Bearing capacity of bored piles. Calculation of the bearing capacity of bored piles

Pile-grillage foundation on bored piles is a combined type of foundation from supporting piles formed in the ground by concreting wells drilled in the ground. The second part of the foundation is a grillage that distributes the load on the pile field. This type of foundation has the highest bearing capacity and can be used to build large houses and private cottages from any materials.

A bored foundation with a grillage allows you to build buildings on difficult soils: viscous, swampy, quicksand, heaving. The foundation on bored piles is indispensable in seismically active areas, areas with extensive networks of underground utilities, as well as in soils with high alkalinity, where it is impossible to use screw supports.

Design advantages:

• increased resistance to vibration;
• the possibility of construction under adverse geological conditions;
• ease of installation;
• lack of large volumes of earthworks;
• relatively low cost.

It is possible to make a bored foundation with a monolithic grillage without the involvement of specialists and professional equipment.

Disadvantages:

• danger of uneven settlement of supports;
• the impossibility of arranging the basement and basement.

Calculation of a bored foundation with a grillage

When calculating, it is necessary to be guided by the data on the characteristics of soils and materials specified in SNiP 2.03.01-84, 11-23-81, 11-25-80, 2.05.03-84 and 2.06.06-85. In total, three settlement operations are carried out:

Calculation of bored piles

During the calculation, the length of piles (depth of occurrence), their cross section, number and layout are determined. The diameter of a bored pile for the construction of a cottage is from 15 to 40 cm. Most often, this parameter is taken equal to 20 cm. concrete and reinforcement:

well drilling

Drilling is carried out with a hand drill, which is deepened to the desired depth. When driving, the soil is not thrown to the surface, compacting along the walls.

During the drilling process, it is necessary to control that the drill enters strictly perpendicularly, without deviating.

After the development of the well, the diameter of which should be 5-7 cm larger than the selected diameter of the piles, the base is carefully rammed. If necessary, a sand and gravel cushion of 10-30 cm is added.

Casing installation

Casing pipes prevent the walls of the well from crumbling and ensure safe work. According to the technology, pipes can not be used on dense clay soils and loams, however, when installing bored piles with your own hands, it is recommended to install them. Inside the pipe it is much easier to mount the reinforcing frame. In addition, the process of pouring and vibrating concrete mixture is simplified.

As casing pipes, you can use plastic, metal or asbestos-cement products of the desired diameter. If financial possibilities allow, then it is better to buy special casing pipes for wells, which have prepared joints with convenient connections. The pipe is strictly vertically installed in the well. If a gap has formed between the pipe wall and the well, then it must be filled with soil with a seal.

Reinforcement

12 mm reinforcement is used to create an armoframe. According to Table 1, when building a cottage, there is no need to use a complex reinforcement plan, 4 or 6 reinforcement bars are enough. The technology of binding the reinforcing frame is very simple: the rods are arranged in a circle, forming a circle with a diameter of 3-5 cm less than the size of the casing. The rods are tied with wire. Clamps can be used to secure. Frame length = casing pipe length + 30 cm. The finished reinforcement cage is installed in the well inside the casing pipe and buried in the ground.

The reinforcement cage must not come into contact with the walls of the casing pipes!

Pouring concrete mix

Concrete used for pouring bored supports must comply with SNiP 2.03.01-84 and be at least class B12.5. For massive houses it is better to use B15 concrete. A funnel is lowered at the wellhead to pour concrete. If you pour the mixture without a funnel, then voids may appear. It is necessary to pour the concrete mixture slowly, each layer 0.5 m thick must be compacted for 5-10 minutes using a deep vibration tool and only after that the next portion is poured. The installation of the grillage can be started after the concrete has gained strength - after 3-7 days.

grillage device

For the foundation of a private house, a reinforced concrete tape grillage is made. Lightweight structures, such as baths, country log houses, allow the use of a wooden grillage. The simplest and least labor-intensive option is a low grillage, which rises 0.2-0.3 m above ground level. A high grillage up to 0.5-0.6 m can be used on wet soils to maximize the rise of the house from the surface.

Stages of construction of a monolithic grillage:

Foundation and formwork

For low grillages, a gravel-sand cushion of 10-20 cm is used, on top of which a footing is laid - a 5 cm layer of lean concrete and waterproofing. Roofing material or hydroisol is used as a waterproofing layer. The formwork is mounted from boards along the entire length of the grillage.

Reinforcement

The technology of reinforcing a strip grillage involves the longitudinal laying of reinforcement bars, which are connected both to each other and to the reinforcement of bored piles. Proper reinforcement ensures a rigid connection of the bored support with the grillage. On the stretched sections, 4 rebars of 20 mm are laid, at the corners - 12-15 mm. To fasten the reinforcement into a single frame, vertical rods of 5-8 mm are used, the distance between them is 25-30 cm.

Pouring concrete

Concrete class B12.5 ... B15 is poured into the formwork and compacted by vibration equipment. At an air temperature of +25 C, concrete must be moistened periodically. To ensure gradual hardening, the grillage must be covered with polyethylene. The final pile-grillage foundation on piles will be ready in 20-25 days.

Insulation of a bored foundation with a grillage

To create a favorable microclimate in the house, it is recommended to insulate the foundation. Piles buried in the ground do not need to be insulated; thermal insulation is necessary for that part of the grillage that is located above the zero level. Warming and waterproofing of the base with a recessed grillage is carried out in the horizontal and vertical planes.

