Conclusion: the studied sample No. 3 is gray, fine, medium-density sand saturated with water with R o = 200 kPa.

2.4 Sample no. 4

The sample was taken from well No. 2, the sampling depth was 8 m.

The name of the soil is determined by the number of plasticity.

The plasticity number is determined by the formula (2.12):

I=41-23=18 – the soil belongs to clays (I>17) in accordance with Table B.11.

The porosity coefficient is determined by the formula (2.10):

,

0 ≤J L ≤0.25 - semi-solid soil in accordance with Table B.14.

According to SNiP 2.02.01-83 * "Foundations of buildings and structures" by double interpolation find

Conclusion: the studied sample No. 4 is brown semi-solid clay with R o = 260.7 kPa.

2.5 Sample no. 5

The sample was taken from well No. 3, the sampling depth was 12 m.

The name of the soil is determined by the number of plasticity.

The plasticity number is determined by the formula (2.12):

I=20-13=7 - the soil belongs to sandy loam (1I7) in accordance with Table B.11.

The porosity coefficient is determined by the formula (2.10):

,

The consistency coefficient is determined by the formula (2.13):

S= = 1

0.25 ≤J L ≤0.5 - hard plastic soil in accordance with Table B.14.

Calculated resistance is determined according to App. 3 R=300kPa.

Conclusion: the studied sample No. 5 is hard plastic grayish-yellow sandy loam with R o = 300 kPa.

3 Collection of loads acting on foundations

The collection of loads is carried out on the cargo area, which is set depending on the static scheme of the structure. In this case, a structural scheme with transverse load-bearing walls located at a modular step of 6.3 and 3.0 m, two longitudinal reinforced concrete walls and flat reinforced concrete floors, forming a spatial system that ensures the seismic resistance of the building and perceives all vertical and horizontal loads.

The values ​​of temporary loads are set in accordance with. Reliability factors for loads g f are also determined by.

Loads are collected from the top of the building to the layout mark.

Figure 3.1 - Cargo area

When calculating live loads, we take the load safety factor equal to 1.4 in accordance with . Collection of temporary loads on interfloor floors, taking into account the reduction factor

, (3.1)

where n is the number of floors from which the load is transferred to the base;

.

Table 3.1 - Collection of loads

4 Selecting the type of base

Judging by the geological section, the site has a calm relief with absolute elevations of 129.40 m, 130.40 m, 130.70 m.

The soil has a sustained bedding of soils. Soils, being in their natural state, can serve as the basis for shallow foundations. For this type of foundation, layer No. 2 will serve as the foundation - silty sand of medium plasticity with R = 150 kPa.

For a pile foundation, it is better to use layer No. 4 as a working layer - fine sand of medium density with R = 260.7 kPa.

5 Choosing a rational type of foundation

The choice of the type of foundations is made on the basis of a technical and economic comparison of the options most commonly used in the practice of industrial construction of foundations:

1 shallow;

2 pile foundations.

The calculation is made for the section with the maximum load - for the section 1-1.

5.1 Calculation of shallow foundation on a natural basis

We set the depth of the foundation sole, depending on the freezing depth, the properties of the soil base and the design features of the structure.

For the city of Komsomolsk-on-Amur, the standard freezing depth is determined by the formula

(5.10)

where L v is the heat of thawing (freezing) of the soil, is found by the formula

, (5.12)

where z 0 - specific heat of phase transformation water - ice,

;

total natural soil moisture, fractions of a unit, ;

the relative (by mass) content of unfrozen water, fractions of a unit, is found by the formula

(5.13)

k w - coefficient taken according to table 1 depending on the plasticity number I p and soil temperature T, ° C;

w p - soil moisture at the boundary of plasticity (rolling), fractions of a unit.

Ground freezing point, °С.

T f,m t f,m - respectively, the average air temperature over long-term data for the period of negative temperatures, °С and the duration of this period, h;

C f - volumetric heat capacity, respectively, of thawed and frozen soil, J / (m 3 × ° С)

l f - thermal conductivity of thawed and frozen soil, respectively, W / (m × ° С)

The estimated freezing depth is determined by the formula

where k h is the coefficient that takes into account the influence of the thermal regime of the structure, ,

0.4 . 2.6 = 1.04 m

Since the laying depth does not depend on the estimated freezing depth, the laying depth is taken for constructive reasons. In our case, the laying depth is set aside from the basement floor structure (see Figure 5.1).

