 kN (1.5 tf). The section size is 0.40x1.00 m, the floor height is 3 m, the lower and upper supports of the wall are hinged and fixed. The wall is designed from four-layer blocks of design grade M50 strength, using mortar of design grade M50.
It is necessary to check the load-bearing capacity of a wall element in the middle of the floor height when constructing 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. We make the calculation using the formula
N ≤ m g R.A. ,
Where N- calculated longitudinal force.
The calculation example given in this Appendix is made according to formulas, tables and paragraphs of SNiP P-22-81 * (given in square brackets) and these Recommendations.
Element cross-sectional area
A= 0.40 ∙ 1.0 = 0.40m.
Design compressive strength of masonry R according to Table 1 of these Recommendations, taking into account the operating conditions coefficient With= 0.8, see paragraph, 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 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 = Z m.
The flexibility of the element is
.
Elastic characteristics of masonry , adopted according to these “Recommendations”, is equal to
Buckling coefficient determined from the table.
The coefficient taking into account the influence 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.
Calculated load-bearing capacity of the wall section N cc 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).
Design longitudinal force N less N cc :
N= 165 kN< N cc= 268.6 kN.
Therefore, the wall satisfies the load-bearing capacity requirements.
II example of calculating the heat transfer resistance of building walls made of four-layer thermally efficient blocks
Example. Determine the heat transfer resistance of a 400 mm thick wall made of four-layer thermally efficient blocks. The inner surface of the wall on the room side 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 - = 0.92 W/m ∙ 0 C;
Outer layer - porous expanded clay concrete 80 mm thick, density 1800 kg/m 3 - = 0.92 W/m ∙ 0 C;
Thermal insulation layer - polystyrene 170 mm thick, - 0.05 W/m ∙ 0 C;
Dry plaster made from gypsum sheathing sheets 12 mm thick - = 0.21 W/m ∙ 0 C.
The reduced heat transfer resistance of the outer wall is calculated based on the main structural element that is 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 heat transfer resistance of the wall is determined according to SNiP 02/23/2003 " Thermal protection buildings”, based on 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 building walls (stage II)
GSOP = (20 + 3.6)∙213 = 5027 deg. days
Total heat transfer resistance R o the adopted 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 C and 23 W/m 2 ∙ 0 C
respectively;
R 1 ,R 2 ...R n- thermal resistance of individual layers of block structures
n- layer thickness (m);
n- thermal conductivity coefficient of the layer (W/m 2 ∙ 0 C)
= 3.16 m 2 ∙ 0 C/W.
Determine the reduced heat transfer resistance 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 premises plasterboard sheets the heat transfer resistance of the wall increases by
R PC.
=
= 0.571 m 2 ∙ 0 C/W.
The thermal resistance of the wall will be
R o= 3.808 + 0.571 = 4.379 m 2 ∙ 0 C/W.
Thus, the design of an external wall made of four-layer heat-efficient blocks 400 mm thick with an internal plaster layer of 12 mm thick plasterboard sheets with a total thickness of 412 mm has a reduced heat transfer resistance equal to 4.38 m 2 ∙ 0 C/W and satisfies the requirements for the thermal insulation qualities of external enclosing structures of buildings in the climatic conditions of Moscow and the Moscow region.
Necessity of calculation brickwork when building a private house, it is obvious to any developer. In the construction of residential buildings, clinker and red bricks are used, finishing brick used to create an attractive appearance of the outer surface of walls. Each brand of brick has its own specific parameters and properties, but the difference in size between different brands minimal.
The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick.
Clinker bricks are used for the construction of luxury houses. It has a large specific gravity, attractive appearance, high strength. Limited use due to the high cost of the material.
The most popular and in demand material is red brick. It has sufficient strength with a relatively small specific gravity, is easy to process, and is little affected by environment. Disadvantages - sloppy surfaces with high roughness, the ability to absorb water at high humidity. IN normal conditions operation this ability does not manifest itself.
There are two methods for laying bricks:
- pinned;
- spoon
When laying using the butt method, the brick is laid across the wall. The wall thickness must be at least 250 mm. Outside surface the wall will consist of end surfaces material.