Thermal insulation is carried out with foam boards or other foam insulation. It is impossible to use heat insulators based on mineral wool, because. they intensively absorb moisture from the soil and quickly become unusable. The algorithm for creating hydro and thermal insulation of the grillage is simple:

1. Waterproofing is carried out: a layer of bitumen or rolled roofing material. The upper and side parts of the grillage are waterproofed.
2. Insulation plates are glued with glue and fastened with dowel-nails.
3. Sealing of joints and corners is carried out using mounting foam or liquid polyurethane foam.
4. The side walls of the grillage are finished with plaster or other decorative material.

Simultaneously with thermal insulation, a blind area is made, which also helps to retain heat and remove moisture from the foundation.

A properly executed pile-grillage foundation on bored piles will last at least 100 years. The design is maintenance-free and affordable.

A characteristic indicator of the strength of a pile foundation is the bearing capacity of a single pile. This characteristic affects the total number of piles in the perimeter of the foundation - by adjusting the frequency, you can increase the load limit that the foundation will be able to withstand. The number of bored piles and the bearing capacity of a single pile column are interrelated characteristics, the optimal ratio of which is determined by simple calculations.

Preparation for calculation

The initial data that will be needed to calculate the bearing capacity of a bored pile is obtained as a result of geological surveys and the calculation of the total expected load of the building. These are the mandatory stages of the calculation, the implementation of which is justified by the theory of calculating the strength characteristics of bored foundations.

Such indicators as the freezing depth, the level of groundwater, the type of soil and its mechanical characteristics are very important for obtaining an accurate result. Information on the depth of soil freezing is in SNiP 2.02.01-83 *, the data is divided by climatic regions, presented cartographically and in the form of tables.

Do not rely on geological and hydrogeological survey data obtained in neighboring areas. Even within the perimeter of one land allotment, the state of foundation soils can change dramatically. Three to four control wells at the control points of the perimeter will give accurate information about the condition of the soil.

The calculation of the mass of the building is carried out taking into account the climatic region, the location of the building relative to the rhumb of the winds, the average amount of precipitation in the winter, the mass of building structures and equipment. This indicator is most significant in the design of the foundation - the data for this part of the calculation, as well as the scheme and calculation formulas can be found in SNiP 2.01.07-85.

Conducting geology

Conducting geological surveys is a responsible event, and in mass production construction, this is done by geologists. In individual housing construction, an independent assessment of the state of the soil is often carried out. Without experience in conducting surveys of this level, it is very difficult to assess the real state of affairs. The work of a competent specialist for the most part consists in a visual assessment of the state of the strata.

To begin with, suffers are arranged on the site - vertical excavations of soil of rectangular or circular cross-section, with a depth of two meters and a width sufficient for visual inspection of the base of the walls of the pit. The purpose of the shuffer is to open the soil in order to access the strata hidden under the top layer of soil. Geologists measure the depth of the layers, take a soil sample from the middle of each layer, and subsequently monitor the accumulation of water at the bottom of the face. Instead of shufers, round wells can be arranged, from which a core is taken using a special device or local samples are taken.

Shufry shelter for a while - two or three days - limiting the ingress of precipitation. After that, the water level that has risen in the well cavity is estimated - this mark, counted from the upper boundary, will be the level of groundwater occurrence.

All the data obtained are entered into a summary table. In addition, a profile of the soil section is compiled, which makes it possible to predict the state of the soil at points where drilling has not been carried out. When self-assessing the bases, one should be guided by the information provided in SNiP 2.02.01-83 * and GOST 25100-2011, where the relevant sections present soil classifications with descriptions, methods for visually determining soil types and characteristics in accordance with types.

How to use geological exploration data

After the geology of the area has been carried out - independently or by hired specialists - you can begin to determine the initial geometric characteristics of the piles.

We are interested in the type of soil, the coefficient of soil heterogeneity, the depth of freezing and the level of groundwater. The scheme for calculating the bearing capacity of a bored pile for various types of soil is in the appendices of SP 24.13330.2011.

The depth of the pile should be at least half a meter below the freezing depth in order to prevent the impact of frost heaving of the soil on the supporting part of the column. The average freezing depth in the central strip of Russia is 1.2 meters, which means that the minimum length of the pile should be 1.7 meters in this case. The value varies for individual regions.

Not only relative humidity, but also the relative position of the lower mark of soil freezing and the depth of groundwater. In the cold season, high-lying frozen groundwater will exert strong lateral pressure on the body of the pile column - such soils are highly deformed and are considered heaving.

Some soils, characterized as weak, high heaving and subsidence, are not suitable for pile foundations - strip or slab foundations are more suitable for them. To determine the type of soil, as well as the type of compatible foundation, means to exclude the rapid destruction of structures. The indicators of soil heterogeneity indicated in the tables of the above regulatory documents are used in further calculations.

Calculation of the total load

The collection of loads allows you to determine the mass of the building, which means the force with which the building will act on the foundation as a whole and on its individual elements. There are two types of loads acting on the supporting structure - temporary and permanent. Permanent loads include:

• Mass of wall structures;
• The total mass of floors;
• The mass of roof structures;
• Mass of equipment and payload.

You can calculate the mass of structures by determining the volume of structures, and multiplying it by the density of the material used. An example of calculating the mass for a one-story building with reinforced concrete floors, a roof made of ceramic tiles and walls of 600 mm reinforced concrete, dimensions 10 by 10 meters in plan, a floor height of 2 meters:

• We calculate the volume of the walls, for this we multiply the cross-sectional area of ​​\u200b\u200bthe wall by the perimeter. We get V walls = 20 ∙ 2 ∙ 0.6 = 24 m3. We multiply the obtained value by the density of heavy concrete, which is equal to 2500 kg / cm3. The total mass of wall structures is multiplied by the safety factor, for concrete equal to k = 1.1. We get the mass M of the wall = 66 tons.
• Similarly, we consider the volume of floors (basement and attic), the mass of which, with a thickness of 250 mm, will be equal to Mpc = 137.5 tons, taking into account a similar safety factor.
• We calculate the mass of roof structures. The mass of the roof for 1 m2 of metal tiles is 65 kg, for soft roofs - 75 kg, for ceramic tiles - 125 kg. The area of ​​a gable roof for a building of such a perimeter will be approximately 140 m2, which means that the mass of structures will be Mcr = 17.5 tons.
• The total size of the permanent load will be equal to Mpost = 221 tons.

