The initial data for calculating the quantity is the type of foundation (slab, strip, columnar) and its configuration. The type of foundation and parameters are selected depending on the load on the foundation. In one of the previous articles, an example was given of calculating the load ("") on the foundation of a house measuring 6 m by 6 m with one inner wall. This article below provides examples of calculating the consumption of reinforcement for a foundation for the same house.
How much rebar and tie wire is needed for a slab foundation?
First of all, you need to decide on the class and diameter of the reinforcement bar: for this you need to use only reinforcement with a ribbed surface and a diameter of at least 10 mm. The strength of the entire structure depends on the diameter of the reinforcement: the thicker the reinforcement, the stronger. When choosing its thickness, you should focus on the weight of the house and. If the soil is non-porous and dense, i.e. has a good bearing capacity, then under load from the house it will deform less and less stability is required from the plate. The second factor is the weight of the house. The larger it is, the greater the load on the plate and the greater its deformation. If you are building a light wooden house on good soil, then reinforcement with a diameter of 10 mm will be enough to reinforce the slab. If a heavy house is on soft ground, then the reinforcement should be used thick 14-16 mm. Grid spacing reinforcing cage the plate is usually 20 cm, with such a step on our foundation of 6x6 m, 31 bars must be laid along and the same number across, for a total of 62 bars. There are two reinforcement belts at the slab - upper and lower, so the total number of bars will be 124 pieces, with a bar length of 6 m, we get a consumption of 124 x 6 m = 744 running meters of reinforcement. In addition, the upper reinforcement mesh must be connected to the lower one; this connection is made at the intersection of the longitudinal and transverse reinforcement bars. There will be 31 x 31 = 961 such connections. If the slab is 20 cm thick and the rebar frame is 5 cm from the surface, then each joint requires a 10 cm rebar (20 cm thickness minus 5 cm top and bottom). All connections will require 0.1 x 961 = 96.1 meters of reinforcement. The total amount of reinforcement for the entire slab foundation will be 744 m + 96.1 m = 840.1 linear meters.
To calculate how much binding wire, it is necessary first of all to determine the connection method: first, the longitudinal and transverse bars of the reinforcement of the lower chord are connected, then the vertical bars are attached to them, and then the longitudinal and transverse bars of the upper chord are attached to them. Thus, in every place where two horizontal bars and one vertical bar intersect, there are two binding wire connections. There are 961 such places in the lower belt and the same number in the upper one. To knit one intersection of the bars, you need 15 cm of knitting wire bent in half, that is, 0.3 m of net length. The total consumption of knitting wire for the slab foundation will be 0.3 m x 961 x 2 = 576.6 m.
How much reinforcement is needed to reinforce the strip foundation?
Concrete is a non-plastic material and cracks easily under tensile stress. Under the influence of heaving forces from the side of the soil or with an uneven load from the side of the building, the foundation is slightly deformed. With any deformation in the material, a compression zone is created on the one hand, and a tensile zone on the other. It is in the tensile zone that cracks occur. To prevent the appearance of cracks, the foundation must be reinforced.The calculation of reinforcement for a foundation is an important stage in its design, therefore it must be carried out taking into account the requirements of SNiP 52-01-2003 for choosing a reinforcement class, section and its required quantity.
First you need to understand why metal reinforcement is needed in a monolithic concrete base. Concrete, after gaining industrial strength, has a high compressive strength, and a much lower tensile strength. An unreinforced concrete base is prone to cracking when the soil swells, which can lead to deformation of the walls and even destruction of the entire building.
Calculation of reinforcement for slab foundation
Calculation example
A house made of aerated concrete blocks is installed on a 40 cm thick slab foundation on medium-heavy loams. Overall dimensions of the house - 9x6 meters.
Calculation of reinforcement for a strip foundation
In the main tensile load falls along the tape, that is, it is directed longitudinally. Therefore, for longitudinal reinforcement, a bar with a thickness of 12-16 mm is chosen, depending on the type of soil and wall material, and for transverse and vertical bonds it is allowed to take a bar of a smaller diameter - from 6 to 10 mm. In general, the principle of calculation is similar to the calculation of reinforcement slab foundation, but the step of the reinforcing grid is selected 10-15 cm, since the tensile forces strip foundation may be significantly larger.
