Reinforcement monolithic slab floors – mandatory technological process. Fittings included concrete structure takes on the load, increases the strength characteristics of the element.

Buildings with complex architecture have a plan non-standard shape, far from the square. In these conditions, factory-made hollow-core floor slabs are replaced with monolithic structures. Fine linked into the system other load-bearing elements of the building, transfer the load walls and foundation.

Reinforcement is called principle of sharing two materials to strengthen strength. The overall work of monolithic concrete and metal makes it possible to construct durable structures of complex shapes and large sizes.

Advantages of floor slab reinforcement

Reinforcement increases the ability of a structural element to absorb loads, withstanding deformations greater than the calculated values. The total load per square meter of flooring, taking into account temporary and permanent, is 400-450 kg.

Overlap working hard. At the same time upper part the floor slab is compressed, and the lower one, on the contrary, is stretched. Concrete easily tolerates compressive deformations and tensile loads accepts metal fittings.

Important!

Without a reinforcing frame, the structure collapses at the bottom of the slab.

The technology of pouring reinforced monolithic slabs does not require the use construction machines during installation and transportation. All work is carried out on site and is suitable for DIY.

In a private house with a span of up to 6 m and the slab resting on the walls along the contour, the recommended mesh size is 20 x 20 cm. The cross-sectional diameter of the lower mesh is 12 mm, the upper mesh is 10 mm.

What is reinforcement of a monolithic floor slab?

Reinforcement of the structure is carried out by installing reinforcing bars and meshes inside the walls/sides of the formwork before filling the form with concrete. The reinforcement is laid directly on the formwork according to the design.

General principle of slab reinforcement:

metal rods binding wire tied into nets;
form frame on the walls. The ceiling will rest on the load-bearing wall, this size is called the support area and is determined by the thickness and material of the wall;
bottom mesh rises from the plane formwork with 25-30 mm clamps;
the upper grid is located relative to the concrete level, retreating 25-30 mm;
at the junction of the slab with the walls is added support reinforcement.

Support reinforcement is needed for preventing concrete cracking. The supporting area of the floor slab is taken according to the design, but not less than 80 mm.

The distance from the edge of the formwork to the mesh is retreated to form a protective layer of concrete. Steel without a protective layer exposed to air and moisture corroded.

In places with a weakened cross-section, a large number of holes for laying utilities, the frame is strengthened. To do this, lay several additional straight rods 0.4-1.5 m long.

How to independently reinforce a concrete floor slab

Reinforcing concrete is an easy job to do yourself. Based on the project data, they select section, length and required quantity fittings, soft knitting wire. Welding connections are not used. Weld seam does not have the necessary plasticity, can be damaged by vibration when compacting concrete.

Clamps for lifting the mesh above the formwork can be made independently from scraps of reinforcement.

What kind of reinforcement is laid

The diameter of the rods depends on the accepted design load, is calculated during the design stage of the building.

The diameter of the reinforcement used in private individual buildings is usually 8-14 mm.

The corrugated periodic cross-section of steel rods increases the strength of the product and adheres well to the hardening concrete mixture.

Please note

The class of reinforcement intended for the construction of buildings and structures is ASH.

Reinforcement consumption

When purchasing, reinforcement is measured in weight units - kilograms, tons. When calculating the consumption of reinforcement, the sides of the floor are divided by the grid pitch and one rod is added. By doubling the quantity, you get the total number of rods of different lengths located along the length and width of the ceiling.

Concreting, carried out at high speed over large areas, is the most economical way to level the old floor and install a new one. Once cured, the smooth surface becomes an ideal base for any topcoat. To increase the service life of the screed, reinforcement of the concrete floor is used. This process is carried out using various materials and designs from them.

According to the functions performed and location, the screed is divided into the following types:

rough - rests on the ground;
multilayer - includes heat and sound insulating gaskets;
leveling - laid on a rough layer, serves as the basis for a floor covering or tubular insulation;
construction - lies on the floor slab.

It is advisable to reinforce a self-leveling concrete floor when constructing rough and multi-layer screeds (in the absence of a monolithic support, the effect of tensile and bending loads is enhanced), as well as to reduce the design layer of concrete.

Types of reinforcing structures and materials

1. Frame made of rods. Most often it is laid in two layers, made from rods with a diameter of 6 to 40 mm. Used when coating thickness is more than 8 cm.

