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Examples of strip foundation reinforcement. Foundation reinforcement: calculation of reinforcement, laying and tying. Foundation strip reinforcement

The calculation of reinforcement for the foundation occurs already at the design stage and is its most important component. It is produced taking into account SNiP 52 - 01 - 2003 in matters of choosing the class of reinforcement, its quantity and cross-section. Reinforcement of monolithic structures is carried out in order to improve the tensile strength of a concrete structure. After all, unreinforced concrete can collapse when the soil swells.

Calculation of reinforcement for slab foundations

Slab foundations are used for the construction of cottages and country houses, as well as other buildings without a basement. This base is a monolithic concrete slab, which is reinforced with a rod in two perpendicular directions. The thickness of such a foundation is more than 20 cm, and the mesh is knitted both from above and from below.

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First, the type of reinforcement bar is determined. For a slab monolithic foundation, which is carried out on strong, dense and non-heaving soils with a very low probability of horizontal shear, it is possible to use ribbed reinforcing rod with a diameter of 10 mm or more, having class A-I. If the soil is quite weak, heaving or the building is designed on a slope, reinforcement must be taken with a thickness of at least 14 mm. For vertical connections between the lower and upper rows of the reinforcing mesh, it will be enough to use a smooth 6 mm rod class A-I.

Foundation with reinforcement

The material of the future walls of the building is also very important. After all, the load on the foundation has significant differences between frame houses, as well as wooden houses and buildings made of brick or aerated concrete blocks. As a rule, for light buildings it is possible to use a reinforcement bar with a diameter of 10-12 mm, and for walls made of bricks or blocks - at least 14-16 mm.

The gaps between the rods in the reinforcing mesh are usually about 20 cm in the longitudinal as well as in the transverse direction. This circumstance assumes the presence of 5 reinforcing bars per 1 meter of the length of the foundation wall. The intersections of perpendicular rods are connected to each other with soft wire using a device such as a hook for knitting reinforcement.

Foundation reinforcement scheme

Useful advice! If the volume of construction is very large, then you can purchase a special gun for tying reinforcement. It is capable of automatically connecting rods together at a very high speed.

Example of real calculation

Let's assume that we need to calculate the reinforcement for the foundation of a private house made of lightweight aerated concrete blocks. It is designed to be installed on a slab foundation, which has a thickness of 40 cm. Geological survey data indicate that the soil under the foundation is loamy with medium heaving. Dimensions of the house – 9x6 m:

Rebar frame

  • Since we have planned a fairly large thickness of the foundation, we will need to pour two horizontal grids into it. The block structure on medium-heaving soils requires horizontal rods to have a diameter of 16 mm and ribbing, and vertical rods can be smooth with a thickness of 6 mm;
  • To calculate the required amount of longitudinal reinforcement, take the length of the longest side of the foundation wall and divide it by the lattice pitch. In our example: 9/0.2 = 45 thick reinforcing bars, which have a standard length of 6 meters. We calculate the total number of rods, which is equal to: 45x6 = 270 m;

Foundation reinforcement options

  • in the same way we find the number of reinforcement bars for transverse ligaments: 6/0.2 = 30 pieces; 30x9 = 270 m;
  • by multiplying by 2 we obtain the required amount of horizontal reinforcement in both grids: (270+270) x 2 = 1080 m;
  • vertical ligaments have a length equal to the entire height of the foundation, that is, 40 cm. Their number is calculated by the number of perpendicular intersections of the longitudinal rods with the transverse ones: 45X30 = 1350 pcs. Multiplying 1350x0.4, we get a total length of 540 m;
  • it turns out that to build the required foundation you will need: 1080 m of A-III D16 rod; 540 m of rod A-I D6.

Use of reinforcement in foundation construction

Useful advice! In order to calculate the mass of all reinforcement, it is necessary to use GOST 2590. According to this document, 1 l.m. reinforcing bar D16 weighs 1.58 kg, and D6 - 0.22 kg. Based on this, the total mass of the entire structure: 1080x1.58 = 1706.4 kg; 540x0.222 = 119.9 kg.

To construct the reinforcement, you also need knitting wire. Its quantity can also be counted. If you knit with a regular crochet, then one knot will take approximately 40 cm. One row contains 1350 connections, and two - 2700. Therefore, the total consumption of wire for knitting will be 2700x0.4 = 1080 m. At the same time, 1 m of wire with a diameter of 1 mm weighs 6.12 g. So its total weight is calculated as follows: 1080x6.12 = 6610 g = 6.6 kg.