Calculation example
Strip foundation wooden house, the width of the foundation is 0.4 m, the height is 1 meter. The dimensions of the house are 6x12 meters. The soil is heaving sandy loam.
- To perform a strip foundation, two reinforcing meshes are required. The lower reinforcing mesh prevents rupture of the foundation tape during subsidence of the soil, the upper one - when it heaves.
- The mesh spacing is 20 cm. For the construction of the foundation tape, 0.4 / 0.2 = 2 longitudinal bars are required in each layer of reinforcement.
- The diameter of the longitudinal bar for a wooden house is 12 mm. To perform a two-layer reinforcement of two long sides of the foundation, 2 12 2 2 = 96 meters of rod is required.
- For short sides 2 6 2 2 = 48 meters.
- For cross-links, we select a bar with a diameter of 10 mm. Laying step - 0.5 m.
- We calculate the perimeter of the strip foundation: (6 + 12) 2 = 36 meters. The resulting perimeter is divided by the laying step: 36 / 0.5 \u003d 72 transverse bars. Their length is equal to the width of the foundation, therefore, the total number is 72 0.4 = 28.2 m.
- For vertical ties, we also use a D10 bar. The height of the vertical reinforcement is equal to the height of the foundation - 1 m. The number is determined by the number of intersections, multiplying the number of transverse bars by the number of longitudinal ones: 72 4 \u003d 288 pieces. With a length of 1 m, the total length will be 288 m.
- Thus, to perform the reinforcement of the strip foundation, you will need:
- 144 meters of class A-III D12 bar;
- 316.2 meters of A-I D10 bar.
- According to GOST 2590 we find its mass. A running meter of D16 bar weighs 0.888 kg; bar meter D6 - 0.617 kg. We calculate the total mass: 144 0.88 \u003d 126.72 kg; 316.2 0.617 = 193.51 kg.
Calculation of binding wire: the number of connections can be calculated by the number of vertical reinforcement, multiplying it by 2 - 288 2 = 576 connections. Wire consumption for one connection is 0.4 meters. The wire consumption will be 576 0.4 = 230.4 meters. The mass of 1 meter of wire with a diameter of d = 1.0 mm is 6.12 g. For knitting foundation reinforcement, 230.4 6.12 \u003d 1410 g \u003d 1.4 kg of wire will be required.
Comments:
Concrete foundation for a house without fail reinforced. The calculation of reinforcement for the foundation is carried out in accordance with SNiP. When building a house on your own, this is one of the most important stages of work. Accurate determination of the type and number of reinforcing elements will allow you to create a foundation that tolerates deformation loads well. If the concrete in the base takes on compressive loads, then the metal elements resist tension. The second essential point in determining the required amount of reinforcement is the calculation of the cost of the project.
Calculation for a strip base
In accordance with the requirements of building codes, the content of reinforcing elements in the strip base should be 0.001% of its cross-sectional area. The calculated cross-sectional area of the profile and the theoretical mass of 1 rm can be taken from the table (image 1).
Information on which rod to use can be found in the design guide. So, with a side length of more than 3 m, it is allowed to lay longitudinal reinforcement with a diameter of 12 mm or more. To balance the load resistance, two reinforcement belts are created.
For transverse reinforcement, there are the following restrictions: for a frame up to 0.8 m high, a rod from 6 mm is used, for a frame over 0.8 m high - more than 8 mm. Moreover, its diameter must be at least ¼ of the diameter of the longitudinal rods.
- tape length - 10x2 + (6-2x0.4)x3 = 35.6 m;
- sectional area - 60x40 \u003d 2400 square meters. cm.
Thus, the total cross-sectional area of the reinforcing belt must be at least 2400x0.001 \u003d 2.4 square meters. cm. This area corresponds to two rods with a section of 14, 3 - with a section of 12 or 4 - with a section of 10 mm. Considering that the length of the wall is more than 3 m, it would be optimal to use a rod with a diameter of 12 mm. To evenly distribute the load, it is placed in 2 belts of 2 rods.
The total length in the longitudinal direction when laying 4 rods, taking into account launches (10 m), will be:
35.6x4 + 10 \u003d 152.4 m.