Figure 5.1 Foundation depth

2.72 - 1.2 = 1.52 m

All subsequent calculations are performed by the method of successive approximations in the following order:

Preliminarily determine the area of ​​​​the base of the foundation by the formula

, (5.15)

R o - design soil resistance under the base of the foundation, R 0 \u003d 150 kPa;

h is the depth of the sole, 1.52 m;

k zap - fill factor (taken equal to 0.85);

g is the specific gravity of the foundation materials (taken equal to 25 kN / m 3).

According to table 6.5, we select a slab of the FL 20.12 brand, having dimensions: 1.18 m, 2 m, 0.5 m and wall blocks of the FBS 12.4.6 brand, having dimensions: 1.18 m, 0.4 m, 0.58 m, wall blocks of the FBS 12.4.3 brand, having dimensions: 1.18 m, 0.4 m, 0.28 m.

According to table 2 of appendix 2 for dusty sand of medium plasticity with e = 0.67 we find 29.2 o and 3.6 kPa

According to table 5.4, interpolating by the angle of internal friction φ n, we find the values ​​of the coefficients: 1.08, 5.33, 7.73.

We determine the value of the calculated resistance R by the formula

where g c1 and g c2 are the coefficients of working conditions, taken according to Table 5.3

g c1 = 1.25 and g c2 = 1.2;

k is a coefficient taken equal to 1.1 if the strength characteristics

soil (c and j) are taken according to Table. 1.1;

M g , M q , M c - dimensionless coefficients taken according to Table. 1.3;

k Z is the coefficient taken at b< 10 м равным 1;

b - width of the base of the foundation, b = 2 m;

g II - the calculated value of the specific gravity of soils lying below the sole

foundations (in the presence of groundwater, it is determined taking into account the weighing effect of water), kN / m 3;

g 1 II - the same, lying above the sole, kN / m 3;

C n - the calculated value of the specific adhesion of the soil lying directly under the base of the foundation, kPa;

d 1 - the depth of the internal and external foundations from the basement floor m, determined by the formula

, (5.17)

where h S is the thickness of the soil layer above the base of the foundation from the basement side, m,

h cf is the thickness of the basement floor structure, h cf =0.12m;

g cf - the calculated value of the specific gravity of the basement floor structure, kN / m 3,

for concrete g cf \u003d 25 kN / m 3.

The depth to the basement floor is determined by the formula

d b =d-d 1 , (5.18)

d b \u003d 1.52-0.67 \u003d 0.85 m

The calculated value of the specific gravity of soils lying below the base of the foundations is determined by the formula

gII , (5.19)

where γ n is the specific gravity of the soils of the corresponding layers, kN/m 3 ;

h n is the thickness of the corresponding layers, m.

In the presence of groundwater, the calculated value of the specific gravity of soils is determined taking into account the weighing effect of water according to the formula

where γ s is the specific gravity of solid soil particles, kN/m 3 ;

γ w is the specific gravity of water, kN/m3;

γ 1 \u003d 1.83 × 9.8 \u003d 17.93 kN / m 3

γ 2 \u003d 1.9 × 9.8 \u003d 18.62 kN / m 3

γ 3 \u003d 2 × 9.8 \u003d 19.6 kN / m 3

Figure 5.2—Geological section for well No. 2

The calculated value of the specific gravity of soils lying above the base of the foundations is determined by the formula:

Check the value of the average pressure under the base of the foundation according to the formula

, (5.21)

where N f - foundation weight, kN;

N g is the weight of the soil on the edges of the foundation, kN;

b – foundation width, m;

l \u003d 1 m, since all loads are given per linear meter.

Since ∆<10%, следовательно, фундамент запроектирован, верно.

5.2 Calculation of the pile foundation

The design of pile foundations is carried out in accordance with. For a centrally loaded foundation, the calculations are performed in the following order:

a) Determine the length of the pile:

The thickness of the grillage is taken equal to 0.5 m.

To determine the area of ​​the conditional foundation, the weighted average angle of internal friction is determined by the formula:

, (5.28)

where j i is the angle of internal friction of the i-th layer; O

h n - thickness of the n-th layer of soil, m;.

Then find the width of the conditional foundation according to the formula:

b conditional = 2tgah + b 0 , (5.30)

where, h – pile length, m;

b 0 - the distance between the outer faces of the extreme rows of piles, m.