Example of foundation reinforcement

How to correctly calculate the need for reinforcement for a strip foundation

Peculiarities strip foundation are such that its rupture is most likely in the longitudinal direction. Based on this, the need for reinforcement for the foundation is calculated. The calculation here is not very different from the previous one, which was made for a slab type of foundation. Therefore, the thickness of the rod can be 12-16 mm for longitudinal fastening, and 6-10 mm for transverse and vertical fastening. In the case of a strip foundation, choose a step of no more than 10-15 cm in order to avoid longitudinal rupture, since the load in it is much greater.

For example, let’s calculate a strip-type foundation as applied to wooden house. Let's assume that its width is 40 cm and its height is 1 m. The geometric dimensions of the building are 6x12 m. The soil is heaving sandy loam:

  • In the case of a strip foundation, it is mandatory to install two reinforcing mesh. The lower one prevents physical rupture of the monolithic tape during ground subsidence, and the upper one during soil heaving;
  • The optimal grid pitch seems to be 20 cm. Therefore, for the correct arrangement of the tape of such a foundation, you need 0.4/0.2 = 2 longitudinal rods in both layers of reinforcement;
  • For wooden house The diameter of the reinforcing bar is 12 mm. To perform two-layer reinforcement on the longest sides of the base, you need 2x12x2x2 = 96 m of rod. Short sides require 2x6x2x2 = 48 m;

Strip foundation reinforcement

  • for the crossbars we take a 10 mm rod. Its installation step is 50 cm.

Perimeter of the building: (6+12) x 2 = 36 m. Divide it into steps: 36/0.5 = 72 reinforcing transverse bars. Since their length is equal to the width of the foundation, the total requirement is 72x0.4 = 28.2 m;

  • For vertical connections we can also use D10 rod. Since the height of the vertical component of the reinforcement is equal to the full height of the foundation (1 m), the required quantity is determined by the number of intersections. To do this, multiply the number of transverse rods by the number of longitudinal ones: 72x4 = 288 pcs. For a height of 1 m, the total length will be 288 m;
  • that is, to complete the full reinforcement of our strip foundation it is necessary: ​​144 m of rod A-III D12; 316.2 m of rod A-I D10.

Useful advice! In accordance with the same GOST 2590, it is possible to determine the mass of all reinforcement based on the fact that 1 l.m. rod D16 weighs 0.888 kg; D6 – 0.617 kg. Hence the total mass: 144x0.8 = 126.7 kg; 316.2x0.62 = 193.5 kg.

The examples of calculating reinforcement for the foundation will help you navigate the needs of materials in any case. To do this, you just need to substitute your data into the formulas.

Calculation of reinforcement for the foundation: how to do it correctly


Calculation of reinforcement for the foundation: how it is carried out for strip and slab types. Detailed example calculating the need for reinforcing bars for the foundation

The foundation is the basis of a building for any purpose; it is the most important part of any building. There is a load on it that is transferred to the ground. There are certain types of foundations; they need to be reinforced in their own way. However, we will talk below about the strip base.

The need for reinforcement

The foundation will be strong only when iron is built into the concrete structure. Thanks to technology, strip foundations are durable and allow the construction of even monolithic houses on their surface. If you have a construction vibrator, then you can create a strong foundation, which will not depend on the thickness of the walls of the house.

Selection of fittings

The rules for reinforcing strip foundations provide for a special approach to the selection of base material. It is important to pay attention to the designation. Thus, the index “C” indicates that this is what is being welded. If the material is designated by the letter “K,” then the reinforcement has qualities of resistance to cracking and corrosion. Such phenomena may well occur under voltage. If the reinforcement is not designated by any of the listed indices, then it is not suitable for use in the construction of the foundation.

Due to the fact that the process of welding 12-mm rods is very labor-intensive, the electric arc method is not used; in addition, the rods may be burned out in the process. Arc welding cannot be used for 35GS either. The overlap should be about 30 diameters, and the elements should be installed in such a way that they do not touch the formwork. This space is called a protective layer and protects the material from atmospheric and temperature influences, as well as corrosion.

Features of reinforcement

Reinforcement of a monolithic strip foundation requires compliance with certain rules. The base is a concrete solution, which is prepared from water, sand and cement. Physical characteristics building material do not ensure the absence of deformation of the building. In order to withstand shifts and negative factors such as temperature fluctuations, the presence of metal in the structure is necessary. It is quite plastic, but guarantees a strong fixation, so the process of laying reinforcement is considered an important step.