Now let's do the calculation for the cross grid. The height of the frame, taking into account the indentation from the edges of 50 mm, will be:
600-2x50 = 500.
Since the frame height is less than 0.8 m, a profile with a diameter of 6 mm can be used. Let's check if it meets the second condition:
12/4=3<6, требование выполняется.
The size of one horizontal rod in millimeters, taking into account two indents from the edges, will be:
400-2x50 = 300,
and the size of the vertical:
600-2x50 = 500.
For one bundle, you will need 2 horizontal and vertical rods with a total length:
2x300 + 2x500 = 1600 mm = 1.6 m.
Such ligaments with a distance between them of 30 cm and a total length of the foundation of 35.6 m will be:
We calculate the total length of the transverse grid:
199x1.6 = 190.4 m.
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Calculation for a pile foundation
Let's calculate the amount of reinforcement for the foundation of piles for a similar house. With a distance between supports of 2 m, the foundation will require 16 piles 2 m long and 20 cm in diameter. How much rod will be required?
Each pile will take 4 rods, each of which has a length equal to the length of the pile plus 350 mm of launch for connection with the grillage frame. Total:
4x(2+0.350) = 9.4 m.
We have 16 such piles, so the total length of the periodic profile will be equal to:
16x9.4 = 150.4 m.
To connect the vertical profile that forms the frame of the column, we use smooth rods with a cross section of 6 mm. The connection is made at three levels. The size of one bar will be equal to:
3.14x200 = 628 mm.
For one pile you need 3 strappings:
3x628 = 1884 mm (rounded 1.9 m).
The total length of connecting elements for 16 points:
16x1.9 = 30.4 m.
The calculation of longitudinal reinforcement for a grillage is similar to the calculation for a strip foundation. A total of 152.4 m is needed. But the transverse rod, taking into account the height of the grillage of 400 mm, will require somewhat less. The total length of four profiles for one bundle will be:
4x(400-2x50) = 1200 mm = 1.2 m.
For 119 connections you need:
119x1.2 = 142.8 m.
For piles with a cross-sectional diameter of less than 200, 3 rods can be taken. With an increase in this size, the amount of reinforcement required increases.
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Calculation for a monolithic base
A monolithic reinforced concrete slab is laid under the entire area of \u200b\u200bthe building.
Among all types of foundations, slab foundations are the most materially costly, this applies to both concrete and reinforcement.
Laying a monolithic base is justified on soft and moving soils.
It provides maximum stability and best resists heaving forces. With any movement of the soil, the entire slab is lowered or raised, preventing distortions and cracking of the walls. Because of this, the monolithic base was called floating.
We will calculate the reinforcement for a slab foundation for a building of 10x6 m. The thickness of the slab is determined by calculating the load on the base. In our example, it will be 30 cm. Reinforcement is performed by two belts with a grid spacing of 20 cm. It is easy to calculate what is needed for each belt:
1000/200 \u003d 50 transverse rods 6 m long,
6000/200 = 30 longitudinal rods 8 m long.
The total length for 2 belts will be:
(50x6+30x8)x2 = 1200 m.
The connection of the belts is made by smooth profile reinforcement. We count in total.
As you know, any construction begins with the calculation and laying the foundation. The durability and strength of the building directly depends on how accurately this calculation is made. Being the basis of the building, the foundation takes on the load and redistributes it to the ground. The upper plane of the structure, which is the basis for external and internal walls, is called the edge, and the lower one, which performs the function of load distribution, is called the sole.
Characteristics of the strip foundation
The most common in private construction are reinforced concrete strip foundations. 
This is due to the relative simplicity of laying - with its construction, you can do without the use of lifting and special construction equipment. It is important to correctly calculate not only the calculation of the section and depth, but also the calculation of the reinforcement for the strip foundation.
This type of foundation is especially popular due to the fact that it is suitable for almost any soil and has the longest service life - up to 150 years.
Such durability is provided not only by the physical characteristics of concrete, but also by the choice of the correct reinforcement scheme. Despite the apparent strength, concrete is a rather fragile material and can burst even with slight shifts in the soil. To give it some plasticity, reinforcement is used. It is made using a metal rod. Moreover, most of it should have a ribbed surface. This is necessary to improve adhesion to concrete.