Reliability factors for various materials are in the seventh section of SP 20.13330.2011. When calculating, the mass of partitions, facade cladding materials and insulation should be taken into account. The volume occupied by window and door openings is not subtracted from the total volume for ease of calculation, since it is an insignificant part of the total mass.

Calculation of live loads

Grill on screw piles

Live loads are calculated in accordance with the climatic region and the instructions of the set of rules "Loads and effects". Temporary loads include snow and payloads. The payload for residential buildings is 150 kg per 1 m2 of floor, which means that the total payload will be Mpol = 15 tons.

The mass of equipment that is supposed to be installed in the building is also summed up in this indicator. For a certain type of equipment, a safety factor is applied, located in the above set of rules.

There are various types of special loads that also need to be taken into account in the design. These are seismic, vibrational, explosive and others.

where ce is the snow drift coefficient equal to 0.85;

ct is a thermal coefficient equal to 0.8;

m - conversion factor, for buildings in terms of less than 100 m, taken according to table D of the above joint venture;

St is the weight of the snow cover per 1 m2. Accepted according to table 10.1, depending on the snow area.

The indicators of temporary loads are summed up with constant ones and a quantitative indicator of the total load of the building on the foundation is obtained. This number is used to calculate the load per pile column and compare the tensile strength. For the convenience of calculation and clarity of the example, we will take the temporary loads Mvr = 29 t, which, in total with the constants, will give Mtotal = 250 t.

Determination of the bearing capacity of the pile

The geometrical parameters of the pile and the tensile strength are interrelated quantities. In this example, the load per meter of foundation will be 250/20 = 12.5 tons.

The calculation of the limit of the load limit on a single bored pile is carried out according to the formula:

where F is the limit of the bearing capacity; R - relative soil resistance, an example of the calculation of which is in SNiP 2.02.01-83 *; A is the sectional area of ​​the pile; Eycf, fi and hi are the coefficients from the above SNiP; y is the perimeter of the section of the pile column, divided by the length.

Watch the video on how to check the bearing capacity of a pile using professional equipment.

For a pile of one and a half meters in length with a diameter of 0.4 meters, the bearing capacity will be 24.7 tons, which allows increasing the pitch of pile columns to 1.5 meters. In this case, the load on the pile will be 18.75 tons, which leaves a fairly large margin of safety. By changing the geometric characteristics, as well as the pitch of the pile columns, the bearing capacity is regulated. This table, presented below, shows the dependence of the bearing capacity of a one and a half meter pile on the diameter:

The dependence of the bearing capacity on the width of the pile

There are many services that allow you to calculate the bearing capacity of a pile online. You should use only trusted portals with good reviews.

It is important not to exceed the allowable load on the pile and leave a margin of safety - few services are able to plan load distribution, so you should pay attention to the calculation algorithm.

The calculation of the pile foundation is carried out depending on its type. It is important to understand that the calculation of bored piles will differ from the calculations for screw piles. But in all cases, it is required to perform preliminary preparation, which includes the collection of loads and geological surveys.

Studying the characteristics of the soil

The bearing capacity of a bored pile will largely depend on the strength characteristics of the foundation. First of all, it is worth finding out the strength indicators of soils on the site. For this, two methods are used: manual drilling or excerpt of pits. The soil is developed to a depth of 50 cm more than the estimated foundation mark.

Collection of loads

Before calculating the bored foundation, it is also necessary to collect loads from all overlying structures. Two separate calculations are required:

This is necessary because the calculation of the pile foundation grillage and the characteristics of the piles will be performed separately.

When collecting loads, it is necessary to take into account all elements of the building, as well as live loads, which include the mass of snow cover on the roof, as well as the payload on the ceiling from people, furniture and equipment.

To calculate the pile-grillage foundation, a table is compiled in which information about the mass of structures is entered. To calculate this table, you can use the following information:

Design
Frame wall with insulation, 15 cm thick	30-50 kg/sq.m.
Wooden wall 20 cm thick	100 kg/sq.m.
Wooden wall 30 cm thick	150 kg/sq.m.
Brick wall 38 cm thick	684 kg/sq.m.
Brick wall 51 cm thick	918 kg/sq.m.
Plasterboard partitions 80 mm without insulation	27.2 kg/sq.m.
Plasterboard partitions 80 mm with insulation	33.4 kg/sq.m.
Interfloor ceilings on wooden beams with insulation	100-150 kg/sq.m.
Interfloor floors made of reinforced concrete 22 cm thick	500 kg/sq.m.
Pie roofing using a coating of
sheets of metal tiles and metal	60 kg/sq.m.
ceramic tiles	120 kg/sq.m.
shingles	70 kg/sq.m.
Live loads
From furniture, people and equipment	150 kg/sq.m.
from the snow	determined according to the table. 10.1 SP "Loads and impacts" depending on the climatic region

The self-weight of the foundations and the grillage is determined depending on the geometric dimensions. First you need to calculate the volume of the structure. The density of reinforced concrete is assumed to be 2500 kg/m3. To get the mass of an element, multiply the volume by the density.

Each component of the load must be multiplied by a special coefficient, which increases reliability. It is selected depending on the material and method of manufacture. The exact value can be found in the table:

Pile calculation

At this stage of the calculations, it is necessary to determine the following characteristics:

• pile pitch;
• the length of the pile to the edge of the grillage;
• section.

