Using standard methods will facilitate the planning and calculation of foundations, an example of calculation of the foundation will simplify calculations. Based on the recommendations given in the article, it is possible to avoid errors in the construction of the selected structure (columnar, pile, tape or slab type).
Pillar base
For example, a one-story building is used with parameters in terms of 6x6 m, as well as with walls made of timber 15x15 cm (volumetric weight is 789 kg / m³), finished on the outside with clapboard on roll insulation. The basement of the building is made of concrete: height - 800 mm and width - 200 mm (volumetric mass of concrete materials - 2099 kg / m³). It is based on a reinforced concrete beam with a section of 20x15 (volume indicators of reinforced concrete - 2399). The walls are 300 cm high, and the slate roof is distinguished by two slopes. The plinth and attic are made of boards located on beams with a section of 15x5, and are also thermally insulated with mineral wool (bulk weightinsulation is 299 kg).
Knowing the norms of loads (according to SNiP), you can correctly calculate the foundations. An example of a foundation calculation will allow you to quickly perform calculations for your own building.
Load rates
- Plinth - 149.5 kg/m².
- To the attic - 75.
- The norm of snow load for the area in the middle zone of the Russian Federation is 99 kg / m² relative to the roof area (in a horizontal section).
- Different loads exert pressure on the bases along different axes.
Pressure on each axis
Accurate indicators of structural and standard loads allow you to correctly calculate the foundations. An example of calculating the foundation is given for the convenience of beginner builders.
Constructive pressure along axis "1" and "3" (end walls):
- From the log house of the wall slab: 600 x 300 cm=1800 cm². This figure is multiplied by the thickness of the vertical overlap of 20 cm (including exterior finishes). It turns out: 360 cm³ x 799 kg / m³ \u003d 0.28 tons.
- From a rand beam: 20 x 15 x 600=1800 cm³ x 2399 ~ 430 kg.
- From plinth: 20 x 80 x 600=960 cm³ x 2099 ~ 2160 kg.
- From the base. The total mass of the entire overlap is calculated, then 1/4 of it is taken.
Lags with 5x15 sides are placed every 500mm. Their mass is 200 cm³ x 800 kg/m³=1600 kg.
It is necessary to determine the mass of the floor covering andsheeting included in the calculation of foundations. An example of a foundation calculation indicates a 3 cm thick insulation layer.
The volume is 6 mm x 360 cm²=2160 cm³. Further, the value is multiplied by 800, the total will be 1700 kg.
Mineral wool insulation is 15 cm thick.
Volumetric indicators are 15 x 360=540 cm³. When multiplied by the density of 300.01, we get 1620 kg.
Total: 1600, 0 + 1700, 0 + 1600, 0=4900, 0 kg. We divide everything by 4, we get 1.25 t.
- From the attic ~ 1200 kg;
- From the roof: the total weight of one slope (1/2 of the roof), taking into account the weight of the rafters, grating and slate flooring - only 50 kg / m² x 24=1200 kg.
Load rate for columnar structures (for axes "1" and "3" you need to find 1/4 of the total pressure on the roof) allows you to calculate the pile foundation. An example of the construction in question is ideal for stuffed construction.
- From base: (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 snow: (100 kg/m² x 360 cm²) /2=1800 kg.
As a result: the total indicator of constructive loads is 9.2 tons, standard pressure - 4.1. Each axle "1" and "3" has a load of about 13.3 tons.
Design pressure along axis "2" (middle longitudinal line):
- From the log house of wall slabs, rand beams and the basement surface of the load are similar to the values \u200b\u200bof the axis "1" and "3": 3000 +500 + 2000=5500 kg.
- From the basement and the attic they have double indicators: 2600 +2400=5000 kg.
Below is the normative load and calculation of the base of the foundation. Example used in approximate values:
- From plinth: 2800 kg.
- From the attic: 1400.
As a result: the total design pressure is 10.5 tons, standard loads - 4.2 tons. Axle "2" has a weight of about 14,700 kg.
Pressure on axes "A" and "B" (cross lines)
Calculations are made taking into account the structural weight from the log house of wall slabs, rand beams and plinth (3, 0, 5 and 2 tons). The pressure on the foundation along these walls will be: 3000 + 500 +2000=5500 kg.
Number of poles
To determine the required number of pillars with a cross section of 0.3 m, soil resistance (R) is taken into account:
- With R \u003d 2.50 kg / cm² (often used indicator) and the base area of \u200b\u200bthe shoes is 7.06 m² (for ease of calculation, a smaller value is taken - 7 m²), the bearing capacity of one column will be: P \u003d 2, 5 x 7=1.75 t.
- An example of calculating a columnar foundation for soil with resistance R=1.50 takes the following form: P=1.5 x 7=1.05.
- When R=1.0, one column is characterized by the bearing capacity P=1.0 x 7=0.7.
- The resistance of watery soil is 2 times less than the minimum values of tabular indicators, which are 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.
The selected house will require a volume of 0.02 m³ of reinforced concrete.
Placement points
- For wall slabs: along lines "1" and "3" with a weight of ~ 13.3 t.