Choice of rod diameter
The calculation of the load on the foundation of a residential building, and, consequently, the choice of the diameter of the reinforcement is carried out by specialists during the development of the project. Most often, reinforcement with a diameter of 10 or 12 mm is used, much less often 14 mm. And only for small, light buildings on non-rocky soils, it is permissible to use a rod with a diameter of 8 mm.
Foundation reinforcement scheme
To ensure the strength of the foundation, it is necessary to strengthen both its lower part and its upper part. For this, two horizontal rows of steel bars are used, interconnected by vertical jumpers.
The main load in the tensile zones of the foundation is assumed by the longitudinal horizontal rods, while the vertical and transverse horizontal rods are used mainly as a frame, as well as to give the foundation shear strength. As a rule, the laying of four horizontal longitudinal steel ribbed bars is considered sufficient - two along the top and two along the bottom.
Vertical jumpers can be located at a distance of 30 to 80 cm from one another and are often made of a smooth rod of a smaller diameter, which is quite acceptable.
It should be remembered that the distance between the longitudinal reinforcement bars should not exceed 0.3 m, and to protect steel from corrosion, the bar should be buried in concrete by at least 5 cm.
Calculation of reinforcement for the foundation
When the decision on the foundation reinforcement scheme is made, it is important to correctly calculate the required amount of material so as not to pay twice for delivery if it turns out that it is not enough. Yes, and it is unlikely that someone will want to spend money on surpluses.
First you need to calculate how much ribbed reinforcement you will need. To do this, you need to calculate the perimeter of your house, add to this number the length of the internal walls under which the foundation will be laid, and multiply all this by the number of rods in the scheme.
As an example, let's calculate the amount of reinforcement required for laying a 5/6 m foundation with one internal wall 5 m long. Let's assume that the reinforcement scheme provides for 4 longitudinal rods with a diameter of 12mm. So:
(5+6)*2=22 - building perimeter
22+5=27 - total length of the foundation
27*4= 108 - total length of reinforcement
If you were unable to purchase a rod of the required length, and you plan to connect the segments, this must be done with a large overlap - at least 1 meter. Take this into account in your calculations. We will assume that each longitudinal rod of our frame will have one connection.
4 (number of rods in the scheme) * 5 (number of walls) \u003d 20
In total, we get 20 connections, which means that an additional 20 meters of reinforcement will be required. We add to the previous value and get:
Now we calculate the required amount of a smooth rod with a diameter of 8 mm for vertical racks and horizontal transverse jumpers.
Let's take the distance between the jumpers equal to 0.5 m. Then, dividing the total length of the foundation by this value, we will get the number of reinforcing "rings".
27 / 0.5 \u003d 54 - total number of reinforcing rings
If the height of the reinforcement grid is 0.5 m, and the distance between the bars is 0.25 m, then the reinforcement calculation will look like this:
(0.5 + 0.25) * 2 \u003d 1.5 - the perimeter of one "ring";
54 * 1.5 \u003d 81m - the total length of the rod.
In the calculations, it is also necessary to take into account possible trimmings and overlaps. It will not be possible to calculate their exact number, so experts advise adding about 10% to the resulting length.
We round up and get 90m.
Rarely enough, a rod or rebar is sold per footage. Much more often, or rather almost always, we pay not for the length, but for the weight of the product. In order to determine the exact quantity, a reinforcement calculation table is needed. Most large enterprises for the production of rolled metal products are required to comply with the requirements of GOST 5781-82, where the mass of one meter of one or another type of product is indicated. There is also GOST 2590-88, which regulates the weight of the steel circle. It should be noted that the figures in both documents are the same, and the only difference is that the step of the diameters of the circle is much less than the step of the diameters of the bar reinforcement. For bar reinforcement, these values are:
Rod diameter Weight in kg/m
Based on this table, it is possible to calculate the mass of reinforcement required for pouring our foundation:
128 * 0.888 = 113.664 kg - the required amount of ribbed reinforcement with a diameter of 12 mm
90 * 0.395 = 35.55 kg - the required amount of a smooth rod with a diameter of 10 mm
Of great importance is also the method of connecting the details of the structure. Many people mistakenly believe that the stronger the bars are connected to each other, the more durable the foundation will be and choose welding for mounting the frame. However, during the welding process, the structure of the metal is disturbed, which leads to its premature destruction. Experts advise connecting the reinforcement with knitting wire. The easiest way to do this is to crochet, like this:
Self-filling floor
Unfortunately, the cost of finished reinforced concrete structures is quite high. Therefore, quite often trying to save money, they are made independently. Overlappings, like any other reinforced concrete structures, require reinforcement. As a rule, a grid with a cell of 15/15 cm is used for this. With a ceiling thickness of up to 15 cm, one reinforcing mesh is sufficient. With an increase in the thickness of the plate, the number of gratings increases.
