Calculation of foundations. Example of calculating the foundation of pile, tape, columnar, monolithic, slab. Calculation of the foundation foundation: an example. Calculation of the foundation for the rollover: an example

The use of typical methods will facilitate the planning and calculation of foundations, an example of calculating the foundation will simplify calculations. Based on the recommendations given in the article, errors can be avoided when constructing the selected structure (columnar, pile, tape or plate type).

Column base

For example, a one-story structure with parameters in the plan of 6x6 m is used, and also with walls from a bar of 15x15 cm (the bulk weight is 789 kg / m³), trimmed from the outside by the lining of roll insulation. The basement of the building is made of concrete: height - 800 mm and width - 200 mm (bulk weight of concrete materials - 2099 kg / m³). It is based on a reinforced concrete beam with a cross section of 20x15 (volume indicators of reinforced concrete - 2399). The walls have a height of 300 cm, and the slate roof is distinguished by two slopes. The socle and the attic are made of boards, located on beams with a section of 15x5, and also insulated with mineral wool (the bulk insulation weight is 299 kg).

Knowing the norms of loads (according to SNiP), it is possible to correctly calculate the foundations. An example of calculating the foundation will allow you to quickly perform calculations for your own building.

Norms of loads

• On the plinth - 149.5 kg / m².
• In the attic - 75.
• The snow load norm for the terrain in the central strip of the Russian Federation is 99 kg / m² relative to the roof area (in horizontal section).
• Different loads are applied to the bases on different axes.

Pressure on each axis

Accurate indicators of structural and regulatory loads allow you to correctly calculate the foundation. An example of calculating the foundation is given for the convenience of beginning builders.

Constructional pressure along the axis "1" and "3" (extreme walls):

• From the log building wall covering: 600 x 300 cm = 1800 cm². This indicator is multiplied by the thickness of the vertical overlap of 20 cm (including external trim). It turns out: 360 cm³ x 799 kg / m³ = 0,28 tons.
• From the handball: 20 x 15 x 600 = 1800 cm³ x 2399 ~ 430 kg.
• From the socle: 20 x 80 x 600 = 960 cm³ x 2099 ~ 2160 kg.
• From the socle. The total mass of the entire overlap is counted, then 1/4 of it is taken.

Logs with sides 5x15 are placed every 500 mm. Their mass is 200 cm³ x 800 kg / m³ = 1600 kg.

It is necessary to determine the mass of floor covering and binder, included in the calculation of foundations. An example of calculating the foundation indicates a layer of insulation thickness of 3 cm.

The volume is 6 mm x 360 cm² = 2160 cm³. Further, the value is multiplied by 800, the total is 1700 kg.

Mineral wool insulation has a thickness of 15 cm.

The volume indices are equal to 15 x 360 = 540 cm³. When multiplying by a density of 300.01, we obtain 1620 kg.

Total: 1600.0 + 1700.0 + 1600.0 = 4900.0 kg. We divide by 4, we get 1.25 tons.

• From the attic ~ 1200 kg;
• From the roof: the total mass of one slope (1/2 roof), taking into account the mass of rafters, gratings and slate flooring - only 50 kg / m² x 24 = 1200 kg.

The load norm for columnar structures (for the axis "1" and "3" it is required to find 1/4 of the total pressure on the roof) allows the calculation of the pile foundation. An example of this design is ideal for printed construction.

• From the socle: (600.0 x 600.0) / 4 = 900.0 x 150.0 kg / m² = 1350.0 kg.
• From the attic: 2 times less than from the basement.
• From the snow: (100 kg / m² x 360 cm²) / 2 = 1800 kg.

As a result: the total indicator of structural loads is 9.2 tons, the standard pressure is 4.1. On each axis "1" and "3" there is a load of about 13.3 tons.

