How to make a reinforced concrete floor with your own hands. Lightweight monolithic ceiling on a profiled sheet - constructive, experience of portal participants. Caring for concrete after pouring

Konstantin, Novosibirsk asks a question: Hello. I had a slight hitch while building my house. Please tell me how to properly fill the floor slab yourself and what is required for this. Nowadays, reinforced concrete floors are used in construction, since structures of this kind have a very high degree of strength and can withstand large load-bearing loads. How can you fill a slab of such a floor yourself? What is needed for this and what is the sequence of actions? The expert answers:

Hello. To learn how to properly fill a floor slab yourself, read the recommendations below. Filling the floor is carried out in several stages. Moreover, if the technology for filling the floor is violated, this can lead to very disastrous and unpredictable consequences.

To fill the floor yourself, prepare materials such as wooden beams or boards to make formwork. To fasten them you will need screws and, of course, a screwdriver to fasten the formwork parts. You will definitely need chipboards (chipboards) or sheets of metal so that the ceiling has a flat surface. Please note that the accuracy of the formwork directly affects the durability and strength of the floor itself. Therefore, if necessary, still turn to professionals for help.

Next, boards should be laid across the entire room, supported by load-bearing walls and additional installed supports. They need to be laid on edge to ensure greater strength. Supports must be installed strictly vertically to ensure maximum load-bearing capacity. Verticality can be checked using a plumb line. The distance between the boards should be about 1 meter. On top of them is a roll made of sheets of iron or chipboard. The sheets are attached to the boards with screws or nails. The main purpose of the boards is to prevent sagging and destruction of the floor itself when installing reinforcement. We advise you to use metal pipes to make the supporting structure.

For even greater strength, the floor slab must be reinforced. For this purpose, steel reinforcement is used, the cross-section of which must be strictly coordinated with the design of the house being built. The reinforcement bars should be laid longitudinally and transversely at a distance of about 20 cm from each other. The rods are fastened together with twisted wire. The ends of the reinforcement must extend beyond the edges of the load-bearing walls of the building.

After the formwork and other stiffening elements are installed and securely fastened, you need to fill the slab. For this purpose, concrete grade M200 is used, which is mixed with sand and crushed stone. Concrete is poured using a pump directly from the mixer. It is important to take into account that pouring should begin from the farthest corner of the ceiling, gradually moving towards the outer edge. The poured concrete is carefully leveled and compacted using a vibrator.

After formation, the overlap is left to dry for some time until it completely hardens. But due to the large thickness of the layer, the concrete dries unevenly and cracks may appear on the surface. To avoid this, you need to evenly moisten the surface of the stove twice a day using a hose with a sprinkler.

During the entire period of pouring the floor, you must constantly check the design documentation.

How to pour concrete - technologies for forming a monolith

The construction of buildings and structures is connected with concrete everywhere and inextricably - there is no modern capital structure in which there would not be a concrete part, at least at the base. Concrete and reinforced concrete can be used in different ways - in the form of finished parts or a monolith, but in any case it will need to be poured into a pre-prepared form, formwork.

There are several ways to direct the solution into the mold and distribute it evenly there.

How concrete is poured into different forms

Pouring concrete into the formwork must be carried out so that the monolith or building part meets the requirements of strength, frost resistance and water resistance, and this can only be achieved if the form is uniformly filled with mortar and there is time for the main processes - setting and hardening of concrete.

Ceilings made of concrete: installation of floors, assembly of formwork, reinforcement, pouring scheme

The first task to be solved during the pouring process is the distribution of the solution throughout the entire volume of the formwork.

To achieve the first goal, several methods are used:

• direct pouring, filling - the solution is poured directly into the formwork, first filling corners and difficult places, then the center is filled, from which the solution is distributed to the sides;
• pouring under pressure is used in cases where the volume of the mold is large, but the penetration of the solution is limited by the frequency of the reinforcement and the presence of complex cavities - after the formation of a small initial layer, the outlet of the hose is placed under the surface of the solution;
• in the most difficult cases, when it is necessary to form a monolith close to groundwater, the monolith is formed separately - a layer of filler (crushed stone) is laid, onto which a sand-cement mixture is supplied;
• the most accurate, labor-intensive technology is channel pouring or extrusion with concrete, which is performed under pressure through small holes if the shape of the cavity does not allow it to be filled from above by gravity or vibration.

To create foundations and medium-strength monoliths, M300 concrete is used, the most common of custom mortars, suitable for private and low-rise construction.

In large projects, concrete of this grade is used to fill parts of the structure that accept part of the loads, but do not determine the entire strength of the structure. Continuity of concrete supply is ensured using mobile and stationary concrete pumps.

Gravity and vibration compaction of concrete

The final strength properties of concrete are formed at the stage of pouring and compaction due to the influence of gravity, mechanical and chemical factors on the solution.

Gravity filling of the mold does not always allow you to fill all the cavities and obtain reliable and complete adhesion of the mortar to the reinforcement. To enhance the effect, vibration is applied, which can be set in three ways.

Deep vibratory compaction

Deep vibration - vibrators are immersed in the solution mass, which force the future monolith to be evenly distributed throughout the entire volume, expel air and promote compaction and shrinkage of concrete.

With this compaction method, the effect of high quality volumetric structures is achieved, in which the distribution of the solution is hampered by the frequent arrangement and complex configuration of the reinforcement. In private construction, vibration is sometimes replaced by piercing the poured mortar with a rod to the bottom of the formwork.

Sealing from surface

Surface vibration - vibrating planks and vibrating platforms affect only the surface of the concrete if a large-area monolithic slab is created.

After a few hours, the solution is deeply compacted, a strong and well-connected structure of aggregate, cement and sand is formed in the absence of air.

Shape vibration

Vibration of the entire mold is used in the manufacture of individual concrete parts. This method requires sophisticated equipment, so it is practically not used on construction sites.

Chemical additives to concrete - improving the quality of pouring

For monolithic work, when increased demands are placed on the strength of the structure or individual parts, M400 concrete, which is sensitive to vibration, is used.

The structure and ability of concrete to harden is affected by the speed and completeness of the process of cement hydration and its interaction with water, therefore the concrete solution is sensitive to external temperature.

Already at -5 C, a gradual slowdown in hydration begins, and this leads to the fact that the monolith hardens slowly, its structure is formed with sedimentation and subsidence of the filler, and the bonds between sand and cement are incomplete. To compensate for loss of strength in frosty weather, concrete is poured with special salt additives that prevent water from freezing.

Structure and quality of concrete monolith

When working with large volumes and monoliths of complex shape, it is necessary to achieve structural unity of the structure, therefore the pouring process can be organized continuously or divided into technological stages with the formation of hot and cold seams.

In the first case, when stopping the pouring, a pause is made for 12 hours or less so that the setting process begins, and a new layer of solution is applied on top. In the second case, it is necessary to wait for the monolith to partially harden and continue pouring with a cold seam after a break of at least a day.

Why is so much attention paid to techniques and methods for filling formwork? Is it possible to abandon any stage or operation without compromising the quality of the structure? Concrete is not an initially homogeneous medium with evenly distributed components; it is a more complex structural mass, which must be given certain properties.

All techniques and methods for pouring concrete mortar into formwork are technological operations, described many times, subject to standards, therefore the use of any method must be reflected in the project and technological maps.

Ignoring possible changes in the characteristics of a monolith is dangerous; this leads to a violation of the integrity of the structure, cracking of concrete, and destruction of the building.

Calculation of the thickness of the base plate: monolithic foundation of aerated concrete house

Regarding the functionality/cost ratio for the layout of this type of foundation, it is preferable to consider the more familiar analogues - tape or pile.