Most often, the dimensions of the section are determined in advance, and the remaining indicators are selected based on their available data. Thus, the result of the calculation should be the distance between the piles and their length.

The entire mass of the building obtained at the previous stage must be divided by the total length of the grillage. In this case, both external and internal walls are taken into account. The result of the division will be the load on each running meter of foundations.

The bearing capacity of one element of the foundation can be found by the formula:
P = (0.7 R S) + (u 0.8 fin li), where:

• P is the load that one pile can withstand without destruction;
• R - soil strength, which can be found in the tables below after studying the composition of the soil;
• S - sectional area of ​​the pile in the lower part, for a round pile the formula is as follows: S = 3.14*r2/2 (here r is the radius of the circle);
• u - the perimeter of the foundation element, can be found by the formula for the perimeter of a circle for a round element;
• fin - soil resistance on the sides of the foundation element, see table for clay soils above;
• li is the thickness of the soil layer in contact with the side surface of the pile (find for each soil layer separately);
• 0.7 and 0.8 are coefficients.

The step of the foundations is calculated using a simpler formula: l = P / Q, where Q is the mass of the house per linear meter of the foundation, found earlier. To find the distance between the bored piles in the light, the width of one foundation element is simply subtracted from the found value.

Reinforcement of bored piles is carried out in accordance with regulatory documents. Reinforcing cages consist of working reinforcement and clamps. The first takes on the bending effects, and the second ensures the joint work of individual rods.

Frames for bored piles are selected depending on the load and section dimensions. The working reinforcement is installed in a vertical position; steel rods D from 10 to 16 mm are used for it. In this case, a material of class A400 (with a periodic profile) is chosen. For the manufacture of transverse clamps, you will need to purchase class A240 smooth fittings. D = minimum 6-8 mm.

The frames of bored piles are installed so that the metal does not extend beyond the edge of the concrete by 2-3 cm. This is necessary to provide a protective layer that prevents corrosion (rust on the reinforcement).

The dimensions of the grillage and its reinforcement

The element is designed in the same way as the strip foundation. The height of the grillage depends on how much you need to raise the building, as well as on its mass. You can independently calculate an element that is flush with the ground, or slightly buried in it. The basis for calculating the hanging option is too complicated for a non-specialist, so this work should be entrusted to professionals.

An example of the correct knitting of a reinforcing cage

The dimensions of the grillage are calculated as follows: B \u003d M / (L R), where:

• B is the minimum distance for supporting the tape (width of the strapping);
• M is the mass of the building, excluding the weight of the piles;
• L - strapping length;
• R is the strength of the soil near the surface of the earth.

The reinforcement cages of the strapping are selected in the same way as for a building on a strip foundation. In the grillage, it is required to install working reinforcement (along the tape), horizontal transverse, vertical transverse.

The total cross-sectional area of ​​the working reinforcement is selected so that it is not less than 0.1% of the tape section. To select the cross section of each rod and their number (even), use the assortment of reinforcement. It is also necessary to take into account the instructions of the joint venture for the smallest sizes.

Calculation example

To better understand the principle of performing calculations, it is worth studying an example calculation. Here we consider a one-story building made of brick with a hip roof made of metal. The building is supposed to have two floors. Both are made of reinforced concrete with a thickness of 220 mm. The dimensions of the house in terms of 6 by 9 meters. The thickness of the walls is 380 mm. Floor height - 3.15 m (from floor to ceiling - 2.8 m), the total length of the internal partitions - 10 m. There are no internal walls. Hard-plastic sandy loam was found at the site, the porosity of which is 0.5. The depth of this sandy loam is 3.1 m. From here, according to the tables, we find: R = 46 tons / sq.m., fin = 1.2 tons / sq.m. (for calculations, the average depth is taken equal to 1 m). Snow load is taken according to the values ​​of Moscow.

We collect loads in the form of a table. At the same time, we do not forget about the reliability coefficients.

Type of load	Payment
Brick walls	wall perimeter = 6+6+9+9 = 30 m; wall area = 30 m * 3m = 90 m2; wall mass \u003d (90 m2 * 684) * 1.2 \u003d 73872 kg
Partitions made of plasterboard, not insulated, 2.8 m high	10m*2.8*27.2kg*1.2 = 913.92kg
Ceiling from reinforced concrete slabs 220 mm thick, 2 pcs.	2pcs*6m*9m*500 kg/m2 *1.3 = 70200 kg
Roof	6 m * 9 m * 60 kg * 1.2 / cos30ᵒ (roof slope) = 4470 kg
Load from furniture and people on 2 floors	2*6m*9m*150kg*1.2 = 19440 kg
Snow	6m*9m*180kg*1.4/cos30° = 15640 kg
TOTAL:	184535.92 kg ≈ 184536 kg

We pre-assign a grillage 40 cm wide, 50 cm high. The length of the pile is 3000 mm, D section = 500 mm. We use an approximate pile pitch of 1500 mm.
To calculate the total number of supports, you need to divide 30 m (grillage length) by 1.5 m (pile pitch) and add 1 pc. If necessary, the value is rounded down to the nearest whole number. We get 21 pcs.

The area of ​​one pile \u003d 3.14 0.52 / 4 \u003d 0.196 sq.m., perimeter \u003d 2 3.14 0.5 \u003d 3.14 m.

Let's find the mass of the grillage: 0.4 m 0.5 m 30 m 2500 kg / m3. 1.3 = 19500 kg.

Let's find the mass of piles: 21 3 m 0.196 sq.m. 2500 kg/cu.m. 1.3 = 40131 kg.

Let's find the mass of the entire building: the sum from the table + the mass of the piles + the mass of the grillage = 244167 kg or 244 tons.