The sand is fine, of medium density with e 0 \u003d 0.66 with n \u003d 1.8 kPa and φ n \u003d 31.6 about;

1.3; Mg=6.18; M c = 8.43.

,

Therefore, the foundation is designed correctly.

Figure 5.6 - Calculation scheme of the pile foundation

5.3 Feasibility comparison of options

For strip and pile foundations, their cost is compared according to aggregated indicators. Cost estimation, comparison of the main types of work in the construction of foundations is carried out for 1 running meter.

The volume of the pit is found by the formula

(5.30)

where, a, b - the width of the pit below and, respectively, on top of the pit, m;

u is the depth of the pit, m;

l is the length of the pit, m;

For shallow foundations, the volume of the pit will be equal to

For a pile foundation it will be equal to:

Comparison of the cost of foundations is given in tabular form (Table 5.1).

Table 5.1 - Feasibility comparison of options

Conclusion: according to a preliminary estimate of the cost of the main types of work in the construction of foundations, out of 2 options, a shallow foundation is more economical and efficient.

6 Calculation of foundations of the accepted type

6.1 Calculation of shallow foundations in section 2 - 2

We determine the main dimensions and calculate the design of the prefabricated strip foundation for the inner wall. The depth of the footing is taken similarly to the depth of the wall in section 1-1 (see Section 5.1). We determine the approximate dimensions of the foundation in terms of the formula (5.15)

According to the table 6.5 and 6.6 we select a slab of the brand FL 14.12, having dimensions L = 1.18 m, b = 1.4 m, h = 0.3 m and wall blocks FBS 12.4.3 and FBS 12.4.6

According to the table 2 adj.2 for dusty sand of medium plasticity with a porosity coefficient e = 0.67 we find φ n = 29.2 0 and C n = 3.6 kPa.

According to the table 5.4, ​​interpolating over φ II, we find the values ​​of the coefficients:

1.08; M g = 5.33; M c = 7.73.

The depth to the basement floor is determined by the formula (5.18):

d b \u003d 1.32-0.47 \u003d 0.85 m

According to the formula (5.16) we determine the calculated value of the resistance R:

Checking the value of the average pressure under the sole of the foundation

Р=156.9 kPa< R=171,67 кПа, приблизительно на 8,9%, значит фундамент запроектирован верно.

Because two-way filtering use the 0-1 case.

1) The total stabilized draft is determined by the formula

, (7.11)

where h e is the thickness of the equivalent layer, m;

m vm – average coefficient of relative soil compressibility, MPa -1 ;

2) determine the thickness of the equivalent layer by the formula

h e = A wm b, (7.12)

where A wm is the coefficient of the equivalent layer, depending on the Poisson's ratio, the shape of the sole, the rigidity of the foundation, taken according to Table. 6.10,

A wm \u003d 2.4 (for silty clay soils);

h e \u003d 2.4 × 2 \u003d 4.8 m

H \u003d 2 h e \u003d 2 × 4.8 \u003d 9.6 m

Figure 7.4

3) determine the average relative compressibility coefficient by the formula:

, (7.13)

where h i is the thickness of the i-th layer of soil, m;

m n i - coefficient of relative compressibility of the i-th layer, MPa -1;

z i is the distance from the middle of the layer of the i-th layer to a depth of 2h e, m.

4) According to the formula (7.11.) we find the draft

5) Determine the consolidation ratio by the formula

where g w is the specific gravity of water, kN/m3;

K ft - average filtration coefficient, determined by the formula

where H is the thickness of the compressible stratum, m;

k f i - filtration coefficient of the i-th layer of soil, cm/year.

6) Calculate the time required to compact the soil to a given degree using the formula

(7.16)

year = 0.23N days = 5.52N hours

Given the values ​​of U according to Table V.4, the values ​​of N for the trapezoidal distribution of sealing pressures are determined by the formula

where I is the value of the interpolation coefficients according to Table V.5.

We summarize the data in table 7.4.

Table 7.4

7.5 Calculation of settlement attenuation with time for section 2-2

The calculation is carried out by the equivalent layer method with a layered soil thickness in the following sequence:

1) determine the thickness of the equivalent layer according to the formula (7.12.)

h e \u003d 2.4 × 1.4 \u003d 3.36 m

H \u003d 2 h e \u003d 2 × 3.36 \u003d 6.72 m

Figure 7.5

2) Determine the average relative compressibility coefficient according to the formula (7.13.)