However, in civil engineering construction, a base plate is installed much less frequently. The main reason is that private developers are poorly aware of all the advantages, characteristics and specificity of monolithic construction. This article will fill the knowledge gap and allow you to choose the best version of reliable support for each structure, combined with reasonable savings.

1. Advantages and disadvantages of a monolithic base
2. How to determine the required thickness?
3. Installation technology

There are several names (floating, continuous) and variations to this basis.

It all depends on the version and location of the device. In the design, slabs are known in monolithic, prefabricated, “Swedish”, ribbed, cardboard, reinforced (or without) and many others. Thinking about all the technical solutions is unwise. For an individual builder, an interesting monolithic reinforced concrete slab is most suitable for small private buildings. Therefore, attention will be paid to this, especially since the technology for its production is one of the simplest.

features

Advantages:

Increased load capacity. Due to the uniform distribution of the entire load, the monolithic plate causes little pressure on the floor, regardless of the thickness of the filler. An excellent option for beam houses, cellular concrete, even brick.

2. Spatial rigidity. This eliminates the possibility of clogging in certain areas (eg tape) and cracks in concrete, walls or split joints.

Versatility in use. The panel base is suitable for all floors, including problematic ones.

4. Simplified construction technology. Installation of a monolithic slab does not require extensive excavation, which saves a lot of time.

On a note! This does not apply to the possibility if the project (scheme) provides basement (technological) space. In this case, the cost of monolithic foundations can reach ⅓ - ½ of the total construction estimate.

Possibility of high quality insulation. Options - installation based on polystyrene foam, introduction of special solutions / additives.

6. Reducing concrete consumption. Although this is only true when deploying unlocked monolithic plates.

flaws:

Many of them are relative, but worth mentioning.

Complexity of calculations. This concerns the thickness of the future disk. If it is a basement building, it is better to choose another basement option. First, construction costs will increase dramatically. Secondly, calculations for a monolithic plate will become much more complex.

2. High costs. Here much depends on the specific design, but it cannot be denied that with such a design savings are achieved in other materials.

If the base slab is shallow, with little thickness, it can be impressive.

3. Work intensity. The question is how well the construction work is organized. For example, using a “vehicle mixer” greatly simplifies the technology of mixing concrete mixture and saves time.

The same applies to the accuracy of calculating the thickness of a monolithic base.

4. Some problems with individual projects. First of all, when implementing a scheme with a basement and during the construction process, there is a relief on the floor.

Calculation of panel thickness

Initial data for calculating the thickness of the foundation:

• Soil type
• Configuration of underground aquifers.
• Soil freezing level.
• Availability of a drainage system on site and its layout (if installed).

What is indicated:

Thickness of concrete reinforcing elements (rod, mesh).

2. The size of the anchor cells and the spacing between layers in the monolith.

The distance of the rod from the upper and lower cuts of the base.

Advice. If you saved something, just don't calculate it. The instructions on thematic sites devoted to this topic give only general recommendations on the optimal concrete thickness in the range from 200 to 400 mm. But this does not take into account the specifics of laying a monolithic foundation for a specific structure in a given area.

The difference in this basic parameter for the same type of structure can be significant.

For example, the thickness of a panel for a wooden house varies over fairly large boundaries and depends on the characteristics of the floor, although it is a relatively light structure on 1-2 floors.

*Dimensions are in “mm”.

• Section 12.
• Two levels of reinforcement, the interval between which is 70.
• The reinforcement distance from monolithic concrete parts is 50.

Calculation: 12 x 2 + 70 + 50 x 2 = 194.

Rounded - 20 cm.

For example, this is the smallest thickness of a slab for a house made of aerated concrete. But subject to the construction of monolithic foundations for shallow burial on good, dense soil. Therefore, all calculations are desirable for training a specialist.

Installation procedure

In addition, only the main stages of construction of a monolithic structure will gradually be taken into account without taking into account the features of the terrain and the structures themselves.

Marking the territory.

It is produced after complete removal in accordance with the construction scheme and the most acceptable method - “golden triangle”, diagonals, etc.

2. Excavations.

The depth of the notch is determined by the total thickness of the base plate and the "cushion". For the latter, this parameter is selected within 350 mm. If additional insulation of the base from Penoplex is expected, the amount of soil extracted will be increased accordingly.

Opinions about the structure of the “pillow” vary greatly.

There are recommendations for sleeping ASG; some recommend that sand be used alternately with crushed stone. It should be borne in mind that the coating absorbs moisture from the ground as little as possible; more will be the foundation. It follows from this that it is preferable for the coarse sand under the monolith to compress its layer and from the upper gravel, which is also compressed.

On a note!

Before placing the “pillows” it is necessary to accumulate as much soil as possible in the hole. The reliability of the monolithic structure depends on this. In addition, it is advisable to place a bottom with geotextile underneath.

3. Installation of formwork.

If the foundation is slab, you may want to limit yourself to narrow planks that are positioned around the perimeter of the excavation and sunk into one structure.

As an option, foamed polystyrene panels are available in the form of detachable panels.

Thermal insulation layer.

Not necessarily, but when laid under the Penopolix monopoly, the floors of the 1st floor will be much warmer.

Reinforcement.

The first network is not installed on waterproofing (insulation), but on special devices called “concrete protection”. Their height determines the thickness of the layer from the reinforcement to the bottom cut of the plate. There are different versions of this support, so it's not difficult to choose (or do it yourself).

Filling out the solution.

There is nothing difficult in this operation if something is planned in advance.

• When choosing concrete, you should focus not only on your brand (at least 300), but also on the size of the aggregate fractions.

  Do-it-yourself monolithic hanging devices

  More, later, it will be more difficult to reduce the decision. And given the small thickness of the panel, this will need to be addressed.

• You can't leave work the next day.

  The monolith smoothly merges together. Therefore, at least one assistant will be required, although the foundation is small and size.

During the construction of houses, garages, cottages, and other structures, a stage comes when it is necessary to perform floors. Floors can be interfloor or ceiling, made of wood, using wooden beams, using concrete slabs or by pouring concrete. Each of these floor installation methods has its own legal right to exist, supported by the economic feasibility of using a specific option in each individual case.

In this article, we wanted to talk about a specific case, namely the pouring of concrete interfloor (ceiling) floors. Before we talk about the methods of installing these floors, we wanted to touch on the topic of the use and installation of poured concrete floors, let's talk about their feasibility and advantages relative to other similar floors.

Advantages of poured concrete floors (monolithic concrete floors)

First of all, monolithically poured concrete floors should be considered as an alternative to slab floors.

Wooden floors are too different from concrete-monolithic floors, first of all in price, monolithic ones are much more expensive, secondly, in strength, they are much stronger, thirdly, in durability and other not so significant differences.

That is why it is worth comparing, first of all, with slab floors. Thus, in some cases, monolithic (concrete) floors are cheaper, which is an undeniable advantage, and at the same time they have similar strength properties. Another important advantage is that poured monolithic concrete floors can be made of any complex shape, almost anywhere, which is sometimes impossible for standard, factory-made concrete products.

Example of installation of concrete, monolithic floors

Do-it-yourself floor slabs. Drawing and cost of making the plate

In this case, this is a particular example; we will describe possible improvements that could be made to improve the quality of the overlap as alternative solutions. So, first of all, it is necessary to build a support for the poured concrete mixture and formwork.

After this, it is necessary to install the fittings.

It is best to carry out installation using mounting wire and lay two layers of grating.

One reinforcing grid should be at the bottom, the second, laid through the “frogs,” should be at the top.

Such a monolithic floor will more correctly perceive the bending load, due to the operation of the reinforcement in the most stressed places, which will significantly increase the strength of the floor.