The calculation will require a load per linear meter grillage = Q = 244 t/30 m = 8.1 t/m.

Pile calculation. Example

We find the allowable load on each element according to the formula indicated earlier:
P \u003d (0.7 46 tons / sq.m. 0.196 sq.m.) + (3.14 m 0.8 1.2 tons / sq.m. 3 m) \u003d 15.35 tons.
The pile spacing is assumed to be P/Q = 15.35/8.1= 1.89 m. Rounded up to 1.9 m. If the pitch is too large or small, you need to check a few more options, while changing the length and diameter of the foundations.

For frames, rods D = 14 mm and clamps D = 8 mm are used.

Grillage calculation. Example

It is necessary to calculate the mass of the building excluding piles. Hence M = 204 tons.
The width of the tape is taken equal to M / (L R) \u003d 204 / (30 75) \u003d 0.09 m.
Such a grillage cannot be used. The overhangs of the walls of a brick building from the foundation should not exceed 4 cm. We designate the width as 400 mm. The height remains at 500 mm.

Reinforcement of the pile foundation grillage:

• Working 0.1% * 0.4 * 0.5 \u003d 0.0002 sq.m. = 2 sq. cm. Here, 4 rods with a diameter of 8 mm will be enough, but according to regulatory requirements, we use the minimum possible diameter of 12 mm;
• Horizontal clamps - 6 mm;
• Vertical collars - 6 mm.

The calculation will take a certain amount of time. But with their help, you can save money and time in the construction process.

You can also calculate the foundation using an online calculator. Just click on the Calculate Column Foundation link and follow the instructions.

The construction of any foundation begins with design. Calculations and drawings can be performed without the involvement of specialists, on their own. Of course, these calculations will not be highly accurate and will represent a simplified version of the calculation, but they can give an idea of ​​​​how to ensure the bearing capacity of the foundation. Further, bored piles and an example of their calculation are considered.

Design work is carried out in the following order:

• study of soil characteristics;
• collection of loads on the foundation;
• bearing capacity calculations, determination of the distance between piles and their sections.

About each item in order.

Geological surveys

During mass construction, the characteristics for the calculators are prepared by geologists. They take soil samples, conduct laboratory tests and give accurate values ​​for the bearing capacity of a particular layer, the location of soils with different characteristics. If bored piles are used for private housing construction, it is not economically profitable to carry out such activities. The work is done independently in two ways:

• pits;
• manual drilling.

Important! Characteristics are studied at several points, all of them located under the build-up patch of the building. One is always in the lowest part of the earth's surface. The depth of soil development in the study of soil characteristics is assigned 50 cm below the expected mark of the base of the foundation.

Pit - a pit of a rectangular or square shape, the soil is studied by analyzing the soil of the walls of an open pit. When drilling, soil analysis is performed on the blades of the drill. After reviewing, determine the type of soil. For some types of substrates, it will be necessary to determine the consistency or moisture content. Table 1 will help with this question.

External signs and methods	Consistency
Clay bases
If the soil is compressed or hit, it crumbles into pieces.	Semi-hard or hard ground
The sample is difficult to knead, when trying to break the bar, before breaking into two parts, it is strongly bent	hard-plastic
Retains molded shape, easy to mold	soft plastic
Wrinkles hands without difficulty, but does not retain the sculpted shape	fluid plastic
If the sample is placed on an inclined surface, then it will slowly slide down (drain)	Fluid
sandy foundations
Disintegrates when squeezed in the hand, has no external signs of moisture	Dry
The check is carried out with filter paper, it must remain dry or damp after a period of time. When squeezed in the palm of the hand, the sample gives a feeling of coolness.	low humidity
The sample is placed on filter paper and a wet spot is observed. When compressed, a feeling of moisture is created. Able to maintain shape for some time	Wet
Shake the sample in the palm of your hand, it should turn into a cake	Saturated with moisture
Spreads or spreads without external mechanical action (at rest)	waterlogged

Having determined by external signs the type and consistency of the base with the use and tables, they begin to determine the standard resistances. These values ​​are needed to calculate the bearing capacity of the foundation and calculate the distance between the piles.

Bored piles transfer the load not only on the layer of soil on which they rest, but also on the entire side surface. This increases their effectiveness.

Table 2 shows the standard resistance of the bases, in places where the soles of bored piles rest on them.

Priming	Regulatory resistance, taking into account additional tests, t / m 2
Clay bases
—	Porosity factor	Solid consistency		Semi-hard		hard-plastic		soft plastic
sandy loam	0,50	47		46		43		41
	0,70	39		38		35		33
Loam	0,50	47		46		43		41
	0,70	37		36		33		31
	1,00	30		29		24		21
Clay	0,50	90		87		78		72
	0,60	75		72		63		57
	0,80	45		43		39		36
	1,10	37		35		28		24
sandy foundations
—	Dense				medium density
	wet		low-moisture		wet		low-moisture
Large fraction	70		70		50		50
Medium faction	55		55		40		40
Fine fraction*	37		45		25		30
Dusty*	30		40		20		30
Coarse clastic bases
Crushed stone with sand	90
Gravel formed from crystalline rocks	75
Gravel formed from sedimentary rocks	45

The soil porosity coefficient is the ratio of the volume of voids to the total volume of the rock. To calculate the pore sizes of cohesive rocks (clayey), such quantities as specific and volumetric gravity are used.

Also, when calculating the bearing capacity of bored piles, it is necessary to take into account the resistance along the lateral surface. Values ​​for shale formations are presented in Table 3.

Having found out all the necessary data related to soil resistance, proceed to the next point in the calculation of the bearing capacity of the foundation.