With the spoon method, the brick is laid lengthwise. The side surface appears outside. Using this method, you can lay out half-brick walls - 120 mm thick.
What you need to know to calculate
The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick. The result obtained will be approximate and overestimated. For a more accurate calculation, the following factors must be taken into account:
- masonry joint size;
- exact dimensions of the material;
- thickness of all walls.
Manufacturers quite often, for various reasons, do not maintain standard product sizes. Red masonry brick according to GOST it should have dimensions of 250x120x65 mm. To avoid unnecessary mistakes material costs It is advisable to check with suppliers about the sizes of available bricks.
Optimal thickness external walls for most regions is 500 mm, or 2 bricks. This size ensures high strength of the building, good thermal insulation. The disadvantage is the large weight of the structure and, as a result, pressure on the foundation and lower layers of masonry.
The size of the masonry joint will primarily depend on the quality of the mortar.
If you use coarse-grained sand to prepare the mixture, the width of the seam will increase; with fine-grained sand, the seam can be made thinner. The optimal thickness of masonry joints is 5-6 mm. If necessary, it is allowed to make seams with a thickness of 3 to 10 mm. Depending on the size of the seams and the method of laying the brick, you can save some of it.
For example, let’s take a seam thickness of 6 mm and a spoon laying method. brick walls. If the wall thickness is 0.5 m, you need to lay 4 bricks wide.
The total width of the gaps will be 24 mm. Laying 10 rows of 4 bricks will give a total thickness of all gaps of 240 mm, which is almost equal to the length of a standard product. The total area of the masonry will be approximately 1.25 m2. If the bricks are laid closely, without gaps, 240 pieces fit in 1 m2. Taking into account the gaps, the material consumption will be approximately 236 pieces.
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Calculation method for load-bearing walls
When planning the external dimensions of a building, it is advisable to choose values that are multiples of 5. With such numbers it is easier to carry out calculations, then carry them out in reality. When planning the construction of 2 floors, you should calculate the amount of material in stages for each floor.
First, the calculation of the external walls on the first floor is performed. For example, you can take a building with dimensions:
- length = 15 m;
- width = 10 m;
- height = 3 m;
- The thickness of the walls is 2 bricks.
Using these dimensions you need to determine the perimeter of the building:
(15 + 10) x 2 = 50
3 x 50 = 150 m2
By calculating the total area, you can determine the maximum amount of bricks for building a wall. To do this, you need to multiply the previously determined number of bricks for 1 m2 by the total area:
236 x 150 = 35,400
The result is inconclusive, the walls must have openings for installing doors and windows. Quantity entrance doors may vary. Small private houses usually have one door. For buildings large sizes It is advisable to plan two entrances. The number of windows, their sizes and location are determined internal layout building.
As an example, you can take 3 window openings per 10-meter wall, 4 per 15-meter walls. It is advisable to make one of the walls blank, without openings. Volume doorways can be determined by standard sizes. If the sizes differ from standard ones, the volume can be calculated using overall dimensions, adding to them the width of the installation gap. To calculate, use the formula:
2 x (A x B) x 236 = C
where: A is the width of the doorway, B is the height, C is the volume in the number of bricks.
Substituting standard values, we get:
2 x (2 x 0.9) x 236 = 849 pcs.
Volume window openings is calculated similarly. With window sizes of 1.4 x 2.05 m, the volume will be 7450 pieces. Determining the number of bricks per temperature gap is simple: you need to multiply the length of the perimeter by 4. The result is 200 pieces.
35400 — (200 + 7450 + 849) = 26 901.
Purchase required amount should be done with a small margin, because errors and other unforeseen situations are possible during operation.
Picture 1. Calculation diagram for brick columns of the designed building.
A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea is to lay out columns from clay brick, and especially the walls of a house, is far from new and all possible aspects of the calculations of brick walls, piers, pillars, which are the essence of the column, are described in sufficient detail in SNiP II-22-81 (1995) “Stone and reinforced stone structures". This is exactly what normative document and should be used as a guide when making calculations. The calculation below is nothing more than an example of using the specified SNiP.
