Picture 1. Calculation scheme for brick columns of the designed building.
A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea 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 masonry structures”. It is this regulatory document that should be used as a guide when making calculations. The calculation below is nothing more than an example of using the specified SNiP.
To determine the strength and stability of columns, you need to have quite a lot of initial data, such as: the brand of brick in terms of strength, the area of support of the crossbars on the columns, the load on the columns, the cross-sectional area of the column, and if none of this is known at the design stage, then you can proceed in the following way:
An example of calculating a brick column for stability under central compression
Designed:
Terrace dimensions 5x8 m. Three columns (one in the middle and two at the edges) made of facing hollow brick with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The strength grade of the brick is M75.
Calculation prerequisites:
.With this design scheme, the maximum load will be on the middle lower column. This is exactly what you should count on for strength. The load on the column depends on many factors, in particular the construction area. For example, in St. Petersburg it is 180 kg/m2, and in Rostov-on-Don - 80 kg/m2. Taking into account the weight of the roof itself 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
Since the current loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab It’s not exactly planned, but it is assumed that the ceiling will be wooden, from separate edged boards, then to calculate the load from the terrace, you can take a uniformly distributed load of 600 kg/m2, then the concentrated force from the terrace acting on the central column will be:
N from terrace = 600 5 8/4 = 6000 kg or 6 tons
The dead weight of columns 3 m long will be:
N from column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons
Thus, the total load on the middle lower column in the section of the column near the foundation will be:
N with rev = 3000 + 6000 + 2 650 = 10300 kg or 10.3 tons
However, in 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 time, 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. Eventually design 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, value given 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 floor structure (separately lying boards) does not provide sufficient rigidity in specified plane. There are 4 possible ways out of this situation:
1. Apply a fundamentally different design scheme
for example - metal columns, rigidly embedded in the foundation, to which the floor beams will be welded; then, for aesthetic reasons, the metal columns can be covered with facing bricks of any brand, since the entire load will be carried by the metal. In this case, it is true that the metal columns need to be calculated, but the calculated length can be taken l 0 = 1.25H.
2. Make another overlap,
for example from sheet materials, which will allow us to consider both the upper and lower supports of the column as hinged, in this case l 0 = H.
3. Make a stiffening diaphragm
in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower supports of the column as hinged, but in this case it is necessary to additionally calculate the stiffness diaphragm.
4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l 0 = 2H
In the end, the ancient Greeks erected their columns (though not made of brick) without any knowledge of the 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 inertia is, in principle, not difficult; you need to divide the moment of inertia of the section by the cross-sectional area, and then take the square root of the result, but in this case there is no great need for this. Thus λ h = 2 300/25 = 24.
Now, knowing the value of the flexibility coefficient, you can finally determine the buckling coefficient from the table:
table 2. Buckling coefficients for masonry and reinforced masonry structures (according to SNiP II-22-81 (1995))
In this case, the elastic characteristics of the masonry α determined by the table:
Table 3. Elastic characteristics of masonry α (according to SNiP II-22-81 (1995))
As a result, the value of the longitudinal bending coefficient will be about 0.6 (with the elastic characteristic value α = 1200, according to paragraph 6). Then the maximum load on the central column will be:
N р = m g φγ with RF = 1х0.6х0.8х22х625 = 6600 kg< N с об = 9400 кг
This means that the adopted cross-section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, it is most optimal to increase the cross-section of the column. For example, if you lay out a column with a void inside of one and a half bricks, measuring 0.38x0.38 m, then not only will the cross-sectional area of the column increase to 0.13 m2 or 1300 cm2, but the radius of inertia of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the coefficient value φ = 0.8. In this case, the maximum load on the central column will be:
N r = m g φγ with RF = 1x0.8x0.8x22x1300 = 18304 kg > N with rev = 9400 kg
This means that a section of 38x38 cm is sufficient to ensure the stability of the lower central centrally compressed column and it is even possible to reduce the grade of brick. For example, with the initially adopted grade M75, the maximum load will be:
N r = m g φγ with RF = 1x0.8x0.8x12x1300 = 9984 kg > N with rev = 9400 kg
That seems to be all, but it is advisable to take into account one more detail. In this case, it is better to make the foundation strip (united for all three columns) rather than columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of such columns will be 2601 cm2.