Collection of loads

Here it is necessary to take into account the mass of all structures. These include:

• walls and partitions;
• overlaps;
• roof;
• temporary loads.

The first three loads are permanent. They depend on what materials the house will be built from. To calculate the mass of walls, ceilings or partitions, they take the density of the material from which they are planned to be made, and multiply by the thickness and area. When calculating the roof, everything is a little more complicated. You need to take into account:

• filing;
• lower and upper crate;
• rafter legs;
• insulation (if any);
• roofing.

You can give average values ​​​​for the three most common types of roofing:

1. weight of 1 m2 of a roof pie with a metal tile coating - 60 kg;
2. ceramic tiles - 120 kg;
3. bituminous (flexible) tiles - 70 kg.

Temporary loads include snow and useful. Both are accepted. Snow depends on the climatic region, which is determined by the joint venture "Construction climatology". Useful is assigned depending on the purpose of the building. For residential - 150 kg / m² of floors.

It is not enough to calculate all the loads, each of them needs to be multiplied by the reliability factor.

• the coefficient for calculating permanent loads depends on the material and method of manufacturing the structure and is taken according to table 7.1;
• coefficient for snow load - 1.4;
• the coefficient for useful in a residential building is 1.2.

All values ​​​​are added up and proceed to the calculation of bored piles for bearing capacity.

Formulas for calculations

P = Rosn + Rbok. pov-ti,

where P is the bearing capacity of the pile, Rosn is the bearing capacity of the pile at the base, Rbok. pov-ti - bearing capacity of the side surface.

Rosn \u003d 0.7 * Rn * F,

where Rn is the standard bearing capacity from Table 2, F is the area of ​​the base of the bored pile, and 0.7 is the coefficient of soil uniformity.

Rbok. rep = 0.8 * U * fin * h,

where 0.8 is the coefficient of working conditions, U is the perimeter of the pile along the section, fin is the standard soil resistance at the side surface of the bored pile according to Table 3, h is the height of the soil layer in contact with the foundation.

Q \u003d M / U at home,

where Q is the load per linear meter of the foundation from the building, M is the sum of all loads from the building structures calculated earlier, Uhome is the perimeter of the building.

Important! If the house has a large area and it is planned to install internal walls under which the foundation will be built, their length is added to the perimeter to calculate the distance between the bored piles of the foundation.

where P and Q are the previously found values, and L is the maximum distance between piles.

The calculation to calculate the distance between the foundation piles is usually carried out several times. In this case, different sections and depths are selected.

Important! Due to the fact that not only the supporting part of the bored foundation works, the bearing capacity increases with increasing depth in most cases (depending on the characteristics of the base for the foundation). When designing a support for a future home, it is recommended to consider several examples, changing the cross section and the depth of the foundation. The distance between piles and their number is calculated. After that, the estimate is “pretending” (exact calculations can be time consuming, therefore, approximate values ​​\u200b\u200bare enough), and the most economical option is selected.

Before calculating, you need to familiarize yourself with. According to the requirements of this standard, bored piles up to 3 meters long are recommended to be provided with a diameter of 30 cm or more.

Calculation example

Initial data:

• Geological conditions of the area: at a depth of 2 meters from the soil surface, loams are hard-plating, then hard clays with a porosity coefficient of 0.5 are located throughout the entire depth of the study.
• It is required to design a foundation for a one-story house with an attic. The dimensions of the house in terms of plan are 4 by 8 meters, the roof is covered with metal tiles and is hipped (the height of the outer wall is the same on all sides), the walls are made of bricks 0.38 m thick, the partitions are plasterboard, the ceilings are reinforced concrete slabs. The height of the walls within the first floor is 3 meters, on the attic floor the outer walls are 1.5 meters high. There are no internal walls (only partitions).

Collection of loads:

1. wall mass \u003d 1.2 * (24 m (house perimeter) * 3 m (ground floor) + 24 m * 1.5 m (attic)) * 0.38 m * 1.8 t / m³ (brickwork density) \u003d 88.65 t (1.2 - load safety factor);
2. mass of partitions = 1.2 * 2.7 m (height) * 20 m (total length) * 0.03 t / m² (weight per square meter of partitions) = 2 tons;
3. mass of floors, taking into account the cement screed 3 cm = 1.2 * 0.25 m (thickness) * 32 m² (area of ​​​​one floor) * 2 (first floor floor and attic floor) * 2.5 t / m² = 48 tons;
4. roof weight = 1.2 * 4 m * 8 m * 0.06 t / m² = 2.3 tons;
5. snow load = 1.4 * 4 m * 8 m * 0.18 t/m2 = 8.1 tons;
6. payload = 1.2 * 4 m * 8 m * 0.15 t/m² * 2 (2 floors) = 11.5 tons.

Total: M = 112.94 tons. Building perimeter Uhouse = 24 m, load per linear meter Q = 160.55/24 = 6.69 t / m. We first select a pile with a diameter of 30 cm and a length of 3 m.

According to the formulas for determining the distance between piles

All the necessary formulas are given earlier, you just need to use them in order.

1. F \u003d 3.14 D² / 4 (round pile area) \u003d 3.14 * 0.3 m * 0.3 m / 4 = 0.071 m², U \u003d 3.14 D \u003d 3.14 * 0.3 m = 0.942m; (perimeter of the pile in a circle);

2. Posn \u003d 0.7 * 90 t / m² * 0.071 m2 \u003d 4.47 t;

3. Rbok. pov-ty \u003d 0.8 * (2.8 t / m² * 2 m + 4.8 t / m² * 1) * 0.942 \u003d 7.84 t;

In this formula, 2.8 t / m² is the calculated resistance of the side surface of the pile in refractory loam, 2m is the height of the loam layer in which the foundation is located. The resistance is found according to table 3. The values ​​\u200b\u200bare presented there for the depths of 50, 100 and 200 cm suitable in this case. We take into account the minimum in order to ensure a margin of bearing capacity.