It is necessary to install reinforcing elements in places where tension may occur. The greatest likelihood of stretching is on the surface of the base, which is where the reinforcement should be placed. In order to avoid corrosion of the frame, it should be protected with a layer of concrete. The reinforcement scheme for the strip foundation provides for the location of the rods 5 cm from the surface. Due to the fact that it is impossible to prevent deformation, stretch zones may appear in the lower and upper parts. In the first case, the central part will bend downwards, while in the second the frame will bend upwards. That is why, when drawing up a reinforcement scheme, one should take into account the need to arrange the rods at the top and bottom, the diameter of the elements should be within the limit of 10 to 12 mm. The rods should have a ribbed surface, this will allow them to achieve contact with the concrete.

The technology for reinforcing a strip foundation requires the placement of a skeleton of rods in other parts, and the components may have a smaller diameter and a smooth surface. In this case, the rods should be positioned vertically and horizontally, as well as across. When reinforcing a monolithic foundation with a width of no more than 40 cm, it is permissible to use four elements, their diameter should be in the range from 10 to 16 mm. They should be connected into an 8 mm frame. In order to calculate a strip foundation, it is important to remember that the distance between the horizontally located rods in width should be 40 cm. With an impressive length, the strip foundation has a small width, for this reason longitudinal stretches appear in it. In this case, there will be no transverse ones at all. Transverse vertical and horizontal reinforcement elements, which are thin and smooth, are also necessary to create the frame.

Corner reinforcement

Reinforcement of the corners of the strip foundation is carried out according to a certain method. Quite often there are cases when the deformation occurs precisely in the corner parts and bypasses the middle. When working on creating a reinforcement frame that will be installed in a corner, you need to bend one end of the element and take it into one wall, but the other end should go into the other wall. Reinforcement of the corners of the strip foundation involves connecting the elements with knitting wire. Not every type of reinforcement is made of steel that can be welded. But even if such actions are permissible, problems may arise that can be avoided with the help of wire. Problems can be expressed in overheating of the steel, as well as in changes in properties. The rods may become exhausted, but if this can be avoided, high strength welding seams will not be achieved.

Reinforcement scheme

You can independently draw up a reinforcement scheme for the strip foundation. You need to start work by installing the boards for the formwork; its inner base should be lined with parchment, with its help you will simplify the dismantling of the boards. Drawing up a diagram of the reinforcement frame should be carried out using the following technology. Rods are driven into the ground, the length of which is equal to the depth of the future foundation. In this case, the distances from the formwork must be observed. Supports up to 100 mm high should be installed at the bottom; several threads of the lower row of reinforcement are laid on them. Bricks placed on the edge can be used as stands. In those places where elements intersect, they should be reinforced with wire or welding.

It is important to remember when drawing up a diagram

When drawing up a reinforcement scheme for a strip foundation, it is important to maintain the distance to the outer surfaces of the base. This must be done using bricks. This condition is very important, because the metal structure should not be at the bottom. The distance from the ground should be about 8 cm. Once the reinforcement is installed, you can make ventilation holes and start pouring the solution. The presence of ventilation holes will help improve the shock-absorbing qualities of the base and prevent the occurrence of putrefactive processes.

Determination of material consumption

After the reinforcement scheme for the strip foundation has been drawn up, the material consumption can be calculated. If the foundation is rectangular in shape and its width, length and height are 3.5; 10; 0.2 m, respectively, then the width of the belt will be 0.18 m. Initially, it is necessary to determine the volume of the casting; for this, you should find out the dimensions of the base. If it has the shape of a parallelepiped, then you should perform several simple manipulations: first, determine the perimeter of the base, and then multiply it by the height and width of the casting. However, the calculation of the monolithic foundation is not yet completed. We managed to find out only the base, or rather the casting, which will occupy a volume of 0.97 m 3. Now you need to determine the volume of the inside of the base where the tape is located.

In order to find out the volume of the “filling”, you should multiply the length and width by the height, which will allow you to find out the total volume: 10x3.5x0.2 = 7 m 3. The volume of the casting is calculated as follows: 7 - 0.97 = 6.03 m 3, this figure will become the internal volume of the filler. strip is not yet completed, you can determine the amount of required reinforcement. If its diameter is 12 mm, and the casting has 2 horizontal lines. It is also important to take into account how the elements are arranged vertically. If the distance between them is 0.5 m, and the perimeter is equal, then this value should be multiplied by 2, which will result in 54 m. To calculate the vertical rods, the following calculations must be made: 54 * 2 + 2 = 110, 108 intervals of half a meter and more 2 on the edges. You should add one rod per corner and you will get 114 rods. If we assume that the height of the rod is 70 cm, then by multiplying this parameter by the number of rods we will be able to obtain a footage of 79.8 m.