To determine the strength and stability of columns, you need to have quite a lot of initial data, such as: the brand of brick in terms of strength, the area of support of the crossbars on the columns, the load on the columns, the cross-sectional area of the column, and if none of this is known at the design stage, then you can proceed in the following way:
An example of calculating a brick column for stability under central compression
Designed:
Terrace dimensions 5x8 m. Three columns (one in the middle and two at the edges) made of facing hollow brick with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The strength grade of the brick is M75.
Calculation prerequisites:
.With this design scheme, the maximum load will be on the middle lower column. This is exactly what you should count on for strength. The load on the column depends on many factors, in particular the construction area. For example, in St. Petersburg it is 180 kg/m2, and in Rostov-on-Don - 80 kg/m2. Taking into account the weight of the roof itself is 50-75 kg/m2, the load on the column from the roof for Pushkin Leningrad region may amount to:
N from the roof = (180 1.25 + 75) 5 8/4 = 3000 kg or 3 tons
Because effective loads from the flooring material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab It’s not exactly planned, but it is assumed that the ceiling will be wooden, from separate edged boards, then to calculate the load from the terrace, you can take a uniformly distributed load of 600 kg/m2, then the concentrated force from the terrace acting on the central column will be:
N from terrace = 600 5 8/4 = 6000 kg or 6 tons
The dead weight of columns 3 m long will be:
N from column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons
Thus, the total load on the middle lower column in the section of the column near the foundation will be:
N with rev = 3000 + 6000 + 2 650 = 10300 kg or 10.3 tons
However, in this case it can be taken into account that there is not a very high probability that the temporary load from snow, maximum in winter time, and the temporary load on the floor, maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability coefficient of 0.9, then:
N with rev = (3000 + 6000) 0.9 + 2 650 = 9400 kg or 9.4 tons
The design load on the outer columns will be almost two times less:
N cr = 1500 + 3000 + 1300 = 5800 kg or 5.8 tons
2. Determination of the strength of brickwork.
The M75 brick grade means that the brick must withstand a load of 75 kgf/cm2, however, the strength of the brick and the strength of the brickwork are two different things. The following table will help you understand this:
Table 1. Design compressive strengths for brickwork (according to SNiP II-22-81 (1995))
But that's not all. All the same SNiP II-22-81 (1995) clause 3.11 a) recommends that for the area of pillars and piers less than 0.3 m 2, multiply the value of the design resistance by working conditions factor γ s =0.8. And since the cross-sectional area of our column is 0.25x0.25 = 0.0625 m2, we will have to use this recommendation. As you can see, for M75 brand brick, even when using masonry mortar M100, the strength of the masonry will not exceed 15 kgf/cm2. As a result, the calculated resistance for our column will be 15·0.8 = 12 kg/cm2, then the maximum compressive stress will be:
10300/625 = 16.48 kg/cm 2 > R = 12 kgf/cm 2
Thus, to ensure the required strength of the column, it is necessary either to use a brick of greater strength, for example M150 (the calculated compressive resistance for the M100 grade of mortar will be 22·0.8 = 17.6 kg/cm2) or to increase the cross-section of the column or to use transverse reinforcement of the masonry. For now, let's focus on using more durable facing bricks.
3. Determination of the stability of a brick column.
Brickwork strength and stability brick column- these are also different things and still the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:
N ≤ m g φRF (1.1)
Where m g- coefficient taking into account the influence of long-term load. In this case, we were, relatively speaking, lucky, since at the height of the section h≈ 30 cm, the value of this coefficient can be taken equal to 1.
Note: Actually, with the m g coefficient, everything is not so simple; details can be found in the comments to the article.
φ - longitudinal bending coefficient, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l 0 , and it does not always coincide with the height of the column. The subtleties of determining the design length of a structure are set out separately; here we only note that according to SNiP II-22-81 (1995) clause 4.3: “Calculated heights of walls and pillars l 0 when determining buckling coefficients φ depending on the conditions of supporting them on horizontal supports, the following should be taken:
a) with fixed hinged supports l 0 = N;
b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l 0 = 1.5H, for multi-span buildings l 0 = 1.25H;
c) for free standing structures l 0 = 2H;
d) for structures with partially pinched supporting sections - taking into account the actual degree of pinching, but not less l 0 = 0.8N, Where N- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the clear distance between them."