Constructive pressure along the "2" axis (average longitudinal line):

• From a log-house of wall slabs, randbalki and socle load surfaces are analogous to the values of the axis "1" and "3": 3000 + 500 + 2000 = 5500 kg.
• From the socle and attic they have double indicators: 2600 + 2400 = 5000 kg.

Below is the normative load and calculation of the foundation foundation. The example is used in approximate values:

• From the socle: 2800 kg.
• From the attic: 1400.

As a result: the total indicator of the design pressure is 10.5 tons, the standard load is 4.2 tons. The "2" axis has a weight of about 14700 kg.

Pressure on the "A" and "B" axes (transverse lines)

Calculations are made taking into account the structural weight of the logs of wall slabs, rand bars and base (3, 0.5 and 2 t). Pressure on the foundation along these walls will be: 3000 + 500 +2000 = 5500 kg.

Number of poles

To determine the required number of poles with a cross section of 0.3 m, the ground resistance (R) is taken into account:

• At R = 2.50 kg / cm² (often used indicator) and the reference area of the shoes is 7.06 m² (for simplicity of calculation take a smaller value - 7 m²), the load-bearing capacity of one pillar is: P = 2.5 x 7 = 1 , 75 tons.
• An example of calculating a columnar foundation for a soil with resistance R = 1.50 takes the following form: P = 1.5 x 7 = 1.05.
• For R = 1.0, one column is characterized by a bearing capacity P = 1.0 x 7 = 0.7.
• Resistance of watery soil is 2 times less than the minimum values of the tabulated indicators, amounting to 1.0 kg / cm². At a depth of 150 cm, the average is 0.55. The bearing capacity of the column is P = 0.6 x 7 = 0.42.

For the selected house will require a volume of 0.02 m³ of reinforced concrete.

Placements

• Under the wall slabs: along the lines "1" and "3" with a weight of ~ 13.3 tons.
• On the "2" axis with a weight of ~ 14,700 kg.
• Under the wall overlapping along the axes "A" and "B" with a weight of ~ 5500 kg.

If you need to calculate the foundation for a rollover, an example of calculations and formulas are given for large cottages. For suburban areas they are not used. Particular attention is paid to the distribution of the load, which requires a careful calculation of the number of poles.

Examples of calculating the number of poles for all types of soil

Example 1:

R = 2.50 kg / cm²

For wall slabs along the "1" and "3" segments:

13,3 / 1,75 ~ 8 posts.

On the axis "2":

14.7 / 1.75 ~ 9 pcs.

On the segments "A" and "B":

5.5 / 1.75 = 3.1.

Only about 31 poles. The volume index of the concrete material is 31 x 2 mm³ = 62 cm³.

Example 2:

R = 1.50

On the line "1" and "3" ~ for 12 columns.

On the axis "2" ~ 14.

On the segments "A" and "B" ~ 6.

Total ~ 50 pieces. The volume index of the concrete material is ~ 1,0 m³.

Example 3:

Below you can find out how the calculation of a monolithic foundation is carried out . An example is given for a soil with a table index R = 1.0. It has the following form:

On the line "1" and "2" ~ 19 pieces each.

On the wall "2" ~ 21.

By segments "A" and "B" ~ by 8.

Total - 75 poles. The volume index of the concrete material is ~ 1.50 m³.

Example 4:

R = 0.60

On the line "1" and "3" ~ 32 pieces each.

On the axis "2" ~ 35.

On the segments "A" and "B" ~ 13.

Total - 125 pillars. The volume index of the concrete material is ~ 250 cm³.

In the first two calculations, corner columns are set at the intersection of the axes, and along the longitudinal lines - with the same pitch. Under the socle part of the pillars of poles cast reinforced concrete rake in the formwork.

In example number 3 on the intersecting axes are placed on 3 pillars. A similar number of bases are grouped along the axes "1", "2" and "3". Among the builders, this technology is called "bushes". On a separate "bush" it is required to establish a common reinforced concrete blockhead-grillage with its further placement on poles located on the axes "A" and "B" of the randbals.