2. Steel wire mesh. Used for multi-layer screed on the ground or to strengthen the coating in the garage, hallway, kitchen.

3. Polymer mesh. It does not strengthen the screed, but only prevents cracking during the concrete hardening process. Used for self-leveling floors, reducing cement consumption. The mesh is installed directly on the base or on the heat-insulating layer.

4. Reinforcing fiber for concrete. There are two types: metal and polypropylene. Polymer fiber gives concrete resistance to cracking during shrinkage, temperature fluctuations, and enhances water-repellent properties.

Metal fiber increases the resistance of concrete to vibrations. Replacing reinforcing mesh steel fiber, save time (elements are introduced directly into the mixer), reduce the thickness of the screed. In this case, microcracks lose their ability to expand.

5. Combined reinforcement. In addition to the frame mounted in the lower section of the covering, its upper layer is filled with fiber - this is how the concrete floor screed is protected from cracks. The method is applied to the entire surface or in areas of increased loads (where the floor is adjacent to walls or columns). Fiber should be dosed according to the instructions.

The main stages of concrete pavement reinforcement

The most labor-intensive is the construction of a concrete floor with a soil base. First, according to technology, a gravel-sand mixture is laid, then a foundation slab, parabarrier film, thermal and waterproofing. Next, a reinforced layer of concrete is installed.

1. Depending on the thickness of the screed, in private construction it is reinforced with a frame made of rods or wire mesh. The diameter of the reinforcement for reinforcement is taken from the range from 8 to 20, and the wire - from 4 to 6 mm. The cell size ranges from 10 to 20 cm.

2. A frame made of solid rods is knitted using wire with a diameter of 2 - 3 mm, the layers of the frame are attached to the ribs. If scraps of material are used, they are overlapped with an overlap of half a meter.

Wire mesh is purchased ready-made (cells 5 - 20 cm) or knitted by hand. Purchased products are connected with wire with an overlap of 1 - 2 cells.

3. The finished structure is laid on clamps (“chairs”) at a level of about 3.5 cm from the base. When pouring, the steel elements must be in the middle of the concrete layer - in this case, the loads are evenly distributed over the surface of the coating, its mechanical resistance is ensured, and there is no metal corrosion.

Reinforcement consumption for floor reinforcement

Table 1 shows the data for the mesh version of the reinforcement, and Table 2 shows the data for single reinforcement with steel rods from 10 to 16 mm.

Material, per 1 m2 of floor

1.6. The standards provide average grades of electrodes.

1.7. The costs of installing metal structures and steel cores used as rigid reinforcement should be determined in accordance with the relevant standards of Book 9 "Metal Structures".

1.8. The standards take into account the construction of structures at a height (depth) of up to 15 m from the surface of the earth (with the exception of structures of special structures). When determining the costs of work at elevations above (below) 15 m from the ground surface, wages and labor costs should be adjusted by the coefficients given in (,).

Standards for the construction of special structures (cooling towers, silos, elevators, mine headframes and nuclear power plants) are not subject to adjustment.

1.9. The costs of installing grillages should be determined according to the relevant standards in Table. and for the construction of similar foundations, for example, grillages on single piles or clusters of piles for individual columns - according to the standards for foundations of the corresponding volume for columns; grillages in the form of slabs along a pile field - according to the standards for foundation slabs, grillages in the form of strips along rows of piles - according to the standards for strip foundations, etc.

1.10. The costs of installing anchor bolts and embedded products for fastening building structures should be determined by.

The costs of installing anchor bolts and embedded products for fastening equipment should be determined in accordance with the Guidelines for the Application of Costs for Equipment Installation.

1.11. Standards for the installation of water supply and sewerage tank structures should also be applied when determining the costs of similar technical requirements and construction conditions (tanks for petroleum products, etc.).

Concrete is a very strong material that can easily withstand loads acting on it from above - it is not subject to compression. But during operation, the foundation is also affected by tensile forces, which it cannot withstand. Reinforcement is needed to strengthen concrete base and protect it from stretching and destruction. It is important to correctly calculate the amount of building material that will be required to strengthen the fundamental supporting part, and for this you need to know the consumption of reinforcement per 1 m³ of concrete.