At first glance, our calculation scheme can be considered as satisfying the conditions of point b). i.e. you can take it l 0 = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only in the case when the lower support is really rigid. If a brick column is laid on a layer of roofing felt waterproofing laid on the foundation, then such a support should rather be considered as hinged rather than rigidly clamped. And in this case, our design in a plane parallel to the plane of the wall is geometrically variable, since the structure of the floor (separately lying boards) does not provide sufficient rigidity in the specified plane. There are 4 possible ways out of this situation:
1. Apply a fundamentally different design scheme
For example - metal columns, rigidly embedded in the foundation, to which the floor beams will be welded, then, for aesthetic reasons, metal columns can be covered face brick any brand, since the entire load will be carried by metal. In this case, it is true that the metal columns need to be calculated, but the calculated length can be taken l 0 = 1.25H.
2. Make another overlap,
for example from sheet materials, which will allow us to consider both the upper and lower supports of the column as hinged, in this case l 0 = H.
3. Make a stiffening diaphragm
in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower supports of the column as hinged, but in this case it is necessary to additionally calculate the stiffness diaphragm.
4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l 0 = 2H
In the end, the ancient Greeks erected their columns (though not made of brick) without any knowledge of the resistance of materials, without the use of metal anchors, and even so carefully written building codes and there were no rules in those days, however, some columns still stand to this day.
Now, knowing the design length of the column, you can determine the flexibility coefficient:
λ h = l 0 /h (1.2) or
λ i = l 0 /i (1.3)
Where h- height or width of the column section, and i- radius of inertia.
Determining the radius of gyration is not difficult in principle; you need to divide the moment of inertia of the section by the cross-sectional area, and then extract from the result Square root, however, in this case there is no great need for this. Thus λ h = 2 300/25 = 24.
Now, knowing the value of the flexibility coefficient, you can finally determine the buckling coefficient from the table:
table 2. Buckling coefficients for stone and reinforced concrete stone structures(according to SNiP II-22-81 (1995))
In this case, the elastic characteristics of the masonry α determined by the table:
Table 3. Elastic characteristics of masonry α (according to SNiP II-22-81 (1995))
As a result, the value of the longitudinal bending coefficient will be about 0.6 (with the elastic characteristic value α = 1200, according to paragraph 6). Then the maximum load on the central column will be:
N р = m g φγ with RF = 1х0.6х0.8х22х625 = 6600 kg< N с об = 9400 кг
This means that the adopted cross-section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, it is most optimal to increase the cross-section of the column. For example, if you lay out a column with a void inside of one and a half bricks, measuring 0.38x0.38 m, then not only will the cross-sectional area of the column increase to 0.13 m2 or 1300 cm2, but the radius of inertia of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the coefficient value φ = 0.8. In this case, the maximum load on the central column will be:
N r = m g φγ with RF = 1x0.8x0.8x22x1300 = 18304 kg > N with rev = 9400 kg
This means that a section of 38x38 cm is sufficient to ensure the stability of the lower central centrally compressed column and it is even possible to reduce the grade of brick. For example, with the initially adopted grade M75, the maximum load will be:
N r = m g φγ with RF = 1x0.8x0.8x12x1300 = 9984 kg > N with rev = 9400 kg
That seems to be all, but it is advisable to take into account one more detail. In this case, it is better to make the foundation strip (united for all three columns) rather than columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the most optimal column section would be 0.51x0.51 m, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of such columns will be 2601 cm2.
An example of calculating a brick column for stability under eccentric compression
The outer columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then still, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the outer columns not in the center of the column section. Where exactly the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the modulus of elasticity of the crossbars and columns and a number of other factors, which are discussed in detail in the article "Calculation of the support section of a beam for bearing". This displacement is called the eccentricity of the load application e o. In this case, we are interested in the most unfavorable combination of factors, in which the load from the floor to the columns will be transferred as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be subject to a bending moment equal to M = Ne o, and this point must be taken into account when calculating. IN general case The stability test can be performed using the following formula:
N = φRF - MF/W (2.1)
Where W- section moment of resistance. In this case, the load for the lower outermost columns from the roof can be conditionally considered centrally applied, and eccentricity will only be created by the load from the floor. At eccentricity 20 cm
N р = φRF - MF/W =1x0.8x0.8x12x2601- 3000 20 2601· 6/51 3 = 19975, 68 - 7058.82 = 12916.9 kg >N cr = 5800 kg
Thus, even with a very large eccentricity of load application, we have a more than double safety margin.
