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 be different from the calculations for screw piles. But in all cases, preliminary preparation is required, which includes the collection of loads and geological surveys.
Studying soil characteristics
The bearing capacity of the bored pile will largely depend on the strength characteristics of the base. First of all, it is worth finding out the strength characteristics of soils on the site. Two methods are used for this: manual drilling or a passage of pits. Soil is developed to a depth of 50 cm more than the estimated elevation of the foundation.
Before calculating the pile foundation, it is recommended to familiarize yourself with GOST "Soils. Classification ”Appendix A. There are presented the basic definitions, based on which, the type of soil can be determined visually.
Next, you will need a table indicating the strength of the soil, depending on its type and consistency. All the characteristics necessary for the calculation are shown in the pictures below.
Load collection
Before calculating the bored foundation, it is also necessary to collect the loads from all overlying structures. Two separate calculations will be required:
- load on the pile (including grillage);
- load on grillage.
This is necessary because the grillage of the pile foundation and the characteristics of the piles will be separately calculated.
When collecting loads, it is necessary to take away all the elements of the building, as well as temporary loads, which include the mass of snow 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 | Load |
---|---|
Frame wall with insulation, 15 cm thick | 30-50 kg / sq.m. |
20 cm thick wooden wall | 100 kg / sq.m. |
30 cm thick wooden wall | 150 kg / sq.m. |
38 cm thick brick wall | 684 kg / sq.m. |
51 cm thick brick wall | 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 laying | 100-150 kg / sq.m. |
Reinforced concrete floors with a thickness of 22 cm | 500 kg / sq.m. |
Roof pie using a coating of | |
metal tiles and metal sheets | 60 kg / sq.m. |
ceramic tiles | 120 kg / sq.m. |
shingles | 70 kg / sq.m. |
Temporary load | |
From furniture, people and equipment | 150 kg / sq.m. |
from snow | determined by the table. 10.1 JV “Loads and impacts” depending on the climatic region |
The own weight of the foundations and 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 equal to 2500 kg / cubic meter. To get the mass of an element, you need to multiply the volume by density.
Each component of the load must be multiplied by a special coefficient, which increases reliability. It is selected depending on the material and manufacturing method. The exact value can be found in the table:
Load type | Coefficient |
---|---|
Constant for: - wood - metal - insulation, backfill, screeds, reinforced concrete - manufactured at the factory - manufactured at the construction site | 1,1 1,05 1,1 1,2 1,3 |
From furniture, people and equipment | 1,2 |
From the snow | 1,4 |
Pile calculation
At this stage of the calculation, it is necessary to determine the following characteristics:
- pitch of piles;
- the length of the pile to the edge of the grillage;
- section.
Most often, section sizes 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 in the previous step 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 m 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 breaking;
- R is the soil strength, which can be found in the tables below after studying the composition of the soil;
- S is the cross-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 is the perimeter of the foundation element, can be found by the formula of the perimeter of a circle for a round element;
- fin - soil resistance on the lateral sides of the foundation element, see the table for clay soils above;
- li is the thickness of the soil layer in contact with the side surface of the pile (found for each soil layer separately);
- 0.7 and 0.8 are coefficients.
The pitch of the foundations is calculated by a simpler formula: l = P / Q, where Q is the mass of the house per linear meter of 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.
When performing calculations, it is recommended to consider several options with different element lengths. After that, it will be easy to choose the most economical.
Reinforcement of bored piles is carried out in accordance with regulatory documents. Reinforcing cages consist of working fittings and clamps. The first takes on bending effects, and the second ensure the joint operation of individual rods.
Frames for bored piles are selected depending on the load and section size. The working fittings are 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 selected. For the manufacture of transverse clamps, it will be necessary to purchase smooth reinforcement of class A240. D = minimum 6-8 mm.
Frames of bored piles are installed so that the metal does not reach the edge of concrete by 2-3 cm. This is necessary to provide a protective layer that prevents corrosion (rust on the reinforcement).