Afterwards, we begin pouring concrete.

It is best to buy the planned volume of concrete for this operation in order to carry out the entire pour at one time, since only in this case can you guarantee equal strength of the entire monolithic floor structure.

Also, you should not pour all the concrete into one place to prevent subsidence and collapse of the floor formwork.

It is best to supply the concrete mixture evenly over the entire area; in extreme cases, quickly distribute it over this area using any alternative method.

The final stage will be holding the concrete mixture under certain conditions (temperature and humidity), which will ensure technological hardening of the mixture and its quality.

So, you can read more about the process of hardening the concrete mixture in the article “How to pour a concrete floor screed.”

Afterwards we dismantle the formwork, and our concrete floor is ready for use.

Calculation of formwork holding monolithic, concrete floors during pouring

Someone with a certain amount of construction knowledge can install a concrete floor based on his life experience, or, as they say, “by eye.”

We want to offer you another, albeit not an institute calculation, but which to a high degree will become your successful guarantee of successful work.

Calculation of formwork for this type of floor should be done according to three main parameters:

1. For longitudinal load on the supports holding the formwork. Initially, it is necessary to calculate the cross-section of the supports for the holding formwork. Is this value not so critical? as subsequent parameters, which is why you most likely will not have problems with it.

σ = N/F ≤ Rс where σ are the internal normal stresses arising in the cross section of the compressed beam, kg/cm2; N – mass of our formwork and poured mixture, kg; F is the cross-sectional area of ​​the column cm2; Rc is the calculated compressive resistance of wood at the yield point, kg/cm2.

(For pine, the calculated resistance is 140 kgf/cm2)

2. For bending supports under load. Also, do not forget the factor that the bending rigidity of the beam changes with its length. So, as the length of the holding beam increases, its flexibility also increases, and its rigidity decreases accordingly. In order to take this factor into account, it is necessary to take the cross-sectional area of ​​the beam with a correction factor φ

σ = N/φF ≤ Rc

the coefficient will depend on the ratio of diameter to length; to facilitate calculations, it can be taken from the series below

L/d = 5 10 20 30 40 50
φ = 0.9 0.85 0.5 0.25 0.15 0.08

To ensure the integrity of the formwork base The last thing you should pay attention to is the strength of the retaining formwork onto which the concrete will be poured. So the formwork must withstand not only the static mass of concrete, but also the dynamic load during its pouring.

Also, do not forget about the possible temporary overflow of concrete to a specific local place and the weight of the worker who will distribute the concrete in it. As a result, the permissible thicknesses of plywood formwork, with a margin of 1.5, with a span of no more than 1 m, can be taken from the series below.

Plywood thickness 18 mm 21 mm

Thickness of the poured concrete floor layer up to 9 cm up to 12 cm

Now you can not only pour the concrete floor, but also pre-calculate the auxiliary technological elements for its installation.

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Reinforced concrete floors. Monolithic slab floors.

Monolithic beam floors, ribbed floors.

Monolithic ceiling with liners.

Reinforced concrete floors. Depending on the construction method, they are divided into monolithic and prefabricated. The advantage of such floors is their high load-bearing capacity. The compressive strength of concrete is used here since the dimensions of these floors can be accurately determined by static calculations.

The disadvantage of reinforced concrete floors is their high sound permeability.

Monolithic reinforced concrete floors are made at a construction site in formwork.

Do-it-yourself reinforced concrete monolithic floor

Performing the function of transferring the load from the floor to the load-bearing walls, they also serve as stiffening elements in buildings with a massive frame. To manufacture monolithic reinforced concrete floors, formwork is required, made from a scarce material - wood.

Monolithic reinforced concrete floors are divided by shape into slab, beam, ribbed and liner floors (Fig. 84).

Monolithic slab floors. The simplest design of monolithic floors is the Monier slab, in which the reinforcement is placed in tension areas, i.e. in the lower part of the slab, since steel has 15 times greater tensile strength than concrete.

84. Reinforced concrete floors a - monolithic reinforced concrete slab; b - reinforced concrete monolithic beam floor; 1 - transverse reinforcement of the beam; 2 - beam; 3 - longitudinal main reinforcement of the beam; c - reinforced concrete monolithic ribbed floor

The slab is usually laid on a load-bearing wall, and the length of the surface on which the slab is laid is 10 cm; When using slabs with a thickness of more than 10 cm, the length of the surface on which the slab is laid is equal to the thickness of the slab.

Such floors can have a maximum span of 300 cm (see Fig. 84, a) . For larger spans, the reinforced concrete slab is concreted on steel load-bearing beams spanning the larger span.

Such floors are called slab monolithic reinforced concrete or combined floors with steel load-bearing beams.

Monolithic beam floors. For large spans, floors can have a maximum span of 300 cm.

Reinforced concrete beams are laid on the wall; they are connected to a reinforced concrete slab and reinforced. Such floors, invented by the French engineer Ennabic, are called Ennabic floors. Beams are laid at a distance of 130-500 cm from one another. The length of laying beams on load-bearing brick walls should be 7.5% of the beam span, but not less than 22 cm. Typically, beams are anchored into monolithic reinforced concrete belts with brickwork.

Reinforced concrete beam floors are used in rooms where a flat ceiling is required (basements, warehouses, workshops, etc.).

etc.), since for finishing a flat ceiling the axial distance between the beams of this floor is too large.

The use of beam reinforced concrete floors is cost-effective if there are spans of 6 m (see.

rice. 84, b).

Monolithic ribbed floors. If, when using reinforced concrete floors, it is necessary to make a flat ceiling, the axial distance between the beams should be reduced by 0.5-1 m.

The cross-section of beams is smaller, which is why they are called ribs. To prevent the ribs from bulging, they are reinforced over a span of 6 m with one transverse rib (see Fig. 84, c).

The flat ceiling is finished with hemming and lime-gypsum plaster or reed plaster.

Before concreting a ribbed reinforced concrete floor, pins or wire with a diameter of 10 mm are placed in the reinforcement so that after concreting and stripping they protrude from the sides of the ribs. Planks 2 cm thick are installed on these embedded parts, the lower edge of which protrudes beyond the edge of the lower rib by 1 cm (Fig. 85, a).

85. Retail finishing of hem fastening ribs

a - side mounting; b - slab - base of the filing; c - finishing without slab; 1 — steel rod with a diameter of 8 mm; 2 - mesh

Another method is that when making formwork, the ribs are placed in it before the reinforcement is laid and the plank bottom is secured, after which both ends of the wire are monolied.

To the base made in this way, a sheathing of slabs 12-20 mm thick, nailed, is attached. The joints between the plates should not be wider than 15 mm. Simple plaster is applied to the sheathing or lined with reed mat (Fig. 85, b). Sometimes wire is embedded into the slab and ribs and, after stripping, a chain-link mesh is attached to it and lime-gypsum plaster is applied (Fig.

Monolithic floors with liners. The big disadvantage of ribbed floors and especially floors with a flat ceiling is the complexity of their construction and the high consumption of wood for the manufacture of formwork and lining.

Therefore, floors with liners are more often used. In the places of future gaps between the ribs, liners are placed, which serve as the formwork of the ribs and at the same time as the lower part of the slab formwork. The lower sides of the liners replace the lining with boards and serve as a base for plaster. Inserts are made from various materials of various shapes. The most common are rigid liners made of baked clay, the lower part of which extends to the shelves, forming the lower formwork of the ribs.

The liners are placed in horizontal formwork and, after preparing the reinforcement for the ribs and slabs, they are concreted (Fig. 86).