An example of calculating a brick column for stability under eccentric compression
The outer columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then still, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the outer columns not in the center of the column section. Where exactly the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the modulus of elasticity of the crossbars and columns and a number of other factors, which are discussed in detail in the article "Calculation of the support section of a beam for bearing". This displacement is called the eccentricity of the load application e o. In this case, we are interested in the most unfavorable combination of factors, in which the load from the floor to the columns will be transferred as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be subject to a bending moment equal to M = Ne o, and this point must be taken into account when calculating. IN general 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.
The need to calculate brickwork when building a private house is obvious to any developer. In the construction of residential buildings, clinker and red bricks are used, finishing 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 relatively little specific gravity, easy to process, low impact 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:
- tychkovy;
- spoon
When laying using the butt method, the brick is laid across the wall. The wall thickness must be at least 250 mm. The outer surface of the wall will consist of the end surfaces of the material.
With the spoon method, the brick is laid lengthwise. Outside it turns out side surface. Using this method, you can lay out half-brick walls - 120 mm thick.
What you need to know to calculate
The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick. The result obtained will be approximate and overestimated. For a more accurate calculation, the following factors must be taken into account:
- masonry joint size;
- exact dimensions of the material;
- thickness of all walls.
Manufacturers quite often various reasons do not support standard product sizes. According to GOST, red masonry bricks must have dimensions of 250x120x65 mm. To avoid mistakes and unnecessary 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 the spoon method of laying brick walls. If the wall thickness is 0.5 m, you need to lay 4 bricks wide.
The total width of the gaps will be 24 mm. Laying 10 rows of 4 bricks will give a total thickness of all gaps of 240 mm, which is almost equal to the length 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 maximum amount bricks for building a wall. To do this, you need to multiply the previously determined number of bricks for 1 m2 by the total area:
236 x 150 = 35,400
The result is inconclusive, the walls must have openings for installing doors and windows. 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. The volume of 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.
External load-bearing walls must, at a minimum, be designed for strength, stability, local collapse and resistance to heat transfer. To find out how thick should a brick wall be? , you need to calculate it. In this article we will look at calculating the load-bearing capacity of brickwork, and in subsequent articles we will look at other calculations. So as not to miss the exit new article, subscribe to the newsletter and you will find out what the thickness of the wall should be after all the calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, we will consider all calculations specifically for this category.
Bearing are called walls that take the load from floor slabs, coverings, beams, etc. resting on them.
You should also take into account the brand of brick for frost resistance. Since everyone builds a house for themselves for at least a hundred years, in dry and normal humidity conditions of the premises, a grade (M rz) of 25 and above is accepted.
During the construction of a house, cottage, garage, outbuildings and other structures with dry and normal humidity conditions It is recommended to use hollow bricks for external walls, since its thermal conductivity is lower than that of solid bricks. Accordingly, during thermal engineering calculations, the thickness of the insulation will be less, which will save money when purchasing it. Solid bricks for external walls should be used only when it is necessary to ensure the strength of the masonry.
Reinforcement of brickwork is allowed only if increasing the grade of brick and mortar does not provide the required load-bearing capacity.
Calculation example brick wall.
The load-bearing capacity of brickwork depends on many factors - the brand of brick, the brand of mortar, the presence of openings and their sizes, the flexibility of the walls, etc. The calculation of bearing capacity begins with determining the design scheme. When calculating walls for vertical loads, the wall is considered to be supported by hinged and fixed supports. When calculating walls for horizontal loads (wind), the wall is considered rigidly clamped. It is important not to confuse these diagrams, since the moment diagrams will be different.
Selection of design section.
In solid walls, the design section is taken to be section I-I at the level of the bottom of the floor with a longitudinal force N and a maximum bending moment M. It is often dangerous section II-II, since the bending moment is slightly less than the maximum and is equal to 2/3M, and the coefficients m g and φ are minimal.
In walls with openings, the cross-section is taken at the level of the bottom of the lintels.
Let's look at section I-I.
From the previous article Collection of loads on the first floor wall Let's take the resulting value of the total load, which includes the load from the floor of the first floor P 1 = 1.8 t and the overlying floors G = G p +P 2 +G 2 = 3.7t:
N = G + P 1 = 3.7t +1.8t = 5.5t
The floor slab rests on the wall at a distance of a=150mm. The longitudinal force P 1 from the ceiling will be at a distance a / 3 = 150 / 3 = 50 mm. Why 1/3? Because the stress diagram under the support section will be in the form of a triangle, and the center of gravity of the triangle is located at 1/3 of the length of the support.
The load from the overlying floors G is considered to be applied centrally.