Dimensions 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 lift the building, as well as on its mass. You can independently perform the calculation of an element that rests flush with the ground, or is slightly buried in it. The basis of the calculations of the hanging option is too complicated for the layman, so this work should be entrusted to professionals.
The sizes of grillages are calculated as follows: B = M / (L • R), where:
- B is the minimum distance to support the tape (strapping width);
- M - the mass of the building without taking into account the weight of the piles;
- L is the length of the strapping;
- R is the strength of the soil at the surface of the earth.
Reinforcing wireframes 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 cross-section of the tape. To choose the cross section of each bar and their number (even), use the assortment of reinforcement. It is also necessary to take into account the instructions of the joint venture on the smallest sizes.
Working fittings | tape side length <3m | from 10 mm |
tape side length> 3m | from 12 mm | |
Horizontal clamps | from 6 mm | |
Vertical clamps tape <80 cm high | from 6 mm | |
Vertical clamps with tape height> 80 cm | from 8 mm |
Calculation Example
To better understand the principle of performing calculations, it is worth exploring an example of calculation. Here we consider a single-story brick building 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 are 6 by 9 meters. The wall thickness is 380 mm. The height of the floor is 3.15 m (from the floor to the ceiling - 2.8 m), the total length of the internal partitions is 10 m. There are no internal walls. A refractory sandy loam was found on 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. In this case, do not forget about the safety factors.
Type of load | Calculation |
---|---|
Brick walls | wall perimeter = 6 + 6 + 9 + 9 = 30 m; wall area = 30 m * 3m = 90 m2; wall weight = (90 m2 * 684) * 1.2 = 73872 kg |
Partitions made of drywall not insulated 2.8 m high | 10m * 2.8 * 27.2kg * 1.2 = 913.92 kg |
Overlap of reinforced concrete slabs with a thickness of 220 mm, 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 |
The load of 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 |
First we assign a grillage with a width of 40 cm, a height of 50 cm. The length of the pile is 3000 mm, D section = 500 mm. We use an approximate pitch of piles of 1500 mm.
To calculate the total number of supports, you need to divide 30 m (length of the grillage) by 1.5 m (pitch of piles) and add 1 pc. If necessary, the value is rounded to the nearest integer downward. We get 21 pcs.
The area of one pile = 3.14 • 0.52 / 4 = 0.196 sq.m., perimeter = 2 • 3.14 • 0.5 = 3.14 m.
We find the mass of grillage: 0.4 m • 0.5 m • 30 m • 2500 kg / m3 • 1.3 = 19500 kg.
Find the mass of piles: 21 • 3 m • 0.196 sq.m. • 2500 kg / m3. • 1.3 = 40131 kg.
Find the mass of the entire building: the sum from the table + the mass of piles + the mass of grillage = 244167 kg or 244 tons.
For the calculation, the load on the linear meters of grillage = Q = 244 t / 30 m = 8.1 t / m will be required.
Calculation of piles. Example
We find the permissible loading for each element according to the formula indicated earlier:
P = (0.7 • 46 tons / sq.m. • 0.196 sq.m.) + (3.14 m • 0.8 • 1.2 tons / sq.m. • 3 m) = 15.35 t .
The pitch of the piles is taken equal to P / Q = 15.35 / 8.1 = 1.89 m. We round up to 1.9 m. If the step is too large or small, you need to check a few more options, changing the length and diameter of the foundations.
For frames, rods D = 14 mm and clamps D = 8 mm are used.
Calculation of grillage. 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) = 204 / (30 • 75) = 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 assign a structural width of 400 mm. The height remains 500 mm.
Reinforcement of grillage of pile foundation:
- Working 0.1% * 0.4 * 0.5 = 0.0002 sq.m. = 2 sq. Cm. There will be enough 4 rods with a diameter of 8 mm, but according to regulatory requirements we use the smallest possible diameter of 12 mm;
- Horizontal clamps - 6 mm;
- Vertical clamps - 6 mm.
The calculations will take a certain period of time. But with their help, you can save money and time during the construction process.
You can also calculate the foundation using the online calculator. Just click on the Calculation of the foundation of the column type link and follow the instructions.