It is quite simple to correctly calculate the reinforcement of the ceiling. As an example, let's calculate an overlap of 5/6m in size. It should be borne in mind that the reinforcement should not reach the edge of the slab by 10 cm. Then the width of the fortified section will be 4.8 m. Calculate the required amount of material.
480/15=32 - the number of rods for reinforcing the slab in length. To this value it is necessary to add one more segment - edge. As a result, we get 33 rods with a length of 5.8 m each. Total: 33*5.8=191.4m.
In the same way, we calculate the amount of material for laying in width:
580/15 \u003d 39 (rounded) - the number of rods;
39 * 4.8 = 187.2 m - the length of the reinforcement required for laying in width.
We add both received values:
191.4 + 187.2 = 378.6m - the total length of the required material.
Now it remains only to calculate the mass of such a quantity of reinforcement using the table. As a rule, a rod with a diameter of 10 mm is used for these purposes.
As you can see, the calculation of the amount of reinforcement is quite simple. But still, you should not neglect the help of specialists, especially in the part that concerns the collection of loads on the foundation and determining the type of soil. Everything else you can successfully do on your own.
And its forms. The type and dimensions of the foundation are determined taking into account the calculated loads and. Earlier, as an example, we calculated the loads on the foundation (article) for a house measuring 6 m by 10 m with two internal walls. In this article, we will calculate the amount of reinforcement and knitting wire for the same house.
Calculation of the amount of reinforcement for reinforcing a slab foundation
Based on this type of foundation, we need reinforcement with (class A3 reinforcement) with a diameter of 10 mm or more. The larger the diameter of the reinforcement, the stronger the foundation.
The choice of bar thickness depends on the weight of the house and. If the bearing capacity of the soil is sufficiently high, i.e. the soil is dense and non-porous, then the foundation will deform less and the slab may be less stable. The greater the weight of the house, the greater the load on the foundation, the more stable it should be. During the construction of a light wooden, frame, panel house on the ground with good bearing capacity. Reinforcement with a diameter of 10 mm can be used. And, conversely, for a slab foundation of a heavy house on soft ground, reinforcement with a diameter of 14 mm - 16 mm is required.
How, they do it with a grid step of 20 cm. For a house measuring 6 m x 10 m, it is necessary to lay: (6 / 0.2 + 1) + (10 / 0.2 + 1) \u003d 31 (bars of 6 m each) + 51 ( bars of 10 m each) = 82 bars. There are 2 reinforcement belts in the slab foundation - upper and lower, therefore, we double the number of bars. It turns out:
82 *2 = 164 bars, incl. 62 bars of 6m and 102 bars of 10m each. Total 62*6+102*10= 1392 m of reinforcement.
The upper mesh must be connected to the lower mesh, connections are made at each intersection of the longitudinal reinforcement bars with the transverse ones. The number of connections will be: 31 * 51 = 1581 pcs. With a plate thickness of 20 cm and a distance of the frame to the plate surface of 5 cm, the connection will require rods 20-5-5 = 10 cm or 0.1 m long, the total length of the rods for connection is 1581 * 0.1 = 158.1 m.
The total amount of reinforcement for the slab foundation is: 1392 + 158.1 = 1550.1 m.
Calculation of the amount of knitting wire: at each intersection of the bars, we will have two - the connection of the longitudinal bar with the transverse one and their subsequent knitting with a vertical bar. The number of joints in the upper belt is 31 * 51 = 1581 pieces, in the lower belt the same number. Total connections 1581*2=3162 pcs.