Rice. 86. Monolithic ceiling with liners 1 - plaster; 2 — ceramic liner; 3 - rib reinforcement

The disadvantage of floors with liners is that they are characterized by greater sound permeability than the floors described above, since the liner, after adhesion to reinforced concrete, forms a continuous resonant slab.

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Example of calculation of a square monolithic reinforced concrete slab
with support support

information:

1. solid wall brick 510 mm thick to form a closed space measuring 5x5 m, the walls are built with monolithic reinforced concrete slabs, the width of the supporting surfaces is 250 mm.

Thus, the total panel size is 5.5 x 5.5 m. L 1 = L 2 = 5 m.

2. In addition to the weight directly dependent on the height of the slab, a monolithic reinforced concrete slab must also withstand a certain structural load. Thus, when such a load is known, for example, a flat panel 15 cm thick will have a screed thickness of 5 cm, the screeds will have to determine the thickness of the laminate 8 mm, and the laminate floor will place furniture with the corresponding dimensions along the walls with a total mass of 2000 kg (including contents) , and the middle space will sometimes be a table with corresponding measures weighing 200 kg (with drinks and snacks), and in table 10 a seated person weighing 1200 kg along with chairs.

But this happens very rarely, or rather almost never, since only all the main visionaries can provide all possible options and combinations of loading overlaps. Nostradamus did not leave any comments on this issue, so statistical calculations and probability theory are usually used in calculations.

And these data show that the board in the house can usually be considered a load q v = 400 kg / m2, this load and screed and floor coverings and furniture and guests at the table. This load is usually considered temporary because it can be repaired, converted and other surprises, where one part of the burden is debt and the other part is short.

Since the relationship between long-term and short-term burden is not known to simplify calculations, we simply consider this to be a temporary burden. Since the height of the plate is unknown, then, in advance, for example, H = 15 cm, then the weight of the monolithic plate will be approximately Qp = 0b15h2500 = 375 kg / m2.

Approximately because the exact weight of a square meter of reinforced concrete depends poorly not only on the quantity and diameter of the reinforcement, but also on the size and type of coarse and fine concrete aggregates, the quality of accumulation and other factors.

This load is constant, only anti-gravity technology can change it, but this is not yet available.

Thus, the total distributed load on our board will be:

q = qn + qv = 375 + 400 = 775 kg/m2

3. B20 grade concrete should be used for the panel, which should have the compressive strength of the structure Rb = 11.5 MPa or 117 kgf/cm2 and Class AIII valves with tensile strength Rs = 355 MPa or 3600 kgf/cm2.

required:

Select the cross section of the reinforcement.

solution:

1. Determination of the maximum bending moment.

If our slab relates only to wall 2, so that the plate can be considered as a strand on two joint supports (the width of the bearing surfaces is not yet correct), the beam width for easy calculations is taken to be B = 1 m.

However, in this case our panel supports 4 walls. This means that there is one cross section of the beam relative to the axis X this is not enough because we can take into account our plate and beam according to the axis With. This means that stresses and tensile stresses will not be in the same plane, which is normal to the axis X, but in two planes.

If the load-bearing structure is designed with support brackets with a span L 1 around axis X, then it turns out that the bending moment acts on the beam m1 = q1 L 12/8. In this case, the headlight is carried by a wing with a span L 2 will work exactly the same amount of time as the same range.

But we have one load design:

q = q1 + q2

and if the panel is square, then we can assume that:

q1 = q2 = 0.5 q

m1 = m2 = q1 L 12/8 = q L 12/16 = q L 22/16

This means that the reinforcement is placed parallel to the axis X, and the reinforcement is laid parallel to the axis With, we can count on the same bending moment, at the same time it is half as much as with a panel that rests on two walls.

Thus, the largest bending moment is:

Ma = 775 x 52/16 = 1219.94 kgf m

However, this torque value can only be used for valve design.

Since pressure stresses in two mutually perpendicular planes will work on concrete, it is necessary to take into account the value of the bending moment for concrete:

Mb = (m12 + m22) 0.5 = Ma2 = 1219.94 1.4142 = 1725.25 kgf m

Since we need one moment value for the calculation, we can conclude that the average value between the moment for reinforcement and concrete will be calculated

M = (Ma + Mb) / 2 = 1.207Ma = 1472.6 kgf m

NB:: If you don't like this assumption, you can calculate the reinforcement by the time you work on the concrete.

2. Selecting a reinforcement section.

Calculate the cross section of the reinforcement in both the longitudinal and transverse directions, you can use different methods and the result will be approximately the same.

However, when using any technique, it should be taken into account that the fitting height of the reinforcement will be different, for example, for reinforcement located parallel to the axis X, can be taken in advance h01 = 13 cm, For reinforcement parallel to the axis With, can be taken in advance h02 = 11 cm, because we do not yet know the diameter of the reinforcement.

According to the old method:

A01 = M / bh201Rb = 1472.6 / (1 0.132 1170000) = 0.07545

A02 = M / bh201Rb = 1472.6 / (1 0.112 1170000) = 0.104

Now on the auxiliary table:

Data for the calculation of curved elements of rectangular section,
reinforced by a single reinforcement

we can find η1 = 0.961 and ξ1 = 0.077.

η2 = 0.945 and ξ2 = 0.11. Then the cross section of the reinforcement is required:

Fa1 = M / ηh01Rs = 1472.6 / (0.961 0.13 36000000) = 0.0003275 m2 or 3.255 cm2.

Fa2 = M / ηh02Rs = 1472.6 / (0.956 0.11 36000000) = 0.0003604 m2 or 3.6 cm2.

If longitudinal and transverse reinforcement with a diameter of 10 mm is taken for the combination, and the required part of the transverse reinforcement is recalculated using h02 = 12 cm,

A02 = M / bh201Rb = 1472.6 / (1 0.122 1170000) = 0.087, η2 = 0.957

Fa2 = M / ηh02Rs = 1472.6 / (0.963 0.12 36000000) = 0.000355 m2 or 3.55 cm2.

then, to strengthen 1 linear gauge, 5 bar of longitudinal reinforcement and 5 bar of transverse reinforcement can be used.

This will produce a grid with a cell size of 200x200 mm. The reinforcement cross-section for 1 linear meter will be 3.93 × 2 = 7.86 cmup2. The selection of the reinforcement part is carried out in accordance with Table 2 (see below). The entire panel will require 50 bar, 5.2 to 5.4 meters. Due to the fact that the upper part of the valve section has a reserve, the number of rods in the lower layer can be reduced to 4, then the cross-section of the reinforcing layer 2 is 3.14 or 15.7 cm2 of the total length of the panel.

Section and mass of reinforcing bars

This was a simple calculation, it may be difficult to reduce the number of reinforcements. Since the maximum bending moment only operates in the middle of the panel and when accessing the supports, time on the wall shows that nothing and then the remaining flowmeters from each other can be increased by installing a smaller diameter (the eye size for a 10mm reinforcement diameter does not need to be increased because that our distributed load is quite conditional).

To do this, it is necessary to determine the moment values ​​for each plane under consideration for each subsequent counter and determine the arrays and cell size for each meter of the required compartment. But it makes no sense to use reinforcement with a pitch of more than 250 mm, so the savings on such calculations will not be very good.

NB:: Existing panel design methods are based on contour as prefabricated houses involve the use of an additional factor that takes into account the work of the space plate (as under the influence of the load on the table there will be a strip) and concentration reinforcements in the middle of the panel.

Using this ratio, it reduces the reinforcement by 3-10%, but for concrete slabs that are not produced in a factory and in the field, the use of an additional factor that I do not consider necessary. First, additional strain calculations are required for crack opening, for the percentage of smallest reinforcement. And secondly, the stronger the reinforcement, the less deviation in the middle of the panel, and it will be easier to remove or disguise the finish.