Note: SNiP II-22-81 (1995) “Stone and reinforced masonry structures” recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore I do not present the calculation method recommended by SNiP here.
Brick is quite durable construction material, especially solid, and when building houses of 2-3 floors, walls made of ordinary ceramic bricks As a rule, additional calculations are not needed. Nevertheless, situations are different, for example, it is planned two-storey house with a terrace on the second floor. The metal crossbars, on which the metal beams of the terrace will also rest, are planned to be supported on brick columns made of facing hollow bricks 3 meters high; above there will be columns 3 m high, on which the roof will rest:
A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea of laying columns of clay bricks, and even more so the walls of a house, is far from new and all possible aspects of the calculations of brick walls, piers, pillars, which are the essence of the column, are described in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures." It is this regulatory document that should be used as a guide when making calculations. The calculation below is nothing more than an example of using the specified SNiP.
To determine the strength and stability of columns, you need to have quite a lot of initial data, such as: the brand of brick in terms of strength, the area of support of the crossbars on the columns, the load on the columns, the cross-sectional area of the column, and if none of this is known at the design stage, then you can proceed in the following way:
with central compression
Designed: Terrace dimensions 5x8 m. Three columns (one in the middle and two at the edges) made of facing hollow brick with a cross-section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The strength grade of the brick is M75.
With this design scheme, the maximum load will be on the middle lower column. This is exactly what you should count on for strength. The load on the column depends on many factors, in particular the construction area. For example, snow load for roofing in St. Petersburg is 180 kg/m², and in Rostov-on-Don - 80 kg/m². Taking into account the weight of the roof itself, 50-75 kg/m², the load on the column from the roof for Pushkin, Leningrad region can be:
N from the roof = (180 1.25 +75) 5 8/4 = 3000 kg or 3 tons
Since the current loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but a reinforced concrete slab is definitely not planned, and it is assumed that the floor will be wooden, from separately lying edged boards, then to calculate the load from the terrace you can accept a uniformly distributed load of 600 kg/m², then the concentrated force from the terrace acting on the central column will be:
N from terrace = 600 5 8/4 = 6000 kg or 6 tons
The dead weight of columns 3 m long will be:
N from column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons
Thus, the total load on the middle lower column in the section of the column near the foundation will be:
N with rev = 3000 + 6000 + 2 650 = 10300 kg or 10.3 tons
However, in this case, it can be taken into account that there is not a very high probability that the temporary load from snow, maximum in winter, and the temporary load on the floor, maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability coefficient of 0.9, then:
N with rev = (3000 + 6000) 0.9 + 2 650 = 9400 kg or 9.4 tons
The design load on the outer columns will be almost two times less:
N cr = 1500 + 3000 + 1300 = 5800 kg or 5.8 tons
2. Determination of the strength of brickwork.
The M75 brick grade means that the brick must withstand a load of 75 kgf/cm2, however, the strength of the brick and the strength of the brickwork are two different things. The following table will help you understand this:
Table 1. Design compressive strengths for brickwork
But that's not all. The same SNiP II-22-81 (1995) clause 3.11 a) recommends that for the area of pillars and piers less than 0.3 m², multiply the value of the design resistance by the operating conditions coefficient γ s =0.8. And since the cross-sectional area of our column is 0.25x0.25 = 0.0625 m², we will have to use this recommendation. As you can see, for M75 grade brick, even when using M100 masonry mortar, the strength of the masonry will not exceed 15 kgf/cm2. As a result, the calculated resistance for our column will be 15·0.8 = 12 kg/cm², then the maximum compressive stress will be:
10300/625 = 16.48 kg/cm² > R = 12 kgf/cm²
Thus, to ensure the required strength of the column, it is necessary either to use a brick of greater strength, for example M150 (the calculated compressive resistance for the M100 mortar grade will be 22·0.8 = 17.6 kg/cm²) or to increase the cross-section of the column or to use transverse reinforcement of the masonry. For now, let's focus on using more durable facing bricks.