Following these calculations, it will be possible to obtain how much

Conclusion

When drawing up a diagram, it is important to remember that the metal frame must consist of two rows or more, and they must be located vertically. If we are talking about horizontal elements or transverse stripes, then their number should be determined by the depth of the base. For example, reinforcement involves one such layer.

When building a house on a strip foundation, the question of reinforcement arises. Reinforcement is placed in a concrete structure to increase its flexural strength, since concrete has a very low moment capacity. To prevent problems with the tape poured with your own hands, in the future it is necessary to thoroughly study such an issue as the reinforcement scheme for the strip foundation.

The rods embedded in concrete differ in purpose:

  • Longitudinal horizontal(working fittings). They are located along the belt and absorb bending loads. The diameter is selected by calculation. For any structure whose thickness is 15 cm or less, the reinforcement is laid in one layer. For elements with a thickness of more than 15 cm (strip foundations), a reinforcement cage is used, which most often consists of lower and upper reinforcement. In a strip foundation, the diameters of the longitudinal rods for making frames may differ, but the lower ones are always taken to be larger or equal (for small loads) in diameter.
  • Transverse horizontal(clamps). They ensure the joint operation of longitudinal reinforcement and connect the reinforcing cage into a single whole. When building with your own hands, they are assigned for structural reasons (without calculation).
  • Vertical(clamps). When the thickness of the structure is more than 15 cm, it is necessary to tie not only the longitudinal rods located at the same horizontal level, but also the upper and lower parts of the reinforcement frame. The function is taken over by vertical clamps. The diameter and pitch are determined for design reasons.

For each type of reinforcement, the following is considered separately:

  • diameter;
  • number of rods.
  • steel grade;
  • reinforcement class;
  • protective layer.

Selection of reinforcement material

Basic documents to follow:

  • (clauses 6.2 and 11.2);
  • GOST 5781-82* for steel.

Types of marking of reinforcement products:

  • A - rod (hot rolled);
  • Вр – wire (cold-deformed);
  • K - rope (high strength).

For reinforcement frames of strip foundations, rods of class A400 yield strength are used. There is an outdated marking that is still used by builders - All. When purchasing, it is important to be able to “by eye” distinguish between rods belonging to different classes. It is worth noting that reinforcement cages can be knitted from rods belonging to higher classes, but this is impractical and expensive. To eliminate the possibility of accidentally purchasing a material with a lower yield strength, you need to remember:

  • class A240 (Al) has a smooth surface;
  • class A300 (All) - periodic profile, ring pattern;
  • needed for reinforcing the A400 (Allll) tape, it has a periodic profile with a crescent-shaped pattern (outwardly reminiscent of a herringbone pattern).

When laying reinforcement with your own hands, you should pay attention to the grade of steel. According to GOST, reinforcing bars belonging to class A400 should be made of steel 5GS, 25G2S, 32G2Rps. If steel is purchased in large quantities directly from the plant, then the required grade is indicated in the application. If it is not available, in accordance with GOST, the choice is made by the manufacturer.

Protective layer of concrete

Underneath this phrase lies the distance that the rods should not reach outer surface products, that is, concrete covers the rods from external adverse influences. According to the document “Guide to the design of concrete and reinforced concrete structures made of heavy concrete without prestressing,” the protective layer provides:

  • conditions for the joint work of concrete and reinforcement frame;
  • anchoring and the possibility of making joints of frame elements;
  • protection of steel from corrosion and other negative external influences;
  • protection from high temperatures and direct exposure to fire.

Plastic clamp to create a protective layer of concrete on the sides of the foundation.

According to the above manual, the minimum values ​​of the thickness of the protective layer can be summarized in a table.

In this case, the thickness of the protective layer is taken to be no less than the diameter of the rods.

A plastic cube to create a protective layer of concrete underneath the foundation.

Working reinforcement

When building a house with your own hands, it is not necessary to perform complex calculations on limit states in order to determine the cross-section and number of reinforcement cage bars. As a guide to calculations, use the “Manual for the design of concrete and reinforced concrete structures made of heavy concrete without prestressing” and.