Since the load from the floor slab (P 1) is not applied at the center of the section, but at a distance from it equal to:
e = h/2 - a/3 = 250mm/2 - 150mm/3 = 75 mm = 7.5 cm,
then it will create a bending moment (M) in section I-I. Moment is the product of force and arm.
M = P 1 * e = 1.8t * 7.5cm = 13.5t*cm
Then the eccentricity of the longitudinal force N will be:
e 0 = M / N = 13.5 / 5.5 = 2.5 cm
Since the load-bearing wall is 25 cm thick, the calculation should take into account the value of the random eccentricity e ν = 2 cm, then the total eccentricity is equal to:
e 0 = 2.5 + 2 = 4.5 cm
y=h/2=12.5cm
At e 0 =4.5 cm< 0,7y=8,75 расчет по раскрытию трещин в швах кладки можно не производить.
The strength of the masonry of an eccentrically compressed element is determined by the formula:
N ≤ m g φ 1 R A c ω
Odds m g And φ 1 in the section under consideration, I-I are equal to 1.
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 on 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 this phenomenon? You need to study pp. 6.16...6.20 SNiP II -22-81.
Let's consider the issues of determining the stability of walls using examples.
Example 1. Given a partition made of aerated concrete grade M25 on mortar grade M4, 3.5 m high, 200 mm thick, 6 m wide, not connected to the ceiling. The partition has a doorway of 1x2.1 m. It is necessary to determine the stability of the partition.
From Table 26 (item 2) we determine the masonry group - III. From the tables do we find 28? = 14. Because the partition is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 9.8.
k 1 = 1.8 - for a partition that does not carry a load with a thickness of 10 cm, and k 1 = 1.2 - for a partition 25 cm thick. By interpolation, we find for our partition 20 cm thick k 1 = 1.4;
k 3 = 0.9 - for partitions with openings;
that means k = k 1 k 3 = 1.4*0.9 = 1.26.
Finally β = 1.26*9.8 = 12.3.
Let's find the ratio of the height of the partition to the thickness: H /h = 3.5/0.2 = 17.5 > 12.3 - the condition is not met, a partition of such thickness cannot be made with the given geometry.
How can this problem be solved? Let's try to increase the grade of mortar to M10, then the masonry group will become II, respectively β = 17, and taking into account the coefficients β = 1.26*17*70% = 15< 17,5 - этого оказалось недостаточно. Увеличим марку газобетона до М50, тогда группа кладки станет I , соответственно β = 20, а с учетом коэффициентов β = 1,26*20*70% = 17.6 >17.5 - the condition is met. It was also possible, without increasing the grade of aerated concrete, to lay structural reinforcement in the partition in accordance with clause 6.19. Then β increases by 20% and the stability of the wall is ensured.
Example 2. An external non-load-bearing wall is made of lightweight masonry made of M50 grade brick with M25 grade mortar. Wall height 3 m, thickness 0.38 m, wall length 6 m. Wall with two windows measuring 1.2x1.2 m. It is necessary to determine the stability of the wall.
From Table 26 (clause 7) we determine the masonry group - I. From Table 28 we find β = 22. Because the wall is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 15.4.
We find the coefficients k from tables 29:
k 1 = 1.2 - for a wall that does not bear a load with a thickness of 38 cm;
k 2 = √A n /A b = √1.37/2.28 = 0.78 - for a wall with openings, where A b = 0.38*6 = 2.28 m 2 - horizontal sectional area of the wall, taking into account windows, A n = 0.38*(6-1.2*2) = 1.37 m2;
that means k = k 1 k 2 = 1.2*0.78 = 0.94.
Finally β = 0.94*15.4 = 14.5.
Let's find the ratio of the height of the partition to the thickness: H /h = 3/0.38 = 7.89< 14,5 - условие выполняется.
It is also necessary to check the condition stated in clause 6.19:
H + L = 3 + 6 = 9 m< 3kβh = 3*0,94*14,5*0,38 = 15.5 м - условие выполняется, устойчивость стены обеспечена.
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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 of reinforced concrete underneath.
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 easy preparatory screed- for leveling (it seems like they don’t even put reinforcement in it, although I’m not sure), waterproof it with a primer on top
and then there’s a dilemma - even if you tie reinforcement frames with a width of 150-200mm x 400-600mm in height and lay them in steps of a meter, then you still need to form voids with something between these frames and ideally these voids should be on top of the reinforcement (yes also with some distance from the preparation, but at the same time they will also need to be reinforced on top thin layer under a 60-100mm screed) - I’m thinking of monolithing the PPS slabs as voids - theoretically it would be possible to fill this in one go with vibration.