3. Determination of the stability of a brick column.
The strength of brickwork and the stability of a brick column are also different things and still the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:
N ≤ m g φRF (1.1)
m g- coefficient taking into account the influence of long-term load. In this case, we were, relatively speaking, lucky, since at the height of the section h≤ 30 cm, the value of this coefficient can be taken equal to 1.
φ - longitudinal bending coefficient, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l o, and it does not always coincide with the height of the column. The subtleties of determining the design length of a structure are not outlined here, we only note that according to SNiP II-22-81 (1995) clause 4.3: “Calculation heights of walls and pillars l o when determining buckling coefficients φ depending on the conditions of supporting them on horizontal supports, the following should be taken:
a) with fixed hinged supports l o = N;
b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l o = 1.5H, for multi-span buildings l o = 1.25H;
c) for free-standing structures l o = 2H;
d) for structures with partially pinched supporting sections - taking into account the actual degree of pinching, but not less l o = 0.8N, Where N- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the clear distance between them."
At first glance, our calculation scheme can be considered as satisfying the conditions of point b). i.e. you can take it l o = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only in the case when the lower support is really rigid. If a brick column is laid on a layer of roofing felt waterproofing laid on the foundation, then such a support should rather be considered as hinged rather than rigidly clamped. And in this case, our design in a plane parallel to the plane of the wall is geometrically variable, since the floor structure (separately lying boards) does not provide sufficient rigidity in the specified plane. There are 4 possible ways out of this situation:
1. Apply a completely different design diagram , for example - metal columns, rigidly embedded in the foundation, to which the floor beams will be welded; then, for aesthetic reasons, the metal columns can be covered with facing bricks of any brand, since the entire load will be carried by the metal. In this case, it is true that the metal columns need to be calculated, but the calculated length can be taken l o = 1.25H.
2. Make another overlap, for example, from sheet materials, which will allow us to consider both the upper and lower supports of the column as hinged, in this case l o = H.
3. Make a stiffening diaphragm in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower supports of the column as hinged, but in this case it is necessary to additionally calculate the stiffness diaphragm.
4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l o = 2H. In the end, the ancient Greeks erected their columns (though not made of brick) without any knowledge of the strength of materials, without the use of metal anchors, and there were no such carefully written building codes and regulations in those days, nevertheless, some columns stand and to this day.
Now, knowing the design length of the column, you can determine the flexibility coefficient:
λ h = l o /h (1.2) or
λ i = l o (1.3)
h- height or width of the column section, and i- radius of inertia.
Determining the radius of inertia is, in principle, not difficult; you need to divide the moment of inertia of the section by the cross-sectional area, and then take the square root of the result, but in this case there is no great need for this. Thus λ h = 2 300/25 = 24.
Now, knowing the value of the flexibility coefficient, you can finally determine the buckling coefficient from the table:
table 2. Buckling coefficients for masonry and reinforced masonry structures
(according to SNiP II-22-81 (1995))
In this case, the elastic characteristics of the masonry α determined by the table:
Table 3. Elastic characteristics of masonry α (according to SNiP II-22-81 (1995))
As a result, the value of the longitudinal bending coefficient will be about 0.6 (with the elastic characteristic value α = 1200, according to paragraph 6). Then the maximum load on the central column will be:
N р = m g φγ with RF = 1 0.6 0.8 22 625 = 6600 kg< N с об = 9400 кг
This means that the adopted cross-section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, it is most optimal to increase the cross-section of the column. For example, if you lay out a column with a void inside of one and a half bricks, measuring 0.38 x 0.38 m, then not only will the cross-sectional area of the column increase to 0.13 m or 1300 cm, but the radius of inertia of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the coefficient value φ = 0.8. In this case, the maximum load on the central column will be:
N р = m g φγ with RF = 1 0.8 0.8 22 1300 = 18304 kg > N with rev = 9400 kg
This means that a section of 38x38 cm is sufficient to ensure the stability of the lower central centrally compressed column and it is even possible to reduce the grade of brick. For example, with the initially adopted grade M75, the maximum load will be:
N р = m g φγ with RF = 1 0.8 0.8 12 1300 = 9984 kg > N with rev = 9400 kg
That seems to be all, but it is advisable to take into account one more detail. In this case, it is better to make the foundation strip (united for all three columns) rather than columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of such columns will be 2601 cm2.