According to these documents, using table 5.2 of the manual and clause 10.3.6 of the joint venture, the total cross-section of all longitudinal bars of the reinforcing frame is calculated:

  • when the side of the tape is less than 3 meters - 0.1% of the cross-sectional area of ​​the foundation, the diameter of the rods is at least 10 mm;
  • when the side of the tape is more than 3 meters - 0.1%, the diameter of the rods is at least 12 mm.

Requirements for the minimum diameter of rods, depending on the length, are presented in the manual “Reinforcement of elements of monolithic reinforced concrete buildings”.

The use of rods with a diameter of more than 40 mm is not allowed. The rods are distributed evenly in the upper and lower layers, guided by the assortment of reinforcement. If rods are used for work different diameters(when using leftovers), those that have a larger diameter are placed at the bottom. In this case, the pitch requirements presented in paragraph 10.3.5 and paragraphs 5.9-5.10 of the design manual are taken into account.

The longitudinal rods of the reinforcement frame are placed according to the table.

Important! If laying a large number of rods is required, it is allowed to arrange them in bundles; the distance between them is determined from their total cross-section.

Providing a protective layer and distance between the upper and lower reinforcement is achieved through the use of clamps. To secure individual rods of the lower layer, round-shaped plastic clamps are most often used. The top layer is held in place by vertical clamps. Sometimes they resort to using “chairs” or “frogs” for reinforcement.

The rods are available in standard lengths - 6 and 12 meters. If it is necessary to reinforce longer structures, extensions are made along the length. In this case, the amount of overlap is taken to be at least 20 rod diameters, but not less than 250 mm.

Horizontal cross clamps

When building a foundation with your own hands, these rods are assigned structurally and do not depend on the cross-section. It is necessary to take into account the load from the building elements (for massive ones it is better to provide a reserve). According to the same documents as for longitudinal reinforcement, the minimum diameter of transverse rods is 6 mm, but not less than 0.25 of the diameter of the working reinforcement.

The pitch of the rods is assigned to at least 20 diameters of the working rods. For example, with a cross-section of longitudinal elements of 14 mm, the pitch of horizontal clamps must be at least 280 mm. For ease of installation with your own hands, take a rounded value of 300 mm.

The length of the rods depends on the width of the tape and the required protective layer. Fastening is carried out on top of the working reinforcement. Length joining is usually not required.

Vertical clamps

The diameter is determined depending on the height of the tape:

  • less than 800 mm - from 6 mm;
  • more than 800 mm - from 8 mm, but not less than 0.25 of the diameter of the working rods.

When building foundations with your own hands for massive buildings, it is recommended to lay the rods with a margin. The step is assigned in the same way as for transverse reinforcement. The length of the rods is selected by subtracting the amount of the protective layer at the top and bottom from the height of the foundation strip.

Reinforcement of corners and junctions
According to clause 8.9 monolithic foundations under all walls are rigidly connected to each other and combined into a system of cross strips. In the joint zone, the pitch of transverse reinforcement usually changes and reliable fastening of working rods running in different directions is ensured. There are several methods of reinforcement.

Corner connections

Rigid overlap and “foot”

The free ends of the reinforcement in one direction are bent at right angles and tied to perpendicular rods. In this case, the external ones are connected to each other, and the internal ones are tied to the external one.

The length of the curved section of the “foot”, with the help of which the overlap is ensured, is taken to be 35-50 diameters of the working reinforcement. The pitch of the clamps is set to 3/8 of the height of the foundation strip.

Scheme of reinforcement of the “foot” corner.

L-shaped clamps

To provide reliable connection working rods, the outer rods work together due to an L-shaped clamp placed on them with an overlap of at least 50 diameters of the longitudinal rods. The internal rods are tied to the external ones, as in the previous case:

a. bend the working rods at an angle of 90 degrees, the length of the bend (“foot”) is 50 diameters;

b. attach the legs to the outer rods.

The pitch of the clamps (horizontal and vertical) is 0.75 from the height of the foundation strip.

Reinforcement of the corner with a G-clamp and paws.

U-shaped clamps

In this case, additional reinforcement products are used, bent in the shape of the letter P. For one corner, two such clamps with a length of 50 diameters of the longitudinal rods are required. With this connection, the internal working rods have the same length as the external ones. In the place where the U-shaped clamps overlap, an additional frame of vertical and transverse reinforcement is installed.

Corner reinforcement with P-clamps.