4.8 t/m² is the design resistance of the side surface of the pile in semi-hard clay, 1m is the height of the foundation located in this layer. The last number in the formula is the pile perimeter found in the first paragraph. The values ​​0.7 and 0.8 in paragraphs 2 and 3 are the coefficients from the formulas.

4. Р = 4.47 t + 7.84 t = 12.31 t (full bearing capacity of one pile);

5. L = 12.31 t / 6.69 t/m = 1.84 m - the maximum value of the distance between piles (between centers).

We assign a distance of 1.8 m. the length of our walls is a multiple of 2 m, it is more convenient that the distance between the piles is 2 m, for this you need to slightly increase the bearing capacity of the pile, for example, by increasing its diameter. If the resulting step value is large enough, it is more reasonable to find the minimum, since the greater the distance between the piles, the greater the need for the cross section of the grillage, which will lead to additional costs. By the same principle, calculations are performed for a reduced diameter. The applied amount of material is calculated for several options and the optimal value is selected.

Foundations are an extremely important part of any building. Whether cracks appear on the walls, whether the house will sag over time - it all depends on how well the dimensions and materials for the supporting part are selected. In order to correctly design a bored pile-grillage foundation, it will be necessary to calculate its bearing capacity.

The bearing capacity of the foundation is the load that it can withstand without destruction, deformation or other unpleasant processes. When designing a bored base, you will need to find out the following information:

• element section;
• length;
• distance between individual piles.

The calculation of piles for bearing capacity is often performed with a pre-known section of the foundation. This characteristic depends on the available technology. As initial data it is necessary to prepare:

• composition of soils on the site;
• collection of loads on the support of the house.

Collection of initial data for calculation

Before calculating a bored pile-grillage foundation, it will be necessary to study the properties of the soil at the construction site. This can be done in two ways: extracting pits (deep holes) or drilling with a hand tool. The study of the soil is carried out a little deeper than the intended sole (about 50 cm). When performing work, it is necessary to analyze each soil plate, determine its type.

To get an idea of ​​what soils are, how to distinguish them correctly, it is recommended to read. Annex A deserves special attention, which gives the main definitions.

The next stage in the calculation of a bored pile and grillage is the collection of loads. It is easier to do it in tons. For its implementation, you will need to know the volumes of building structures and the density of the materials from which they are made. To calculate the mass of a building, you need to remember a simple formula from school physics: "We can easily find the mass by multiplying the density by the volume." The collection of loads on foundations includes:

• own weight of the supporting part (appointed approximately);
• a lot of ceilings, walls, partitions (it is better not to subtract openings from the total volume);
• payload on floors (for residential buildings, this load is assigned 150 kg / m 2 of the floor, taken on each floor);
• weight of the roof;
• snow load (depends on the climatic area of ​​construction, the calculation is performed according to).

Advice! To simplify the task, the snow load can be assigned according to a special map or table. That is, without performing complex calculations.

The found mass of each element must be multiplied by the load safety factor. The value of this coefficient depends on the material from which the structure is made. For snow and payloads, the coefficients are constant and are 1.4 and 1.2, respectively.

More information about collecting loads on foundations can be found in the article ".

reference Information

To correctly calculate the bored pile foundation, you will need to know the strength characteristics of the soil. Information about this can be found in VSN 5-71. For convenience, adapted tables from this document are presented below for each type of soil separately.

Table 1. Bearing capacity of clay soils, depending on the consistency and porosity in the bearing area of ​​the pile, t/m2.

Table 2. Bearing capacity of clay soils along the length of a bored pile, t/m2.

Table 3 Bearing capacity of sandy soils, t/m2.

Table 4 Bearing capacity of coarse soils, t/m2.

To calculate the cross-section and distance between piles, one or two (for clays) values ​​must be selected from those given in the table, depending on the results of excavating pits or drilling.

Calculation procedure

After a careful study of all the previous paragraphs for the calculation of the pile-grillage foundation, the following information should be available:

• the mass of the house in tons and the load per linear meter of the grillage;
• bearing capacity of soil in tons per m 2.

To find the load per linear meter of the foundation, you need to divide the mass of the house by the total length of the grillage.

The bearing capacity of one pile is found by the formula:

P = (0.7*R*S) + (u*0.8*fin*li), where

P is the bearing capacity of each foundation pile;

R is the strength of the soil, found according to the table. 1, 3 or 4;

S - sectional area of ​​the pile at the end (the formula for finding is given below);

u - pile perimeter;

fin - soil resistance on the side surface of the bored pile foundation, found from Table. 2;

li is the thickness of the soil layer that resists the side surface;

0.7 and 0.8 are coefficients that take into account the homogeneity of the soil and the working conditions of the pile.

For a pile of circular cross section, the area is found through the diameter or radius: S = 3.14 * D 2 /4 = 3.14 * r 2 /2. Here D and r are the diameter and radius, respectively.

l is the distance between the piles of the bored foundation;

P is the bearing capacity of one pile, found earlier;

Q - load per linear meter of the foundation (weight of the house divided by the length of the grillage).

Advice! Before starting the calculation, you must familiarize yourself with. The minimum diameter of the pile foundation with an element length of less than 3 meters is 30 cm. To find the most rational solution, it is recommended to consider 2-3 options for the geometric dimensions of the piles. For each case, find the distance between the supports and estimate the cost of construction. Choose the most economical option.

A detailed calculation of the distance between piles, considering several examples, can take a long time. But here, the future owner of the house is faced with a choice of what to save: time or money.