An example of calculating a brick column for stability
with eccentric compression
The outer columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then still, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the outer columns not in the center of the column section. Where exactly the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the elastic moduli of the crossbars and columns and a number of other factors. This displacement is called the eccentricity of the load application e o. In this case, we are interested in the most unfavorable combination of factors, in which the load from the floor to the columns will be transferred as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be subject to a bending moment equal to M = Ne o, and this point must be taken into account when calculating. In general, stability testing can be performed using the following formula:
N = φRF - MF/W (2.1)
W- section moment of resistance. In this case, the load for the lower outermost columns from the roof can be conditionally considered centrally applied, and eccentricity will only be created by the load from the floor. At eccentricity 20 cm
N р = φRF - MF/W =1 0.8 0.8 12 2601- 3000 20 2601· 6/51 3 = 19975.68 - 7058.82 = 12916.9 kg >N cr = 5800 kg
Thus, even with a very large eccentricity of load application, we have a more than double safety margin.
Note: SNiP II-22-81 (1995) “Stone and reinforced masonry structures” recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore the calculation method recommended by SNiP is not given here.
To perform a wall stability calculation, you first need to understand their classification (see SNiP II -22-81 “Stone and reinforced masonry structures”, as well as a manual for SNiP) and understand what types of walls there are:
1. Load-bearing walls- these are the walls on which floor slabs, roof structures, etc. rest. The thickness of these walls must be at least 250 mm (for brickwork). These are the most important walls in the house. They need to be designed for strength and stability.
2. Self-supporting walls - these are walls on which nothing rests, but they are subject to the load from all the floors above. In fact, in a three-story house, for example, such a wall will be three floors high; the load on it only from the own weight of the masonry is significant, but at the same time the question of the stability of such a wall is also very important - the higher the wall, the greater the risk of its deformation.
3. Curtain walls- these are external walls that rest on the ceiling (or other structural elements) and the load on them comes from the height of the floor only from the own weight of the wall. The height of non-load-bearing walls should be no more than 6 meters, otherwise they become self-supporting.
4. Partitions are interior walls less than 6 meters high, supporting only the load from its own weight.
Let's look at the issue of wall stability.
The first question that arises for an “uninitiated” person is: where can the wall go? Let's find the answer using an analogy. Let's take a hardcover book and place it on its edge. The larger the book format, the less stable it will be; on the other hand, the thicker the book, the better it will stand on its edge. The situation is the same with walls. The stability of the wall depends on the height and thickness.
Now let's take the worst case scenario: a thin, large-format notebook and place it on its edge - it will not only lose stability, but will also bend. Likewise, the wall, if the conditions for the ratio of thickness and height are not met, will begin to bend out of plane, and over time, crack and collapse.
What is needed to avoid such a phenomenon? You need to study pp. 6.16...6.20 SNiP II -22-81.
Let's consider the issues of determining the stability of walls using examples.
Example 1. Given a partition made of aerated concrete grade M25 on mortar grade M4, 3.5 m high, 200 mm thick, 6 m wide, not connected to the ceiling. The partition has a doorway of 1x2.1 m. It is necessary to determine the stability of the partition.
From Table 26 (item 2) we determine the masonry group - III. From the tables do we find 28? = 14. Because the partition is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 9.8.
k 1 = 1.8 - for a partition that does not carry a load with a thickness of 10 cm, and k 1 = 1.2 - for a partition 25 cm thick. By interpolation, we find for our partition 20 cm thick k 1 = 1.4;
k 3 = 0.9 - for partitions with openings;
that means k = k 1 k 3 = 1.4*0.9 = 1.26.
Finally β = 1.26*9.8 = 12.3.
Let's find the ratio of the height of the partition to the thickness: H /h = 3.5/0.2 = 17.5 > 12.3 - the condition is not met, a partition of such thickness cannot be made with the given geometry.