Reinforcement of obtuse angles

Perform with overlap. The outer rod is bent at the required angle, and the inner ones are connected to the outer ones with an overlap of at least 50 diameters. An additional vertical clamp is provided at the bend point of the outer rod.

Scheme of obtuse angle reinforcement.

Wall connections

Lap joint

The reinforcement of the adjacent wall is bent, the bend length is 50 diameters. Both rods from the adjacent tape are attached to the outer rod of the perpendicular wall. In the connection area, the pitch of vertical and transverse clamps is set to 0.375 times the height of the monolithic tape.

Reinforcement of the junction - “legs”.

L-shaped clamp

Clamps bent at right angles are attached to the rods of the adjacent wall. The rod is bent so that each side is equal to 50 diameters of the working reinforcement. The first side is connected to the rods of the adjacent wall, and the second to the external working rod of the perpendicular tape. The pitch of the clamps (vertical, transverse) at the junction is reduced by half compared to the entire long strip.

Reinforcement of the junction with G-clamps.

U-shaped clamp

The connection is made to the external rod of the working reinforcement with a “claw”. Additional reliability is provided by a rod curved in the shape of the letter P, 2 times the width of the foundation strip.

Reinforcement of the junction with P-clamps.

Common Mistakes

1) knitting rods at right angles;

2) use of longitudinal bent reinforcement without anchoring;

An example of incorrect corner reinforcement.

3) connection of longitudinal rods with viscous crosshairs;

4) lack of connection between external and internal rods.

Another example of incorrect corner reinforcement.

Knitting frames

When building a foundation with your own hands, it is extremely important to ensure that all frame elements are securely fastened together. For convenience, possible questions are summarized in a table.

What and how? For binding, annealed binding wire with a diameter of 0.8-1.0 mm is used. To work you will also need a crochet hook. For large volumes of work, special machines for tying reinforcement (knitting gun) are used.
Why is it better to knit? When constructing foundations with your own hands It is recommended to use knitting. Welding is primarily used for large prefabricated frames. This is due to the fact that in the conditions of a construction site there is a possibility of burning through working reinforcement. In addition, when using welding, the help of a qualified worker will be required, which will increase the cost of construction. In addition, the welding site is a potential point of accelerated corrosion.
When can knitting be replaced by welding? Knitting provides greater reliability under construction site conditions (this does not apply to welded factory-made frames), so it makes sense to replace it only if there is welding machine and experience. Replacing knitting with welding (performed directly at the construction site) is recommended to be done only in straight sections. More details on this issue can be found in GOST 14098-91, Appendix 2 “Assessment of the performance qualities of welded joints under static load”. In this table, one immediately notices a large number of compounds marked ND (inadmissible) or NC (inappropriate).

When designing and constructing foundations, many questions arise. Each of them should be treated carefully to avoid complications during operation.

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The consumption of reinforcement must be determined at the foundation design stage in order to subsequently know exactly the amount of material purchased. Let's look at how to calculate reinforcement for a strip foundation using the example of a shallow foundation 70 cm high and 40 cm thick.

1.2 Reinforcement of a strip foundation (video)


2 Work technology

After the amount of reinforcement has been determined, a strip foundation reinforcement scheme must be selected, according to which the reinforced frame will be assembled. Straight sections of the structure are made of solid rods, while in corner areas additional reinforcement is required with reinforcement curved in a U or L-shape. The use of perpendicular overlap of individual reinforcement bars at corners and junctions is not permitted.

The correct reinforcement of the corners of a strip foundation is shown in the diagram:

Scheme of reinforcement of strip foundations at junctions:

Reinforcing a strip foundation with your own hands involves assembling the frame in a convenient place and then placing it inside the formwork. The technology requires bending reinforcement into rectangular clamps, which can be easily done at home using a homemade device.

On the 20th channel you need to cut out grooves with a grinder, into which the reinforcement is subsequently inserted, and a section is put on the rod steel pipe, used as a lever. The finished rings must be fastened by welding or tied with wire. For rods with a diameter of 10-15 mm, 1.2-1.5 mm wire is used.

Often in the process of preparing for construction, the question arises, what thickness of reinforcement is optimal? On the one hand, the correct calculation of the foundation reinforcement affects its strength, and therefore the reliability and durability of the entire structure. This is especially important considering the funds spent on construction. On the other hand, there is a natural desire not to overpay.

Professional builders, when calculating the parameters of foundation reinforcement, use the provisions of SNiP 52-01-2003 “Concrete and reinforced concrete structures”. In private construction, for calculations, it is more than enough to follow one single rule: in the cross-sectional area of ​​a reinforced concrete structure, the share of the total area of ​​all reinforcing rods should not be less than one thousandth (or 0.1%).