For each knitting of reinforcement, a knitting wire folded in half with a length of 15 cm or 30 cm of net length is required.
The total amount of knitting wire is equal to the number of connections multiplied by the number of knittings in each connection multiplied by the length of the wire per knitting: 3162*2*0.3=1897.2
Strip foundation reinforcement
Calculation of the amount of reinforcement for reinforcing a strip foundation
It is subject to bending to a much lesser extent than the slab foundation, therefore, smaller diameter reinforcement is used to reinforce the strip foundation. In the construction of a low-rise building, reinforcement with a diameter of 10 mm - 12 mm is more often used, less often - 14 mm.
Regardless of the height of the strip foundation, two belts are used for its reinforcement: longitudinal reinforcement bars are laid at a distance of 5 cm from the surface of the strip foundation in its upper and lower parts. Longitudinal bars take on the load on the foundation, so ribbed reinforcement (class A3 reinforcement) is used.
The transverse and vertical bars of the reinforcing frame of the strip foundation do not carry such a load and can be made of smooth reinforcement (class A1 reinforcement).
With a strip foundation width of 40 cm, four longitudinal bars will suffice - two from above and two from below. With a larger foundation width, or when building a foundation on moving soil, as well as building a heavy house, it is necessary to use a larger number of longitudinal bars in each belt (3 or 4) when reinforcing.
The length of the strip foundation under the house 6 m by 10 m with two internal walls will be 6 + 10 + 6 + 10 + 6 + 10 \u003d 48 m
With a foundation width of 60 cm and reinforcement of 6 longitudinal ribbed bars, their length will be 48 * 6 = 288 m.
Transverse and vertical bars can be installed in increments of 0.5 m. With a foundation width of 60 cm, a height of 190 cm and indents of frame bars of 5 cm from the foundation surface, the length of smooth reinforcement with a diameter of 6 mm for each connection will be (60-5-5) * 2 + (190-5-5) * 3 \u003d 640 cm or 6.4 m, there will be 48 / 0.5 + 1 \u003d 97 pieces in total, they will require 97 * 6.4 \u003d 620.8 m of reinforcement.
Each such connection has 6 crossings for knitting reinforcement and will require 12 pieces of knitting wire. The length of the wire per bundle is 30 cm, the total consumption of knitting wire per frame for the strip foundation will be 0.3 m x 12 x 97 = 349.2 m.
Calculation of the amount of reinforcement for a columnar foundation
When reinforcing the foundation columns, it is sufficient to use reinforcement with a diameter of 10 mm - 12 mm. Vertical bars are made of ribbed reinforcement (class A3 reinforcement). Horizontal bars are used only to connect vertical bars into a single frame, they are made of smooth reinforcement of small diameter (6 mm is enough). In most cases, the reinforcing frame of the column consists of 2-6 rods with a length equal to the height of the column, the rods are evenly distributed inside the column. Vertical rods are connected along the height of the column at a distance of 40 cm -50 cm.
To reinforce a column with a diameter of 40 cm and a length of 2 meters, you can limit yourself to four bars of reinforcement with a diameter of 12 mm, located at a distance of 20 cm from each other, tied with smooth reinforcement with a diameter of 6 mm in four places.
The consumption of ribbed reinforcement for vertical bars is 2 m * 4 = 8 m, the consumption of smooth reinforcement is 0.2 * 4 * 4 = 3.2 m.
Thus, for 48 columns you will need ribbed reinforcement 8 m * 48 = 384 m, smooth 3.2 m * 48 = 153.6 m
Each of the four horizontal bars in the column is attached to four vertical ones. For their knitting, 0.3 m * 4 * 4 = 4.8 m of knitting wire is needed. For the entire foundation of 48 pillars, 4.8 m * 48 \u003d 230.4 m of wire will be required.
Calculation of the cost of reinforcement for the foundation
Having calculated the amount of reinforcement in running meters, we can calculate its weight and find out the cost. To do this, we need a table depending on its diameter. Calculation formula: (number of reinforcement in linear meters) * (weight of one linear meter of reinforcement for the corresponding diameter) * (cost of one ton of reinforcement) / 1000.