Those. It looks like a slab of 400-600mm with powerful reinforcement every 1000-1200mm, the volumetric structure is uniform and light in other places, while inside about 50-70% of the volume there will be foam plastic (in unloaded places) - i.e. in terms of consumption of concrete and reinforcement - quite comparable to a 200mm slab, but + a lot of relatively cheap polystyrene foam and more work.
If we somehow replaced the foam plastic with simple soil/sand, it would be even better, but then instead of light preparation, it would be wiser to do something more serious with reinforcement and moving the reinforcement into the beams - in general, I lack both theory and practical experience here.
0 #214 Irina 02.22.2018 16:21
Quote:
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? As I understand it, the stronger the concrete and the larger the surface area of the reinforcement, the better the connection will be, i.e. you need expanded clay concrete with the addition of sand (and not just expanded clay and cement) and thin reinforcement, but more often
Quote:
When independent design brick house there is an urgent need to calculate whether it can withstand brickwork those loads that are included in the project. A particularly serious situation develops in areas of masonry weakened by window and doorways. In case of heavy load, these areas may not withstand and be destroyed.
The exact calculation of the resistance of the pier to compression by the overlying floors is quite complex and is determined by the formulas included in regulatory document SNiP-2-22-81 (hereinafter referred to as<1>). Engineering calculations of a wall's compressive strength take into account many factors, including the configuration of the wall, its compressive strength, the strength of the type of material, and more. However, approximately, “by eye,” you can estimate the wall’s resistance to compression, using indicative tables in which the strength (in tons) is linked to the width of the wall, as well as brands of brick and mortar. The table is compiled for a wall height of 2.8 m.
Table of brick wall strength, tons (example)
| Stamps | Area width, cm | |||||||||||
| brick | solution | 25 | 51 | 77 | 100 | 116 | 168 | 194 | 220 | 246 | 272 | 298 |
| 50 | 25 | 4 | 7 | 11 | 14 | 17 | 31 | 36 | 41 | 45 | 50 | 55 |
| 100 | 50 | 6 | 13 | 19 | 25 | 29 | 52 | 60 | 68 | 76 | 84 | 92 |
If the value of the wall width is in the range between those indicated, it is necessary to focus on the minimum number. At the same time, it should be remembered that the tables do not take into account all factors that can adjust the stability, structural strength and resistance of a brick wall to compression in a fairly wide range.
In terms of time, loads can be temporary or permanent.
Permanent:
- weight of building elements (weight of fences, load-bearing and other structures);
- soil and rock pressure;
- hydrostatic pressure.
Temporary:
- weight of temporary structures;
- loads from stationary systems and equipment;
- pressure in pipelines;
- loads from stored products and materials;
- climatic loads (snow, ice, wind, etc.);
- and many others.
When analyzing the loading of structures, it is imperative to take into account the total effects. Below is an example of calculating the main loads on the walls of the first floor of a building.
Brickwork load
To take into account the force acting on the designed section of the wall, you need to sum up the loads:
When low-rise construction the task is greatly simplified, and many factors of live load can be neglected by setting a certain safety margin at the design stage.
However, in the case of the construction of 3 or more storey structures, a thorough analysis is required using special formulas that take into account the addition of loads from each floor, the angle of application of force, and much more. In some cases, the strength of the wall is achieved by reinforcement.
Load calculation example
This example shows the analysis of the current loads on the piers of the 1st floor. Here, only permanent loads from various structural elements of the building are taken into account, taking into account the unevenness of the weight of the structure and the angle of application of forces.
Initial data for analysis:
- number of floors – 4 floors;
- brick wall thickness T=64cm (0.64 m);
- specific gravity of masonry (brick, mortar, plaster) M = 18 kN/m3 (indicator taken from reference data, table 19<1>);
- the width of the window openings is: W1=1.5 m;
- height of window openings - B1=3 m;
- pier section 0.64*1.42 m (loaded area where the weight of the overlying structural elements is applied);
- floor height Wet=4.2 m (4200 mm):
- the pressure is distributed at an angle of 45 degrees.
- An example of determining the load from a wall (plaster layer 2 cm)
Nst = (3-4Ш1В1)(h+0.02)Myf = (*3-4*3*1.5)* (0.02+0.64) *1.1 *18=0.447MN.