For example, if we use the "Calculation and Design Guide for Precast Solid Tiles in Residential and Public Buildings", then at the bottom of the panel, the room reinforcement for the entire length of the panel is about A01 = 9.5 cm 2 (calculation not shown), which is almost 1 .6 times (15.7 / 9.5 = 1.65) less than the result obtained with us, but it should be noted that the reinforcement should be highest in the center of the range, and therefore it is easy to divide the result, which cannot be achieved by 5 meters.

However, due to this, the value of the cross-sectional area can be an approximation of how well the reinforcement can be preserved due to time-consuming and complex calculations.

Example of calculation of a rectangular monolithic reinforced concrete slab
with support support

To simplify the calculations, all parameters are taken into account, with the exception of the length and width of the room, as in the first case.

Obviously, in the case of rectangular overhead plates, the moments depend on the axis X and in accordance with the axis With, they are not the same.

And the difference between the length and width of the space, the larger the panel, is like a beam on the load-bearing hinges, and when a certain value is reached, the effect of the transverse reinforcement is practically unchanged. Formative experience and experimental data show that with attitude λ = L 2 / L 1 > 3 the transverse moment is five times less than the longitudinal moment.

And if λ ≤ 3, then the connection between the moments can be determined by the following empirical graph:

Graph of moments versus ratio λ:
1 - for plates with hinged support on the periphery
2 - with hinged support on 3 sides

The graph shows the dotted lower limits for selecting reinforcement, and in brackets - λ values ​​for plates are set on three sides (at λ< 0,5 м = λ и для нижних пределов m = λ / 2).

In this case, however, we are interested in curve no. 1, which reflects theoretical values. It shows confirmation of our assumption that the ratio between the moments is equal to unity for a square plate, and from this we can determine the values ​​of the moments for other latitudes.

For example, you need to calculate the board for a room 8m long and 5m wide (for clarity, one of the sizes is the same), the calculated ranges L 2 = 8 m in L 1 = 5 m.

Then λ = 8/5 = 1.6, the ratio between the moments m2 / m1 = 0.49, and then m2 = 0.49m1

Since the total moment is M = m1 + m2, then M = m1 + 0.49m1 or m1 = M / 1.49.

In this case, the value of the total moment is determined on the short side for the simple reason that this is a reasonable solution:

Ma = q L 12/8 = 775 x 52/8 = 2421.875 kgf m

Bending moment of concrete, without taking into account the linear, but definitely stress state

Mb = Ma (12 + 0.492) 0.5 = 2421.875 1.133 = 2697 kg m

then the calculated moment

M = (2421.875 + 2697) / 2 = 2559.43

In this case, the lower (short, 5.4 m long) reinforcements will be counted momentarily:

m1 = 2559.43 / 1.49 = 1717.74 kgf m

and upper (length, length 8.4 m) reinforcement, we will calculate the moment

m2 = 1717.74 x 0.49 = 841.7 kgf m

Thus:

A01 = m1 / bh201Rb = 1717.74 / (1 0.132 1170000) = 0.0888

A02 = m2 / bh201Rb = 841.7 / (1 0.122 1170000) = 0.05

Now, according to supporting table 1, we can find η1 = 0.954 and ξ1 = 0.092.

η2 = 0.974 and ξ2 = 0.051.
Then the cross section of the reinforcement is required:

Fa1 = m1 / ηh01Rs = 1810 / (0.952 · 0.13 · 36000000) = 0.0003845 m2 or 3.845 cm2.

Fa2 = m2 / ηh02Rs = 886.9 / (0.972 · 0.12 · 36000000) = 0.0002 m2 or 2 cm2.

Thus, to strengthen 1 panel sheet, you can use 5 reinforcing bars with a diameter of 10 mm and a length of 5.2 to 5.4 m.

Do-it-yourself monolithic overlay

The intersection of longitudinal reinforcement for 1 linear meter is 3.93 cm2. For transverse reinforcement, you can use four rods with a diameter of 8 mm and a length of 8.2 to 8.4 m. The cross section of the rod for 1 linear meter is 2.01 cm2.

In this case the difference is about 1.26 times.

But again, this is all a simplified version of the calculation.

If you want to further reduce the reinforcement of a section or the grade of concrete or slab height and thus reduce the load, you can explore different loading plate options and calculate whether it will have a certain effect. For example, to facilitate the calculation, the influence of the supporting surfaces is not taken into account, however, if these surfaces of the panels are made on top, the walls are prepared and thus the slabs approach a rigid pinch, when the massive mass of the load wall can be taken into account if the width of the supporting surfaces make up more than half the width of the wall.

When the width of the supporting parts is less than or equal to half the width of the wall, additional calculation of the strength of the wall material will be required, and there is still a possibility that the supporting part of the wall will not carry the weight of the wall load very high.

Consider the case where the width of the base plate segments is about 370 mm to a wall brick width of 510 mm, characterized in that the probability that the full transfer of loads to the walls of part of the base plate is quite high, so that if the wall panel is placed with a width of 510 mm, 2 , height 8 m, and then on these walls there will be at the same time the bottom plate after the ground is a constant load concentrated on the base plate of the measuring instrument part is:

from a solid brick wall 1800 x 2.8 x 1 x 0.51 = 2570.4 kg
from plate height 150 mm: 2500 x 5 x 1 x 0.15 / (2 x 1.49) = 629.2 kg

In this case, it is more relevant to consider that our panel only supports the beam from the console and the concentrated load on the unevenly distributed load on the consoles and closer to the edge of the board, the load is greater, but simplifies the calculations by assuming that the load is distributed evenly on the consoles and is therefore 3199.6 / 0.37 = 8647.56 kg/m.

The moment on the calculated support brackets from this load will be 591.926 kgf m. This means that:

1. The maximum torque M1 in the range is reduced by this amount, and the value m1 = 1717.74 - 591.926 = 1126 kgf m, and therefore the reinforcement section can noticeably reduce or change other parameters of the plate.

2. The bending moment on the supports caused by tensile stresses in the area of ​​the top plate and concrete pull-out work is not calculated, and, therefore, either the plates on the top must be further strengthened, or the width of the supporting part (cantilever beam) must be reduced in order to reduce the load on the support sections.

If there is no additional reinforcement at the top of the plate, cracks will appear in the panel and all of them will turn into a hinge plate without a cantilever.

3. This loading option should be considered in conjunction with the option where the panel already exists but there are no walls, so there is no live load on the panel but no walls or ceiling panels.

The floors are solid reinforced concrete structures. Their use is relevant for increased weight loads, primarily in multi-storey buildings. In private construction, their main advantages include the ability to reduce installation costs by independently performing individual or all stages of work with minimal use of special equipment. The technology is considered labor-intensive; to avoid errors, the calculation of the slab should be entrusted to specialists. The obtained parameters must be taken into account when preparing the main house project.

Conventionally, all are divided into prefabricated (solid or hollow, manufactured at the factory), frequently ribbed (cellular type with sections of lightweight material or empty blocks) and monolithic. The latter are valued primarily for the absence of seams; this option is chosen when concreting multi-story buildings, pouring floors, or demarcating floors in individual buildings. Depending on the design and installation method, they are divided into: beams, beamless (the most popular type in the construction of private houses with a smooth surface), with permanent formwork (at the same time serving as a thermal insulation layer) and laid on a steel flooring. The latter are valued for their reduced labor intensity and the ability to reduce thickness and weight.