Example No. 4 makes it possible to construct "bushes" from 4 poles at the intersection and along the longitudinal part of the lines (1-3), with further installation of grill heads on them. They are placed under the foot of the basement.

Belt base

For comparison, the calculation of the strip foundation is made below. The example is given taking into account the depth of the trench 150 cm (width - 40). The canal will be filled with sand mixture at 50 cm, then it will be filled with concrete to a height of one meter. It will be necessary to develop the soil (1800 cm³), lay the sand fraction (600) and the concrete mixture (1200).

Of the 4-columnar bases for comparison, the third is taken.

Work drill are carried out on an area of 75 cm³ with soil utilization of 1.5 cubic meters, or 12 times less (the rest of the soil is used for backfilling). The need for a concrete mix is 150 cm³, or 8 times less, and in the sand fraction - 100 (it is necessary under the bearing beam). Near the foundation is an exploratory pit, which allows you to know the condition of the soil. According to the tabular data 1 and 2, the resistance is selected.

Important! In the lower rows, this data will allow calculation of the slab foundation - an example is given for all types of soil.

Resistance of sandy soil

Table. 1

Resistance of soil to the base, kg / cm ³

Sand fraction	The density level
	Dense	Medium Density
Large	4.49	3.49
Average	3.49	2.49
Small: mowing / wet	3-2.49	2
Dusty: mauve / wet	2.49-1.49	2-1

Table. 2

Resistance to clay soil

The soil	Level Porosity	Resistance of soil, Kg / cm 3
		Solid	Plastic
Sandy loam	0.50 / 0.70	3.0-2.50	2.0-3.0
Loam	0.50-1.0	2.0-3.0	1.0-2.50
Clay soil	0.50-1.0	2.50-6.0	1.0-4.0

Plate foundation

At the first stage, the thickness of the slab is calculated. Take a combined mass of the room, including the weight of the installation, cladding and additional loads. According to this indicator and the area of the slab, the plan calculates the pressure from the premise on the soil without the weight of the base.

It is calculated what is the mass of the slab for a given pressure on the soil (for fine sand this figure is 0.35 kg / cm², average density is 0.25, hard and ductile sandy loam is 0.5, hard clay is 0.5 and plastic - 0.25).

The area of the foundation must not exceed the conditions:

S> Kh × F / Kp × R,

Where S is the base sole;

Kh - coefficient for determining the reliability of the support (it is 1.2);

F - total weight of all plates;

Kp - factor that determines the conditions of work;

R is the soil resistance.

Example:

• The free mass of the building is 270 000 kg.
• Parameters in the plan - 10x10, or 100 m².
• Primer - loam with a moisture content of 0.35 kg / cm².
• The density of reinforced concrete is 2.7 kg / cm³.

The weight of the slabs lags 80 t - 29 cubes of concrete mixture. For 100 squares, its thickness corresponds to 29 cm, so it takes 30.

The total weight of the slab is 2.7 x 30 = 81 tons;

The total mass of the building with a foundation is 351.

The plate has a thickness of 25 cm: its mass is 67.5 tons.

We get: 270 + 67.5 = 337.5 (the pressure on the soil is 3.375 t / m²). This is sufficient for a gas-concrete house with a cement density of B22.5 compression (plate grade).

Determination of structural overturning

The MU moment is determined taking into account the wind speed and the area of the building to which the impact is effected. Additional fastening is required if the following condition is not fulfilled:

MU = (Q-F) * 17.44

F - the lifting force of the effect of wind on the roof (in the example given it is 20.1 kN).

Q - calculated minimum asymmetric load (by the condition of the task it is equal to 2785.8 kPa).

When calculating parameters, it is important to take into account the location of the building, the presence of vegetation and the construction of a number of structures. Great attention is paid to weather and geological factors.

The above indicators are used to illustrate the work. If it is necessary to build a building on its own, it is recommended to consult with specialists.