Factors affecting material consumption

The consumption of reinforcement per cube of concrete and for the reinforcement of the entire foundation as a whole depends on several important factors:

Density of the solution (composition matters)– the lower the density indicator, the finer the mesh structure in the reinforcing frame should be – the pitch decreases.
Type of structure and its weight– standards for the use of building materials for a specific type of structure are specified in the following regulatory documents: GOST, GESN and FER.
Size (length, width and depth) concrete supporting part determines the number of longitudinal and transverse elements in the reinforcing frame.
Soil type– for stable soils with high bearing capacity, use metal products with a diameter of 10, otherwise – 14–16 millimeters.
Element class, which increases strength, and the cross-sectional area of the rods determines the weight of the future structure and the load on the ground.

The type of foundation also influences - for each type there are approximate (indicative) indicators of the cost of reinforcement per cube of concrete:

For tape sample– 20 kg per 1 cubic meter.
For columnar foundation – 10 kg per 1 cubic meter.
For slab (has two longitudinal belts - upper and lower)– 50 kg per 1 cubic meter.

Norm calculation options

It is not difficult to calculate the consumption of reinforcement per cube of concrete. Between the rows load-bearing structure with stable soil (not subject to buoyancy and swelling), the distance can be 20–30 centimeters. It is necessary to retreat 5 centimeters from all edges so that the solution completely hides the frame and protects it from its influence environment(from corrosion). For the transverse strips of the reinforcing frame, in order to save money, products of the smallest diameter and cost are chosen.

Example of calculations No. 1 (1 m³)

Calculation of consumption of reinforcement with a diameter of 12 millimeters for horizontal rows:

In one concrete cube (that is, in a block with a length, width and height of 100 cm) 4 longitudinal rows will fit (step 30).
Each row will have 4 stripes.
Total: 4*4=16 ninety-centimeter rods (100-2*5).
The total length of the reinforcing elements is 16*90=1440 (14.4 m).

Calculation of reinforcement consumption for transverse horizontal and vertical elements made of material 8 mm thick:

One cross section will accommodate 4 recumbent and standing ninety-centimeter rods (8 in total).
The section is repeated every 0.3 units, which means that it is present 4 times in one cube.
Total: 8 * 4 = 32 ninety-centimeter metal rods, located across the width in one cube of concrete.
The total length of the material is 32*90 = 2880 (28.8 m).

Conclusion: to strengthen a concrete block measuring 1 m³, you will need 14.4 twelve-millimeter and 28.8 meters of eight-millimeter reinforcement.

To calculate the total amount of building material required to strengthen a particular foundation, you need to know its type and exact dimensions.

Example of calculations No. 2 (tape sample)

Calculation of the amount of metal products for strengthening strip foundation width 40, perimeter 3000 (9*6), height 100 centimeters:

The width will fit 2 strips of reinforcement (pitch - 30 cm, thickness - 10 mm).
A 1 meter deep base will accommodate 4 horizontal rows.
Total: 4*2=8 strips, length equal to the perimeter of the supporting parts, that is, 3000 centimeters.
The total length is 8*300=24000 (240 m).
The cross section will fit: 4 horizontal rows of thirty-centimeter rods with a thickness of 6: according to the formula (40–2*5) and 2 vertical ninety-centimeter metal rods (100–2*5).
Total: 4*30+2*90=120+180=300 (3 m) reinforcement in one section under consideration.
The perimeter of the base is 3000, and the transverse section will be repeated every 30 cm, that is, 3000/30=100 times.
The total length is 100*300 = 30000 (300 m).

Conclusion: to strengthen a strip foundation 40 centimeters wide and 100 centimeters deep for a 6*9 house, you will need 240 ten-millimeter and 300 meters of six-millimeter metal products.

Converting linear meters to tons

To convert linear footage into kilograms or tons, you need to have information about how much 1 meter of a given metal product of a certain diameter weighs. The most common types have the following indicators:

16 – 1578.
14 – 1208.
12 – 888.
10 – 617.
8 – 395.
6 – 222.

Mass indicators of the strength-increasing element for 1 m³:

12-14.4*888=12787.2 g (12.787 kg).
8-28.8*395=11376 g (11.376 kg).
The final weight is 12.787+11.376=24.163 kilograms (0.024 tons).