Features and advantages of monolithic flooring

The advantages include:

1. Strength and solidity (no seams), and, as a result, ensuring a uniform load on the foundation and load-bearing walls.

2. Possibility of support on columns. This gives more freedom in the planning process compared to the option of laying prefabricated floor slabs from standard-sized prefabricated elements.

3. Safe arrangement of a balcony without the need for additional supports due to the monolithic nature of the main horizontal structure.

Calculation of the slab, drawing up a reinforcement diagram

Ideally, the design is entrusted to specialists; they will help you choose an option with correctly distributed loads, optimal in terms of “reliability-cost of building materials.” The initial data for independent calculations are the dimensions of the floor with mandatory consideration of the width of the support areas. The thickness of the monolith is selected based on the maximum length of the longitudinal span (the recommended ratio for beamless structures is 1:30, but not less than 15 cm). For floors within 6 m, the minimum is 20 cm; above 6, options with reinforcement with stiffening ribs are considered. In beam-type varieties, the pitch of the supports is taken into account (accordingly, the minimum height is found by dividing it by 30).

The calculation of the slab begins with determining its own weight: the average (2500 kg/m3) is multiplied by the thickness of the floor. The standard temporary load (weight of furniture, equipment and people) for residential buildings is 150 kg/m2, taking into account the 30% reserve it is increased to 195-200. The total, maximum possible load is obtained by adding these values.

To check the cross-section of the reinforcement, the maximum bending moment is calculated, the formula depends on the method of weight distribution. For a standard beamless floor supported on two load-bearing walls M max = (q·l2)/ 8, where q is the total load, kg/cm2, l2 is the span width. This formula is the simplest; in the absence of reinforcement in areas of maximum concrete compression or uneven weight distribution, it becomes more complicated.

To check the cross-section of the reinforcement, a coefficient is calculated that takes into account the design resistance of building materials (reference values ​​depend on the selected mortar strength class and steel grade). The resulting value corresponds to the minimum permissible area of ​​metal in a cross-section of the slab. It is compared with the preliminary one; if it is exceeded, strengthening of the circuit is required (reducing the cell pitch or using rods with a larger diameter).

Due to the complexity, the calculation is usually entrusted to specialists; when it is done, a checkerboard pattern of two grids (lower and upper) with a cell pitch of 20x20 cm and a rod thickness of 10-14 mm (hot-rolled steel) is selected. Provision is made for both reinforcement in the center of the monolithic slab, areas with increased loads and places of contact with supports, as well as a margin for the overlap of the floor on the walls (depending on the strength of the building materials - from 150 mm for brick to 250 for cellular concrete). If possible, longitudinal and transverse rods are laid unbroken; if this condition is violated, they overlap - at least 40 cm.

Main stages of installation

Laying begins with the calculation and purchase of building materials (ideally, project data is used). Formwork structures are prepared: panels made of thick moisture-resistant plywood, metal or plastic, beams and telescopic supports (1 piece/m2), equipment for preparing, feeding and compacting concrete, tools for bending reinforcement and special stands. If necessary, an armored belt is laid around the perimeter of the load-bearing walls; such a need arises when constructing floors in a house made of aerated concrete.

Key steps include:

• Assembly and installation of formwork.
• Placement of reinforced frame.
• Pouring a monolithic slab with concrete, compacting and leveling.
• Humidity maintenance of the solution, covering, dismantling of the formwork after 28 days.

1. Requirements for supports and shields.

Installation involves pouring concrete into a sealed horizontal formwork; preference is given to special prefabricated structures. In principle, it is not difficult to make panels yourself from plywood with a thickness of at least 20 mm (it is better not to use boards due to difficulties in fitting). A prerequisite is the installation of telescopic metal posts (when erecting the ceiling of the first floor of the house, they are replaced by stationary supports). In their absence, replacement with logs with a diameter of at least 8 cm is allowed, but you should be prepared for problems when adjusting the level.

To support the panels, a crossbar is laid - a longitudinal beam with a cross-section of at least 10x10 cm; if necessary, the formwork is reinforced with transverse elements (this situation most often arises when working with homemade products). The boards are laid without gaps, the edges rest tightly against the wall. When installing vertical structures, the amount of overlap on the supporting systems is taken into account. To eliminate the risk of leakage, the bottom is covered with film; sealed factory reusable varieties are lubricated to facilitate the removal process. The stage ends with a level check; deviations are unacceptable.

2. What needs to be taken into account when reinforcing?

Metal reinforcement is the main requirement of the technology. The distance from the edge of the concrete to the metal is at least 25 mm. The joints are tied with wire with a cross-section of 1.2-1.5 mm; welding is not allowed. To install the meshes, pre-prepared clamps are used: made of steel with a thickness of at least 10 mm, with an interval of up to 1 m, similar elements are placed at the ends. Reinforcement of a monolithic reinforced concrete floor is completed by laying connectors that ensure uniform load transfer on the entire system - after 40 cm near the walls, after 70 from it, with a subsequent step of 20.

3. The nuances of concreting.

The main requirement of the technology is process continuity; ideally, the solution is ordered at factories and poured using appropriate equipment. The recommended thickness of the concrete layer is 20 cm, which in most cases coincides with the height of the ceiling itself. The minimum grade is M200; in order to enhance the thermal insulation properties and lighten the weight, part of the coarse high-strength filler can be replaced with expanded clay, but this method requires approval by specialists (strength testing).

Holes for supplying communications and ventilation ducts are laid before pouring begins; drilling a frozen monolithic slab is considered a violation. The stage ends with the mandatory compaction of concrete using deep vibrators. The rules for caring for the surface are generally standard, but you cannot water the structure abundantly; unlike the foundation or vertical walls, it is wetted more carefully.

Prices

The cost of pouring when contacting professional companies varies from 4,000 to 9,000 rubles/m3 (provided that the customer’s formwork is used). The final cost depends on the chosen reinforcement scheme, the height of the future slab (from the ground level or from the level of the previous horizontal support) and its thickness, the method of placement (on columns or load-bearing walls), and the total scope of work. The list of services provided by construction companies includes installation and dismantling of formwork structures, assembly of reinforced frames according to a project prepared in advance (paid separately), continuous concreting and maintenance of the laid mixture: watering, covering, and, if necessary, heating. The advantage of turning to professionals is the mandatory quality control carried out upon completion of the curing process.

The advantages of laying the floor with your own hands include a reduction in the cost of paying for work - up to 30% at least. For pouring, simple building materials are used - concrete and reinforcement; saving on them is unacceptable. The volume of the solution is calculated based on the thickness and area of ​​the slab, the length and weight of the metal is calculated according to a reinforcement scheme drawn up in advance. Renting formwork structures is expensive: the minimum price per m2 is 400 rubles per month (it cannot be removed earlier).

Additional costs when doing the work yourself include the need for special equipment and containers for lifting the solution to the top (shoe buckets and a crane or concrete pump). This is not a problem when installing solid floors on the ground floors of a house, but in other cases it is impossible to do without the appropriate equipment. This is explained by the main requirement of the technology - a continuous concreting process; monolithic floors with separate patches hardened on different days are inferior in quality to those poured at once. The minimum cost when performing all stages independently is 3,200 rubles per 1 m2 with a slab thickness of 20 cm.

During the construction of a house, there is no way to do without installing a ceiling. This design limits the height of the room, insulates it from the penetration of cold air in winter, and takes on loads from the roof or upper floor. In modern technologies, a slab reinforced with reinforcement is most often used.

Description of types, design features, technical requirements

Depending on the location and functional purpose, concrete floors come in several types:

• basement;
• interfloor;
• attic;
• attics.