- Axis "2" with a weight of ~ 14700 kg.
- For wall ceilings along the axes "A" and "B" with a weight of ~ 5500 kg.
If you need to calculate the foundation for overturning, an example of calculations and formulas are given for large cottages. They are not used for suburban areas. Particular attention is paid to load distribution, which requires careful calculation of the number of posts.
Examples of calculating the number of pillars for all types of soil
Example 1:
R=2.50kg/cm²
For wall slabs along the segment "1" and "3":
13, 3 /1, 75 ~ 8 pillars.
Axis 2:
14, 7/1, 75 ~ 9pcs
On segments "A" and "B":
5, 5 /1, 75=3, 1.
There are approximately 31 poles in total. The volumetric index of the concreted material is 31 x 2 mm³=62 cm³.
Example 2:
R=1, 50
On the line "1" and "3" ~ 12 columns each.
Axis 2 ~ 14.
On segments "A" and "B" ~ on 6.
Total ~ 50 pieces. Volumetric index of concreted material ~ 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 soil with a tabular indicator R=1, 0. It looks like this:
On line "1" and "2" ~ 19 pieces each
On the wall "2" ~21.
On segments "A" and "B" ~ on 8.
Total - 75 pillars. Volumetric index of concreted material ~ 1.50 m³.
Example 4:
R=0, 60
On line "1" and "3" ~ 32 pieces each
Axis 2 ~ 35.
On segments "A" and "B" ~ on 13.
Total - 125 pillars. Volumetric index of concreted material ~ 250 cm³.
In the first two calculations, the corner posts are installed at the intersection of the axes, and along the longitudinal lines - with the same step. Reinforced concrete rand beams are cast in the formwork under the basement along the heads of the pillars.
In example 3, 3 columns are placed on the intersecting axes. A similar number of bases are grouped along the axes "1", "2" and "3". Among builders, this technology is called "bushes". On a separate "bush" it is required to install a common reinforced concrete grillage head with its further placement on poles located on the axes "A" and "B" of the rand beams.
Example No. 4 allows you to build "bushes" of 4 pillars at the intersection and along the longitudinal part of the lines (1-3) with further installation of grillage heads on them. Rund beams are placed along them under the basement.
Strip 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 channel will be covered with sand mixture to a depth of 50 cm, then it will be filled with concrete to a height of one meter. Soil excavation (1800 cm³), sand fraction (600) and concrete mix (1200) will be required.
From4-column bases for comparison is taken third.
Drilling is 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 mixture is 150 cm³, or 8 times less, and in the sand fraction - 100 (it is needed under the supporting beam). An exploratory pit is being created near the foundation, allowing you to find out the condition of the soil. According to tabular data 1 and 2, resistance is selected.
Important! In the lower lines, these data will allow the calculation of the slab foundation - an example is indicated for all types of soil.
Sand soil resistance
Tab. 1
| Sand faction | Density level | |
| Tight | Medium heavy | |
| Large | 4, 49 | 3, 49 |
| Average | 3, 49 | 2, 49 |
| Fine: low/wet | 3-2, 49 | 2 |
| Dusty: slightly damp/wet | 2, 49-1, 49 | 2-1 |
Tab. 2
| Soil | Levelporosity | Soil resistance, kg/cm3 | |
| Solid | Plastic | ||
| Supesi | 0, 50/0, 70 | 3, 0-2, 50 | 2, 0-3, 0 |
| Loams | 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 |
Slab Foundation
At the first stage, the thickness of the slab is calculated. The total mass of the room is taken, including the weight of the installation, cladding and additional loads. Based on this indicator and the area of the slab in the plan, the pressure from the placement on the soil without the weight of the base is calculated.
It is calculated what mass of the plate is missing for a given pressure on the soil (for fine sand, this figure will be 0.35 kg / cm², medium density - 0.25, hard and plastic sandy loam - 0.5, hard clay - 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 - coefficient that determines the working conditions;
R – soil resistance.
Example:
- Loose weight of the building is 270,000 kg.
- The parameters in the plan are 10x10, or 100 m².
- Soil - loam with a moisture content of 0.35 kg/cm².
- The density of reinforced concrete is 2.7 kg/cm³.
The mass of the slabs is 80 tons behind - this is 29 cubes of concrete mix. For 100 squares, its thickness corresponds to 29 cm, so 30 is taken.
The total weight of the slab is 2.7 x 30=81 tons;
The total mass of the building with the foundation is 351.
The plate has a thickness of 25cm: 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 enough for an aerated concrete house with a cement density for compression B22.5 (plate brand).
Determining structure overturning
The moment MU is determined taking into account the wind speed and the area of the building that is affected. Additional fastening is required if the following condition is not met:
MU=(Q - F) 17, 44
F is the lifting force of the wind action on the roof (in the given example it is 20.1 kN).
Q is the calculated minimum asymmetric load (according to the condition of the problem, it is 2785.8 kPa).
When calculating the parameters, it is important to take into account the location of the building, the presence of vegetation and structures erected nearby. Much attention is paid to weather and geological factors.
The above indicators are used for clarity of work. If you need to build a building yourself, it is recommended to consult with specialists.