Indicators of the mass of metal products for a strip foundation (from example No. 2):

10-240*617=148080 g (148.08 kg).
6-300*222=66600 (66.6 m).
Total weight – 148.08+66.6=215.4 kilograms (0.216 tons).

Calculating how many materials will be needed to create a reinforcing load-bearing structure of any foundation is not difficult if you know the principles outlined above. This is necessary in order to purchase a sufficient amount of building materials and avoid unnecessary costs.

When building a house, purchasing is always required large quantity building materials. When building a foundation, a larger volume of building materials such as concrete and reinforcement is required. The amount of concrete required for the foundation and the consumption rate of reinforcement per 1 m3 of concrete are the main ones the most important issues everyone who wants to build their own home.

Consumption of reinforcement for reinforcement

First of all, I would like to remind you that saving on building materials when building a house it is not recommended at all, because... this can result in much higher costs for subsequent repairs and a shorter service life.

The consumption of reinforcement per 1 m3 of concrete must be calculated in accordance with accepted standards. If this is a foundation, then its reinforcement is required, since it has to withstand a large load from the house standing on it. Determining the exact amount of reinforcement required is very important for the foundation, because... its strength depends on this.

It is important to understand and understand that reinforcement can be longitudinal and transverse, smooth reinforcement is a reinforcement that does not have reflections, and reinforcement is also divided into prestressed and non-prestressed.

Longitudinal reinforcement resists stretching and prevents the appearance of vertical cracks when stretching a reinforced concrete structure. If the structure is also subject to compressive forces, the reinforcement distributes some of the load and takes it upon itself along with the concrete.

Transverse reinforcement prevents the appearance of various cracks that may appear from stress near the supports, as a rule these are inclined cracks.

Calculation of reinforcement for the foundation of a house

For the foundation of a house type monolithic slab, ribbed reinforcement is used (type A3), the diameter of which must be at least Ø10 mm. The diameter of the reinforcement is the most important characteristic of the reinforcement, because the reliability and strength of the structure being constructed depends on it; the larger the diameter of the reinforcement (the thicker it is), the stronger the structure will be.

When choosing fittings, they look first of all at soil type And weight of the building being erected. For example, if the soil has excellent characteristics and has a small antinode, it means that it will change (deform) less due to the weight of the building, and this in turn means that you can use a foundation with less stability, which can cost an order of magnitude less. The load on the foundation is determined by the weight of the entire building, which must be calculated in the project (to a large extent depends on the materials used for masonry walls, partitions and the type of roof used), the greater the weight of the house, the greater the load on the foundation, and accordingly the deformation will be greater.

For example light wooden house erected on soil with excellent bearing capacity, when constructing a foundation such as a monolithic slab, reinforcement with a diameter of ø10 mm.

When constructing more heavy houses and on soil with less bearing capacity, it is necessary to use reinforcement with a larger diameter ø14-16 mm.

Calculation of reinforcement for a monolithic foundation of a 6x6 house

The reinforcement frame for a foundation such as a monolithic slab is made in steps of 200 mm, taking into account this step, we will need 31 reinforcement rods for laying in parallel and the same amount of reinforcement for laying perpendicularly, so in total we will need 62 steel rods.

For the slab, it is necessary to make two such armored belts, one should be on top, the second on the bottom, respectively, you will need 2 times more rods, i.e. 124 pieces, each 6 m in length, the total amount of required reinforcement is equal to 124 pieces * 6 m = 144 m (linear meters).

It is also necessary to connect these two frames of reinforcement to each other; for this, a connection is made in the places where the longitudinal reinforcement intersects with the transverse reinforcement.

To calculate the number of frame connections you need to multiply 31 by 21 = 961 rods. For example, let’s take a frame 200 mm thick, located 50 mm from the ground surface, in this case we will need a separate reinforcement bar for each connection, each of which is 100 mm in length (200 mm - 50 mm at the top and -50 mm at the bottom). It turns out that to make all the connections of the frame, we need 0.1 * 961 = 96 m of reinforcement.

The total is 744+96 = 840 linear meters of reinforcement we will need to create the frame of a monolithic foundation.

Calculation of reinforcement for strip foundations

A strip foundation is characterized by one feature: its height, as a rule, is always much greater than its width. if we take the width of 35-45 cm, then its height will be 70-80 cm. This is done so that the strip-type foundation reliably resists bending. This allows you to use reinforcement with a smaller diameter for reinforcing the strip foundation, and you can also use smooth reinforcement.