According to another classification, they are divided into solid and prefabricated. The first ones are made independently, pouring the mixture onto a prepared reinforcement frame. This method does not require the use of a crane to lift the slabs, but additional workers will be needed to install the formwork, tie the frame, and pour concrete.

Prefabricated systems are obtained by laying standard panels of the required size. According to their design, they come in three types: monolithic, ribbed, hollow. In private construction, the third option is more often used. Product dimensions: length – up to 7 m, width – 1.5, height – 0.22.

Taking into account the operating conditions, the following requirements are imposed:

• strength and rigidity exceeding the design load (it is defined as the total weight of the slab itself, screed, furniture, and other things);
• high level of sound insulation;
• fire resistance;
• The thickness of the walls for concrete blocks is at least 200 mm.

Concrete has high thermal conductivity, to reduce which it is recommended to insulate it, for example, with mineral wool.

Installation instructions

Preliminary preparation of the structure and slabs helps to lay the finished concrete floor with your own hands.

Scheme of preparatory work

1. To ensure that the panels are in the same plane, the upper end of the load-bearing walls is checked for horizontalness. This can be done in this order: 30-40 cm before the end of the laying, markers are applied to the wall using a laser or liquid level, and then the finishing brick rows are checked with a tape measure. The top row is positioned so that the bricks are oriented towards the inside of the room.

2. Most often, the edge of the box is aligned differently - the perimeter of the upper edge of the walls with reinforcement is concreted. Due to this, the brick or block structure is further strengthened. At a certain level, the masonry leaves an empty space for the armored belt. The depth of support (overlap) of the ceiling depends on the total thickness of the slab together with thermal insulation. Typically the panel extends into the wall by 70-120 mm.

The scheme for pouring an armored belt is similar to laying a foundation: the formwork is installed, a frame of reinforcing bars is made inside it using welding, and the mixture is poured without crushed stone. The design of the belt for the plinth is made faster: simply add additional formwork along the outer edge of the foundation.

3. Before installing the slabs, be sure to close the voids at their ends. If this is not done, when placing the wall of the upper floor on the edge of the ceiling, it may collapse. Concreting the joints between the slabs will not produce results: the mixture will flow into the holes. It is not difficult to close the cavity - half a brick is inserted into it and sealed with mortar.

4. Prepare a site for lifting equipment. This is an area with dense soil, otherwise the crane will get stuck in soft soil. To ensure its stability, temporary placement of road panels on the site is practiced. It is advisable not to place the crane close to the pit to prevent the collapse of the soil or the sliding of heavy equipment.

Laying technology

It will not be possible to install the ceiling yourself; the process usually involves three installers. One worker connects the slabs, the other two correct them as they are lowered.

1. A fairly thick concrete mixture is applied to the reinforced belt (layer thickness is at least 2 cm).

2. The crane operator lowers the panel, holding it by tensioning the sling ropes. In a suspended position, it can be easily moved in the desired direction using a crowbar.

3. Compensation cut. It is necessary if one slab covers several spans. Conventional structures work in bending and must rest on two short ends. If intermediate supports are installed, tensile stresses arise in the upper part of the floors. Since there is no reinforcement there, cracks may appear. To relieve stress, use a grinder to saw through a groove, placing it above the intermediate support. Subsequently, a crack will appear at the site of the slot.

4. Anchoring. This is stitching with reinforcing wire: it is threaded through the mounting lugs, tightened, and then welded. The scheme is usually included in the project; if it is not there, the standard version is used. For load-bearing walls there is at least 1 anchor for every 3 linear meters; for non-load-bearing walls, anchors are removed from all outer loops. The end plates are sewn together with diagonal anchors.

The gaps between the tiles (rusts) are filled with concrete mortar, due to which the structure becomes monolithic and durable.

During installation, sometimes you have to adjust the dimensions. The optimal overlap on the wall is no more than 120 mm, and the maximum permissible value is 250. Increasing this parameter changes the design of the ceiling, and as a result, cracks may appear on it. The panels are shortened in the following way:

• mark a cutting line, place a block under it - its thickness should be such that the edge to be separated is suspended;
• According to the markings, make an incision with a grinder, use a sledgehammer to split the concrete above and below the voids;
• break partitions;
• metal reinforcement is cut with a grinder, leaving a couple of millimeters - this residue is broken with a sledgehammer (otherwise the stressed reinforcement may pinch the disc of the grinder).

If the size is insufficient, the gap near the wall is filled with bricks.

How to make a ceiling yourself?

The frame is made from edged boards (thickness 25-35 mm), plywood (thickness from 20 mm) or rented.

1. Install the formwork. Its dimensions should be such that the edges rest against the walls without gaps. The position of the structure is checked using a level. Cover it with waterproofing film.

2. Reinforced. Typically, rods with a diameter of 12-14 mm are used for the frame. First, longitudinal and then transverse elements are laid (cell 12-15 cm), tied with wire. The upper mesh of the frame is made in the same order, the joints of the rods are arranged in a checkerboard pattern, and the ends of the rods are placed on the load-bearing beams.

3. Preparation of concrete. Volumetric proportions of its components:

• sifted sand - 2 parts;
• crushed stone (gravel) – 1 part;
• cement M400(500) – 1 part;
• water.

Pour enough water so that the solution resembles liquid sour cream in thickness.

4. Filling. All cavities are carefully filled with the mixture, “smoothed” with a shovel, removing air. For the finishing fill, make a thicker mixture and lay it down. The thickness of the layer is 2-3 cm less than the final size of the overlap. After a couple of days, the set composition is covered with a cement-sand mortar of medium thickness and leveled with a rule to an ideal plane.

The solidifying monolith is periodically watered with water and covered with film in hot weather. On the 10th day, the formwork is removed and allowed to gain strength for 3-5 weeks. After this, you can begin the next stage of construction.

It was in October, it was constantly raining, such weather will provide better hydration of concrete, the main thing is not to pour concrete in the rain itself.

The budget was approaching zero, or rather it was almost non-existent, we had to borrow money, in order to save money, a mesh of 14 reinforcement was used for reinforcement, 1 layer instead of two, but we additionally reinforced the frame with crossbars for rigidity.

The first thing to do was knit the crossbars.

For convenience and speed of tying the crossbars, 4 fittings were driven into the wall of the house.

In advance, part of the reinforcement was released from the concrete belt to connect it with the ceiling reinforcement.

In areas of interior walls, belts and additional reinforcement were not made.

Before attaching the boards, a frame was made, logs with supports made of 150x50 mm timber.

(loadposition Adsense)

The first step is to knock out a level in each room and nail a 50x50 mm beam along it perpendicularly under the joists. Nails 100-150 mm. We nailed it directly to the concrete belt while it was damp, the nails went into it well. Additionally, planks were nailed under the 50th. timber about a meter apart with a consumption of about 6 nails per plank, since shell rock walls and nails do not hold very tightly in it, it all depends on the density of the stone.

The supports were also secured by nailing 2 nails diagonally to the joists.

At the bottom, they also made soles for supports and secured them with nails, since they would easily sink in the damp waste filled in.

When the formwork was completed, we began knitting the frame, the frame was made of 14-gauge reinforcement, cells 20x20 cm. Before knitting the frame, the film was laid out on a wooden floor.

A crossbar was tied on top of the finished mesh of reinforcement every meter.

The height and width of the crossbar is 10 cm.

We left space for the future hatch to the second floor; the size was chosen randomly, 1000x2000 mm.

That's it, the monolithic concrete floor is ready. The ceiling was poured in 3 days, the volume of concrete was 14 cubic meters, approximately 800 kg of reinforcement. Concrete was mixed by hand in a trough with a volume of 0.4 cubic meters. You can walk on the fresh floor the very next day, but under no circumstances load it with materials until it has completely hardened, as microcracks may form. The formwork was dismantled in April.