Although the wording may seem a little confusing, the rule is actually not difficult to use. For clarity, we will make, as examples, several practical calculations of the thickness and amount of reinforcement for strip, slab and pile foundations. In the calculations we will need some initial data, we will take them from the table below.

Table of cross-sectional area of ​​reinforcement for reinforcing reinforced concrete structures
(GOST 5781-82)

Rod diameter, mm.Cross area
rod cross-section, cm2		Cross area
rod cross-section, m2
6 0,283 0,0000283
8 0,503 0,0000503
10 0,785 0,0000785
12 1,131 0,0001131
14 1,540 0,000154
16 2,010 0,000201
18 2,540 0,000254
20 3,140 0,000314
22 3,800 0,000038
25 4,910 0,000491
28 6,160 0,000616
32 8,010 0,000801
36 10,180 0,001018
40 12,570 0,001257
45 15,000 0,0015
50 19,360 0,001936
55 23,760 0,002376
60 28,270 0,002827
70 38,480 0,003848
80 50,270 0,005027

Depending on the mechanical properties, reinforcing steel is divided into classes A-I (A240), A-II (A300), A-III (A400), A-IV (A600), A-V (A800), A-VI (A1000).

Reinforcing steel is manufactured in bars or coils. Reinforcing steel of class A-I (A240) is made smooth, classes A-II (A300), A-III (A400), A-IV (A600), A-V (A800) and A-VI (A1000) - periodic profile.

An example of calculating the reinforcement of a strip foundation

A strip foundation with the following section is designed:

  • height 1.8 m;
  • tape width 0.4 m.

Let's calculate the cross-sectional area of ​​the foundation: 1.8 x 0.4 = 0.72 sq.m.

Minimum total cross-section of reinforcement: 0.72 / 1000 = 0.00072 sq. m.

Dividing the obtained value by the cross-sectional area of ​​reinforcement of various diameters (from the table above), we obtain the minimum required number of veins. So for reinforcement with a diameter of 6 mm we have:

0.00072 / 0.0000285 = 25.30580079 pcs.

Rounding the resulting value up (for the safety margin), we get: in order to reinforce a foundation with given dimensions with “six” reinforcement, you will need to install 26 longitudinal rods. Of course - not the best engineering solution.

Continuing the calculation for other reinforcement diameters, we obtain the following options:

  • for rods with a diameter of 6 mm - 26 pcs., by analogy below (mm and pcs are omitted):
  • 8 — 15;
  • 10 — 10;
  • 12 — 7 ;
  • 14 — 5 ;
  • 16 — 4 ;
  • 18 — 4 ;
  • 20 — 3 ;
  • 22 — 3 ;
  • 25 — 2 ;
  • 28 — 2 ;
  • 32 — 2 ;
  • 36 — 1 ;
  • 40 — 1 .

It is easy to see that “our” options are reinforcement bars with a diameter of 16 or 18 mm. Four of them are required for the foundation - two each for the lower and upper tiers.

An example of calculating the reinforcement of a slab foundation

A slab foundation is being designed for a building 8 by 5 meters. The thickness of the slab is 35 cm. The owner has at his disposal fittings with a diameter of 10 mm. It is necessary to determine the parameters of the reinforcement structure.

Cross section. Let's determine its area: 8.0 x 0.35 = 2.8 square meters.

Number of veins: 0.0028 / 0.000079 = 35.5 = 36 pieces

(18 in the top layer and 18 in the bottom).

In total, in the transverse direction, the upper and lower layers each contain 18 reinforcement bars.

Longitudinal section. Let's determine its area: 8.0 x 0.35 = 1.75 square meters.

Minimum total cross-section of reinforcement: 1.75 / 1000 = 0.00175 sq.m.

Number of veins: 0.00175 / 0.000079 = 22.2 = 23 pieces, accept 24 pieces. (12 in the top layer and 12 in the bottom).

In total, in the transverse direction, the upper and lower layers each contain 12 reinforcement bars.

An example of calculating the reinforcement of a pile foundation

Let us determine the most optimal and cost-effective method of reinforcing cast-in-place piles of circular cross-section with a diameter of 20 cm (0.2 m).

Let's determine the cross-sectional area of ​​the pile:

S = PR2 = 3.14 x (0.2 / 2)2 = 0.0314 m kV.