When making a foundation for a private house, reinforcement is used for reinforcement with diameter ø10 or ø12 mm, Sometimes ø14 mm. For a strip foundation, it is typical to use only 2 armored belts and it does not matter what the height of such a foundation will be.

Reinforcement of a strip foundation is carried out 5-7 cm from the top and bottom surfaces of the foundation, where reinforcement frames are located, which help the foundation to withstand loads and resist deformation of the foundation.

Let's calculate the amount of reinforcement required for a strip foundation. Let the width of the foundation be 40 cm, then it will require two pieces of reinforcement for a horizontal arrangement at the top and two at the bottom; sometimes an armored belt consisting of 3 and 4 bars of reinforcement can be used, as a rule, this is only necessary if the soil is sufficiently mobile and the weight of the house big.

So, the foundation strip house 6 by 6, having 1 load-bearing wall, will be 6 m *4 = 24 m + 6 meters for the internal load-bearing wall, total 6*4+6=30 m is the total length of the foundation.

The consumption of type A3 reinforcement to create a reinforcing structure based on 4 rods will be 30 m *4 = 120 meters.

Vertical reinforcement is installed in increments of 0.5 meters. If the width is 0.3 and the height is 0.7 m, the distance from the surface is 50 mm, then for each connection you will need 1.6 meters of reinforcement (smooth) ø6 mm. There will be 61 connections in total, it’s not difficult to count total consumption fittings, 61*1.6= 97 meters. In general, it turns out that each connection will have 4 reinforcement ties. For one such bundle we will need approximately 30 cm of wire for tying the reinforcement. In total, you will need 0.3 * 4 * 61 = 73 meters of wire for dressing.

Calculation of A500C reinforcement for a columnar foundation

When constructing a columnar foundation, you can use ø10 mm reinforcement. Ribbed reinforcement is used for rods that are positioned vertically, and horizontal reinforcement is needed only to ensure reliable ligation, to create a single frame for the post. The reinforcement frame for a columnar foundation usually contains 3-4 in each column; all reinforcement must be longer than the column itself. If pillars for a foundation with a larger diameter than 20 cm are required, then more reinforcement is required and it must be distributed evenly inside the pillar.

If you need standard two-meter poles with a diameter of 200 mm, then 4 bars of reinforcement with a diameter of ø10 mm each located at a distance of 100 mm from each other. Dressing is carried out in 4 places, using smooth fittings with a diameter of ø6 mm.

Total for one pillar you will need 2 m*4= 8 meters ribbed reinforcement and 0.4 m*4 = 1.2 meters of smooth reinforcement.

If necessary 30 pillars for the foundation, then the consumption ribbed reinforcement will be 8m*30= 240 meters, and consumption smooth reinforcement 1.2 m*30= 36 meters.

The cost of reinforcement for the foundation

Having determined the required amount of reinforcement for each type of foundation, it is easy to determine its weight and calculate the cost.

The price per ton of reinforcement averages 24,000-26,000 rubles.

For a monolithic slab type foundation, according to the calculation, we get 840 linear meters of reinforcement; let’s take a diameter of ø14 mm, the weight of such reinforcement is 1.2 kg, the total weight is 1008 kg. This turns out to be a little more than one ton of reinforcement. The price for the fittings will be 25,000 rubles.

Let's calculate the price for a strip-type foundation; according to the calculations above, we got 120 meters of reinforcement; let's take a diameter of ø12 mm for calculations. The weight of one meter of such reinforcement is 0.89 kg, 120 * 0.89 = 106.8 kg of reinforcement. + vertically located reinforcement ø6mm, it needs 97 meters, it weighs 0.22 kg, 97 * 0.22 = 21.34 kg, plus the purchase of dressing wire, the cost of reinforcement for such a foundation is much lower, about 3000-4000 rubles.

Let's calculate the cost of reinforcement for a columnar foundation, we need 240 meters of ribbed reinforcement with a diameter of ø10 mm, the weight of one meter of such reinforcement is 0.62 kg, 240 m * 0.62 = 149 kg. You also need 36 meters of smooth reinforcement with a diameter of ø6 mm, its weight is 0.22 kg, 36 m * 0.22 = 8 kg. The total cost of the fittings is 4000 -5000 rubles.