Width of the loaded area P=Wet*H1/2-W/2=3*4.2/2.0-0.64/2.0=6 m
Nn =(30+3*215)*6 = 4.072MN
ND=(30+1.26+215*3)*6 = 4.094MN
H2=215*6 = 1.290MN,
including H2l=(1.26+215*3)*6= 3.878MN
- Own weight of the walls
Npr=(0.02+0.64)*(1.42+0.08)*3*1.1*18= 0.0588 MN
The total load will be the result of a combination of the indicated loads on the walls of the building; to calculate it, the summation of the loads from the wall, from the floors of the second floor and the weight of the designed area is performed).
Scheme of load and structural strength analysis
To calculate the pier of a brick wall you will need:
- length of the floor (aka height of the site) (Wet);
- number of floors (Chat);
- wall thickness (T);
- width of the brick wall (W);
- masonry parameters (type of brick, brand of brick, brand of mortar);
- Wall area (P)
- According to table 15<1>it is necessary to determine the coefficient a (elasticity characteristic). The coefficient depends on the type and brand of brick and mortar.
- Flexibility index (G)
- Depending on indicators a and G, according to table 18<1>you need to look at the bending coefficient f.
- Finding the height of the compressed part
where e0 is an indicator of extraness.
- Finding the area of the compressed part of the section
Pszh = P*(1-2 e0/T)
- Determination of the flexibility of the compressed part of the pier
Gszh=Vet/Vszh
- Determination according to table. 18<1>fszh coefficient, based on gszh and coefficient a.
- Calculation of the average coefficient fsr
Fsr=(f+fszh)/2
- Determination of coefficient ω (Table 19<1>)
ω =1+e/T<1,45
- Calculation of the force acting on the section
- Definition of sustainability
U=Kdv*fsr*R*Pszh* ω
Kdv – long-term exposure coefficient
R – masonry compression resistance, can be determined from Table 2<1>, in MPa
- Reconciliation
An example of calculating the strength of masonry
— Wet — 3.3 m
— Chat — 2
— T — 640 mm
— W — 1300 mm
- masonry parameters (clay brick made by plastic pressing, cement-sand mortar, brick grade - 100, mortar grade - 50)
- Area (P)
P=0.64*1.3=0.832
- According to table 15<1>determine the coefficient a.
- Flexibility (G)
G =3.3/0.64=5.156
- Bending coefficient (Table 18<1>).
- Height of compressed part
Vszh=0.64-2*0.045=0.55 m
- Area of the compressed part of the section
Pszh = 0.832*(1-2*0.045/0.64)=0.715
- Flexibility of the compressed part
Gszh=3.3/0.55=6
- fsj=0.96
- FSR calculation
Fsr=(0.98+0.96)/2=0.97
- According to the table 19<1>
ω =1+0.045/0.64=1.07<1,45
To determine the effective load, it is necessary to calculate the weight of all structural elements affecting the designed area of the building.
- Definition of sustainability
Y=1*0.97*1.5*0.715*1.07=1.113 MN
- Reconciliation
The condition is met, the strength of the masonry and the strength of its elements are sufficient
Insufficient wall resistance
What to do if the calculated pressure resistance of the walls is insufficient? In this case, it is necessary to strengthen the wall with reinforcement. Below is an example of an analysis of the necessary modernization of a structure with insufficient compressive resistance.
For convenience, you can use tabular data.
The bottom line shows indicators for a wall reinforced with wire mesh with a diameter of 3 mm, with a cell of 3 cm, class B1. Reinforcement of every third row.
The increase in strength is about 40%. Typically this compression resistance is sufficient. It is better to make a detailed analysis, calculating the change in strength characteristics in accordance with the method of strengthening the structure used.
Below is an example of such a calculation
Example of calculation of pier reinforcement
Initial data - see previous example.
- floor height - 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 mortar - cement with sand, brand of bricks - 100, mortar - 50)
In this case, the condition У>=Н is not satisfied (1.113<1,5).
It is required to increase the compression resistance and structural strength.
Gain
k=U1/U=1.5/1.113=1.348,
those. it is necessary to increase the structural strength by 34.8%.
Reinforcement with reinforced concrete frame
Reinforcement is carried out using a B15 concrete frame with a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 with a pitch of 0.150 m.
Section dimensions of the reinforced structure:
Ш_1=1300+2*60=1.42
T_1=640+2*60=0.76
With such indicators, the condition У>=Н is satisfied. The compression resistance and structural strength are sufficient.