How can this problem be solved? Let's try to increase the grade of mortar to M10, then the masonry group will become II, respectively β = 17, and taking into account the coefficients β = 1.26*17*70% = 15< 17,5 - этого оказалось недостаточно. Увеличим марку газобетона до М50, тогда группа кладки станет I , соответственно β = 20, а с учетом коэффициентов β = 1,26*20*70% = 17.6 >17.5 - the condition is met. It was also possible, without increasing the grade of aerated concrete, to lay structural reinforcement in the partition in accordance with clause 6.19. Then β increases by 20% and the stability of the wall is ensured.
Example 2. Dana external curtain wall made of lightweight brick masonry grade M50 on mortar grade M25. Wall height 3 m, thickness 0.38 m, wall length 6 m. Wall with two windows measuring 1.2x1.2 m. It is necessary to determine the stability of the wall.
From Table 26 (item 7) we determine the masonry group - I. From Table 28 we find β = 22. Because the wall is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 15.4.
We find the coefficients k from tables 29:
k 1 = 1.2 - for a wall that does not bear a load with a thickness of 38 cm;
k 2 = √A n /A b = √1.37/2.28 = 0.78 - for a wall with openings, where A b = 0.38*6 = 2.28 m 2 - horizontal sectional area of the wall, taking into account windows, A n = 0.38*(6-1.2*2) = 1.37 m2;
that means k = k 1 k 2 = 1.2*0.78 = 0.94.
Finally β = 0.94*15.4 = 14.5.
Let's find the ratio of the height of the partition to the thickness: H /h = 3/0.38 = 7.89< 14,5 - условие выполняется.
It is also necessary to check the condition stated in paragraph 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
I quote Irina:
profiles will not replace reinforcement
I quote Irina:
Regarding the foundation: voids in the concrete body are permissible, but not from below, so as not to reduce the bearing area, which is responsible for the load-bearing capacity. That is, there should be a thin layer underneath reinforced concrete.
What kind of foundation - strip or slab? What soils?
The soils are not yet known, most likely it will be an open field of all sorts of loam, initially I thought of a slab, but it will be a little low, I want it higher, and I will also have to remove the top fertile layer, so I am leaning towards a ribbed or even box-shaped foundation. I don’t need a lot of bearing capacity of the soil - after all, the house was built on the 1st floor, and expanded clay concrete is not very heavy, freezing there is no more than 20 cm (although according to old Soviet standards it is 80).
I'm thinking about renting upper layer 20-30 cm, lay out geotextiles, cover with river sand and level with compaction. Then a light preparatory screed - for leveling (it seems like they don’t even make reinforcement into it, although I’m not sure), waterproofing with a primer on top
and then there’s a dilemma - even if you tie reinforcement frames with a width of 150-200mm x 400-600mm in height and lay them in steps of a meter, then you still need to form voids with something between these frames and ideally these voids should be on top of the reinforcement (yes also with a certain distance from the preparation, but at the same time they will also need to be reinforced on top with a thin layer under a 60-100mm screed) - I think the PPS slabs will be monolithed as voids - theoretically it will be possible to fill this in 1 go with vibration.
Those. It looks like a slab of 400-600mm with powerful reinforcement every 1000-1200mm, the volumetric structure is uniform and light in other places, while inside about 50-70% of the volume there will be foam plastic (in unloaded places) - i.e. in terms of consumption of concrete and reinforcement - quite comparable to a 200mm slab, but + a lot of relatively cheap polystyrene foam and more work.
If we somehow replaced the foam plastic with simple soil/sand, it would be even better, but then instead of light preparation, it would be wiser to do something more serious with reinforcement and moving the reinforcement into the beams - in general, I lack both theory and practical experience here.
0 #214 Irina 02.22.2018 16:21
Quote:
why fight it? you just need to take it into account in the calculations and design. You see, expanded clay concrete is quite good wall material with its own list of advantages and disadvantages. Just like any other materials. Now, if you wanted to use it for monolithic ceiling, I would dissuade you, becauseIt’s a pity, in general they just write that lightweight concrete (expanded clay concrete) has a poor connection with the reinforcement - how to deal with this? I understand what stronger than concrete and with what larger area surface of the reinforcement - the better the connection will be, i.e. you need expanded clay concrete with the addition of sand (and not just expanded clay and cement) and thin reinforcement, but more often
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