Minimum total reinforcement cross-section:

0.0314 / 1000 = 0.0000314 sq.m.

By dividing the obtained value by the tabulated areas of cuts of reinforcement of various diameters, we obtain:

  • for rods with a diameter of 6 mm - 2 pcs;
  • 8 mm - 1 piece;
  • 10 mm - 1 piece;
  • 12 mm - 1 pc.

The calculation results show that two reinforcement bars with a diameter of 6 mm are quite sufficient. However, reinforcing reinforced concrete products with less than 3 veins is not recommended, as this sharply reduces their strength. In our case, the cheapest solution, but at the same time absolutely meeting the strength requirements, would be 3 rods with a diameter of 6 mm.

An example of calculating the scheme and costs for foundation reinforcement

1. Calculation of longitudinal reinforcement (cross section 7.0 x 0.40).

Sectional area: 7 x 0.4 = 2.8 sq.m.

Minimum total reinforcement cross-section: 2.8 / 1000 = 0.0028 sq. m.

Let's make a calculation for one of the reinforcement diameters, 8 mm;

Number of veins:

0.0028 / 0.0000503 = 55.6 = 56 pieces, or 28 at the bottom and top.

Let's calculate the reinforcement mesh cell in this case:

Subtract the value from the width of the slab minimum distance from the reinforcement to the outer wall (50 mm = 0.05 m), multiplied by two (left and right). On the remaining length we will evenly place the calculated number of rods, namely, divide it by the calculated number of veins minus one. The resulting value is the cell width:

A= (7.0 m – 2 x 0.05 m) / (28 – 1) = 0.26 m = 26 cm.

For longitudinal reinforcement we will need 56 rods 9 m long, so the total length of the reinforcement with a diameter of 8 mm will be:

56 x 9 = 504 meters

According to the reference table, one linear meter of figure eight reinforcement weighs 0.395 kg, which means the total weight will be:

504 x 0.395 = 199 kg.

We carry out similar calculations for other types of reinforcement and get:

  • for 6 mm - 99 pcs, cell 14 cm, total weight: 208 kg;
  • 8 mm - 56 pcs, cell 26 cm, total weight: 199 kg;
  • 10 mm - 36 pcs, cell 41 cm, total weight: 200 kg;
  • 12 mm - 25 pcs, cell 58 cm, total weight: 209 kg;
  • 14 mm - 19 pcs, cell 77 cm, total weight: 202 kg;
  • 16 mm - 15 pcs, cell 99 cm, total weight: 229 kg;
  • 18 mm - 12 pcs, cell 138 cm, total weight: 216 kg;
  • 20 mm - 10 pcs, cell 173 cm, total weight: 223 kg.

2. Calculation of transverse reinforcement (longitudinal section 9.0 x 0.40).

Sectional area: 9 x 0.4 = 3.6 sq.m.

Minimum total cross-section of reinforcement: 3.6 / 1000 = 0.0036 sq.m.

We calculate the values ​​we are interested in for several reinforcement diameters:

  • for 6 mm - 127 pcs, cell 14 cm, total weight: 207 kg;
  • 8 mm - 72 pcs, cell 25 cm, total weight: 199 kg;
  • 10 mm - 46 pcs, cell 40 cm, total weight: 199 kg;
  • 12 mm - 33 pcs, cell 56 cm, total weight: 213 kg;
  • 14 mm - 24 pcs, cell 81 cm, total weight: 188 kg;
  • 16 mm - 19 pcs, cell 99 cm, total weight: 222 kg;
  • 18 mm - 15 pcs, cell 127 cm, total weight: 224 kg;
  • 20 mm - 12 pcs, cell 178 cm, total weight: 208 kg.

Let's consider the obtained values. When making a slab foundation, it is recommended to take the cell size as 40...70 mm. Two diameters fall into this range: 10 and 12 mm.

longitudinal:

  • for 10 mm - 36 pcs, cell 41 cm, total weight: 200 kg
  • for 12 mm - 25 pcs, cell 58 cm, total weight: 209 kg

transverse:

  • for 10 mm - 46 pcs, cell 40 cm, total weight: 199 kg;
  • for 12 mm - 33 pcs, cell 56 cm, total weight: 213 kg.

Total weight for 10 mm diameter: 200+199 = 399 kg; total weight for diameter 12 mm: 209+213 = 422 kg.

Since the cost of fittings is mostly determined by weight, in our case the best option there will be a rod with a diameter of 10 mm. The geometric parameters of the cell are 41 x 40 cm.