Reinforcement of a bored pile

Working reinforcement is located vertically along the pile. As it is used rods of class A400 (Alll) with a diameter of 10-16 mm. The transverse piping is made of smooth reinforcement A240 (Al) with a diameter of 6-8 mm. Each pile must have at least four working vertical bars.

Calculation of the grillage

The calculation of the pile foundation grillage is performed in approximately the same way as the calculations for the tape type of the supporting part of the house. To calculate the width of the tape, you will need to use the formula:

B \u003d M / L * R, where

B - the required width of the grillage;

M is the mass of the house (minus the mass of the piles);

L - grillage length;

R is the bearing capacity of the soil (layer near the surface).

This calculation is suitable for a tape located directly on the ground or with a slight depth. For a hanging grillage, the calculation will be more complicated, it is problematic to carry it out on your own.

Reinforcement grillage

Having chosen the width of the grillage of the bored foundation, it is necessary to correctly reinforce it. You can use the requirements for steel bars from .

As a material for reinforcement, rods of class A400 (Alll) are chosen. The most admissible diameter of working rods - 40 mm. The minimum values ​​are given in the table.

An example of calculating a piled bored foundation

Initial data for calculation:

• one-story brick house with an attic, wall thickness 380 mm;
• dimensions in terms of 7 by 9 meters, no internal load-bearing walls (only partitions), floor height 3 m;
• mansard rafter roofing with a metal tile coating;
• soils on the site - semi-hard clay with a porosity coefficient of 0.6, lies at 3 m, R = 72 t/m2, fin = 3.5 t/m2 (value taken for a depth of 1 m).

It is more convenient to collect loads in tabular form. It is necessary not to forget the coefficients for reliability.

The grillage is preliminarily accepted with a width of 0.4 m and a height of 0.5 m. The length of the bored pile is preliminarily 3 m, the cross section is 40 cm in diameter, and are installed in increments of 1.5 m.

Number of piles = 32 m (L, grillage length) / 1.5 m (pile spacing) +1 = 22 pcs. (round down to the nearest whole number). S \u003d 3.14 * 0.42 / 4 (area formula in terms of diameter, see earlier) \u003d 0.126 m 2.

Grill weight: 0.4 m * 0.5 m * 32 m (length) * 2500 kg / m3 (density of reinforced concrete) * 1.3 (coefficient) = 20800 kg.

Pile weight: 22 pieces * 3 m * 0.126 m2 * 2500 kg / m 3 * 1.3 = 27030 kg.

The total mass of the whole house = 235830 kg = 236 tons.

Load per linear meter = Q = 236 t/32 m = 7.36 t/m.

Pile calculation

Pile calculation option 1.

Bearing capacity of one pile = P = (0.7*R*S) + (u*0.8*fin*li) = (0.7*72 t/m2*0.126 m2) + (1.26 m*0 .8 * 3.5 t / m 2 * 3 m (pile length)) \u003d 16.93 t.

u = 3.14*D = 3.14*0.4 = 1.26 m, where D is the pile diameter.

The distance between the piles = l = P / Q = (16.93 t) / (7.36 t / m) = 2.3 m. The step is large enough, you can reduce the length of the pile to 2 m.

Pile calculation option 2.

In the calculations for the previous case, only one value needs to be replaced. Bearing capacity of one pile \u003d P \u003d (0.7 * R * S) + (u * 0.8 * fin * li) \u003d (0.7 * 72 t / m 2 * 0.126 m2) + (1.26 m * 0.8 * 3.5 t / m 2 * 2 m (pile length)) \u003d 13.41 t.

Distance between piles = l = P/Q = (13.41 t)/(7.36 t/m) = 1.82 m.

Pile calculation option 3.

Consider another option with a pile diameter of 50 cm and a length of 2 m.

S \u003d 3.14 * 0.52 / 4 \u003d 0.196 m 2;

u \u003d 3.14 * D \u003d 3.14 * 0.5 \u003d 1.57 m.

Maximum load of one pile \u003d P \u003d (0.7 * 72 t / m2 * 0.196 m 2) + (1.57 m * 0.8 * 3.5 t / m 2 * 2 m (pile length)) \u003d 18, 67 tons

Distance between supports = l = P/Q = (18.67 t)/(7.36 t/m) = 2.54 m.

It is recommended to choose a pile spacing close to 2 m. In this case, option 2 with foundations of small cross section and length will be optimal. For a more accurate result, you can calculate the material consumption in all cases and compare it.

Since it is planned to build a heavy brick house, we assign larger rods with a diameter of 14 mm as working reinforcement. For the manufacture of transverse clamps, 8 mm reinforcement is used.

Calculation of reinforced concrete grillage
From the mass of the house used in the previous calculations, it is necessary to subtract the mass of the piles. We get a load of 208800 kg = 209 tons.

Grillage width \u003d B \u003d M / L * R \u003d 209 t / (32 m * 72 t / m 2) \u003d 0.1 m. The required grillage width is less than the width of the building wall. We assign a structural value of 0.4 m. The overhangs of the wall from the grillage should not be too large, the maximum value is 0.04 m. We also select the height of the grillage structurally 0.5 m. It remains to assign reinforcement:

• The working is taken 0.001 * 0.6 m * 0.5 m \u003d 0.0003 m2 \u003d 3 cm 2. According to the assortment, 4 rods with a diameter of 10 mm are suitable, but according to the requirements of the joint venture, the minimum value for a grillage side length of 6 m is 12 mm. We accept 4 rods with a diameter of 12 mm (two above and two below).
• Cross reinforcement with a diameter of 6 mm.
• Vertical reinforcement with a diameter of 6 mm (because the tape height is less than 0.8 m).

Performing the calculation will allow the optimal use of materials and labor on the construction site.