5 Comments

Hello Evgeniy! Thank you for paying attention to my pipes! The dimensions of the floor are 8.600x9.200 m, the thickness of the floor is 200 mm, the longest span is 6 meters. If you are interested in any other details, I will be happy to answer.

Hello Vitaly! I was pleasantly surprised by your efficiency. Looking through the photos of the ceiling and the explanations for them, I realized that despite any difficulties, you are optimistic about the future.
And I'm happy for you. I am building a house on my site and this month it is necessary to fill the ceiling - 125 sq.m. I need 17.6 cubic meters of concrete. Now we are laying the frames of the beams, the reinforcement of the main belt, and then the reinforcement mesh of the slabs, but the thickness of the slab is 8 cm, and the beams are 10x25 cm, with a span of 4.4 m, the reinforcement is 8-10 mm. Who told you to choose this type and these sizes? I think that 8 cubic meters of concrete could be enough for you, and this is a big savings, and the load on the shell rock would be reduced by 14.5 tons. But it's done and that's good. And the last question. Will you talk about new stages of construction? All the best Vitaly.

Hello Evgeniy!
I admit honestly at the time of construction, no special calculations were made, they did it “like everyone else.” Regarding the ceiling, 80% of the floor area is a span of 6 meters, if you count from load-bearing wall to load-bearing wall, I don’t take partitions into account, since my black soil is 1-1.5 m and the foundation for the partitions is cast separately with a depth of about 60- 80 cm.
Regarding the thickness of the ceiling, a calculation is usually made where the span length is divided by 25-32, hence the thickness. With a smaller slab thickness, deflections greater than permissible are possible (only from the weight of the slab itself), if the slab is not pre-stressed. The 12.7 cm thick slab of ordinary heavy concrete satisfies most regulatory requirements for fire resistance and fire safety.
I’ll also add that in our area the seismic level is 9 points, so I didn’t skimp on the material, since I plan to put the house into operation in the future.
Yes, of course, I will publish materials in the “With my own hands” thread, there are photos, all that remains is to prepare the text, but at the moment I am busy with work (hacks), doing interior decoration, and closer to the fall I will work more closely on the site. I can also publish your works, making a separate section or “blog” for this; I think your experience will definitely be useful to other users of the site.

Hello Vitaly. I'm thinking about solving a similar problem. House 2nd floor Internal size 9x11. A solid wall of 1 brick divides 11 meters into 4.5 and 6.5 meters. Ceiling after 2nd floor. I also plan to make concrete from reinforcement in 2 layers: the bottom layer is 14 mm, the top layer is 10 mm. The interior walls of the premises have already been bricked out to the floor. The largest room is 4.2 square. If 4200:32=131 mm floor thickness. I would not want to load the foundation any thicker. There are no crossbars. What can you say about the thickness of the ceiling?

It was in October, it was constantly raining, such weather will provide better hydration of concrete, the main thing is not to pour concrete in the rain itself.

The budget was approaching zero, or rather it was almost non-existent, we had to borrow money, in order to save money, a mesh of 14 reinforcement was used for reinforcement, 1 layer instead of two, but we additionally reinforced the frame with crossbars for rigidity.

The first thing to do was knit the crossbars.

For convenience and speed of tying the crossbars, 4 fittings were driven into the wall of the house.

In advance, part of the reinforcement was released from the concrete belt to connect it with the ceiling reinforcement.

In areas of interior walls, belts and additional reinforcement were not made.

Before attaching the boards, a frame was made, logs with supports made of 150x50 mm timber.

(loadposition Adsense)

The first step is to knock out a level in each room and nail a 50x50 mm beam along it perpendicularly under the joists. Nails 100-150 mm. We nailed it directly to the concrete belt while it was damp, the nails went into it well. Additionally, planks were nailed under the 50th. timber about a meter apart with a consumption of about 6 nails per plank, since shell rock walls and nails do not hold very tightly in it, it all depends on the density of the stone.

The supports were also secured by nailing 2 nails diagonally to the joists.

At the bottom, they also made soles for supports and secured them with nails, since they would easily sink in the damp waste filled in.

When the formwork was completed, we began knitting the frame, the frame was made of 14-gauge reinforcement, cells 20x20 cm. Before knitting the frame, the film was laid out on a wooden floor.

A crossbar was tied on top of the finished mesh of reinforcement every meter.

The height and width of the crossbar is 10 cm.

We left space for the future hatch to the second floor; the size was chosen randomly, 1000x2000 mm.

That's it, the monolithic concrete floor is ready. The ceiling was poured in 3 days, the volume of concrete was 14 cubic meters, approximately 800 kg of reinforcement. Concrete was mixed by hand in a trough with a volume of 0.4 cubic meters. You can walk on the fresh floor the very next day, but under no circumstances load it with materials until it has completely hardened, as microcracks may form. The formwork was dismantled in April.

5 Comments

Hello Evgeniy! Thank you for paying attention to my pipes! The dimensions of the floor are 8.600x9.200 m, the thickness of the floor is 200 mm, the longest span is 6 meters. If you are interested in any other details, I will be happy to answer.

Hello Vitaly! I was pleasantly surprised by your efficiency. Looking through the photos of the ceiling and the explanations for them, I realized that despite any difficulties, you are optimistic about the future.
And I'm happy for you. I am building a house on my site and this month it is necessary to fill the ceiling - 125 sq.m. I need 17.6 cubic meters of concrete. Now we are laying the frames of the beams, the reinforcement of the main belt, and then the reinforcement mesh of the slabs, but the thickness of the slab is 8 cm, and the beams are 10x25 cm, with a span of 4.4 m, the reinforcement is 8-10 mm. Who told you to choose this type and these sizes? I think that 8 cubic meters of concrete could be enough for you, and this is a big savings, and the load on the shell rock would be reduced by 14.5 tons. But it's done and that's good. And the last question. Will you talk about new stages of construction? All the best Vitaly.

Hello Evgeniy!
I admit honestly at the time of construction, no special calculations were made, they did it “like everyone else.” Regarding the ceiling, 80% of the floor area is a span of 6 meters, if you count from load-bearing wall to load-bearing wall, I don’t take partitions into account, since my black soil is 1-1.5 m and the foundation for the partitions is cast separately with a depth of about 60- 80 cm.
Regarding the thickness of the ceiling, a calculation is usually made where the span length is divided by 25-32, hence the thickness. With a smaller slab thickness, deflections greater than permissible are possible (only from the weight of the slab itself), if the slab is not pre-stressed. The 12.7 cm thick slab of ordinary heavy concrete satisfies most regulatory requirements for fire resistance and fire safety.
I’ll also add that in our area the seismic level is 9 points, so I didn’t skimp on the material, since I plan to put the house into operation in the future.
Yes, of course, I will publish materials in the “With my own hands” thread, there are photos, all that remains is to prepare the text, but at the moment I am busy with work (hacks), doing interior decoration, and closer to the fall I will work more closely on the site. I can also publish your works, making a separate section or “blog” for this; I think your experience will definitely be useful to other users of the site.

Hello Vitaly. I'm thinking about solving a similar problem. House 2nd floor Internal size 9x11. A solid wall of 1 brick divides 11 meters into 4.5 and 6.5 meters. Ceiling after 2nd floor. I also plan to make concrete from reinforcement in 2 layers: the bottom layer is 14 mm, the top layer is 10 mm. The interior walls of the premises have already been bricked out to the floor. The largest room is 4.2 square. If 4200:32=131 mm floor thickness. I would not want to load the foundation any thicker. There are no crossbars. What can you say about the thickness of the ceiling?