3) According to the formula (7.11.) we find the draft

4) We find the average filtration coefficient according to the formula (7.15.)

,

5) Determine the consolidation ratio by the formula (7.14.):

6) Calculate the time required to compact the soil to a given degree using the formula (7.16)

year \u003d 0.9N days \u003d 21.6N hours,

Calculation of settlement S t is summarized in Table 7.5.

Table 7.5 - Calculation of settlement attenuation

Conclusion: since the settlements in all sections do not exceed the limit values, the dimensions of the foundations and their laying depth are calculated correctly.

Figure 7.7 – Graph of sediment attenuation over time

8 Foundation design

After the geodetic breakdown of the axes of the building, reinforced concrete slabs for strip foundations are installed. Prefabricated foundations consist of a strip assembled from reinforced concrete slabs (FL 20.12) and a wall assembled from concrete blocks. Foundation reinforced concrete slabs are laid entirely along the length of the wall.

The slabs are reinforced with single meshes or flat reinforcing blocks assembled from two meshes: the upper one, which has the marking index K, and the lower one - C. The working reinforcement is a hot-rolled rod of a periodic profile made of steel of class A-III and a wire of a periodic profile of steel of class VR-1. Distribution fittings - a smooth reinforcing wire from steel of a class B-I.

To ensure the spatial rigidity of the prefabricated foundation, a connection is provided between the longitudinal and transverse walls by tying them to foundation wall blocks or laying meshes of reinforcement with a diameter of 8-10 mm into horizontal seams. The walls are protected from surface and groundwater by installing a blind area and laying horizontal waterproofing at a level not lower than 5 cm from the surface of the blind area and not higher than 30 cm from preparing the basement floor. The outer surface of the basement walls is protected by coating insulation in one or two layers.

Protection of ground premises from ground dampness is limited to the device along the leveled surface of all walls at a height of 15-20 cm from the top of the blind area or sidewalk of a continuous waterproof layer of greasy cement mortar or one or two layers of rolled material on bitumen. This layer is one with the concrete preparation of the floor. In places where the floor is lowered, additional insulation is arranged. To protect basements and underground spaces in wet soils, the coating is done on the wall surface plastered with cement mortar.

The surfaces of the basement walls are protected by a horizontal waterproof layer in the wall, reaching the floor of the underground room or basement. The basement floors are insulated at low water levels by the concrete layer itself.

9. Scheme of work

Figure 9.1- Dimensions of the pit

The dimensions of the bottom of the pit in the plan are determined by the distances between the outer axes of the structure, the distances from these axes to the extreme ledges of the foundations, the dimensions of additional structures arranged near the foundations from the outside, and the minimum width of the gap (allowing to start the underground parts of the structure) between the additional structure and the wall of the pit. The dimensions of the pit on top are made up of the dimensions of the bottom of the pit, the width of the slopes or the wall fastening structures and the gap between the faces of the foundations and slopes. The depth of the pit is determined by the foundation mark.

The working layer of the base is protected from violations by a protective layer of soil, which is removed only before the introduction of the foundation. To drain atmospheric precipitation, the surface of the protective layer is made with a slope towards the walls, and along the perimeter of the pit, drainage grooves are arranged with a slope towards the pits, from which water is pumped out as necessary. The arrangement of grooves and sumps and pumping out of the water are carried out in compliance with the requirements of open dewatering.

Descents are provided for the delivery of materials, parts and transportation of mechanisms to the pit. The stability of the walls of the pit is ensured by various types of fasteners or by giving them appropriate slopes. The fixing method depends on the depth of the excavation, the properties and stratification of soils, the level and flow of groundwater, the conditions for the production of work, the distance to existing buildings.

The erection of foundations and underground elements, as well as backfilling of the sinuses of pits, should be carried out immediately after excavation

Pit pits with natural slopes are arranged in low-moisture stable soils. With a pit depth of up to 5 m, the walls can be made without fastening, but with a slope and steepness of slopes, which are indicated in Table.

The pits are fastened with sheet pile walls. Wooden sheet piling (board and block) is used for fastening shallow pits (3 ... 5 m). Plank pile is used for fastening shallow pits (3 ... 5 m). Plank piles are made from boards up to 8 cm thick, block piles are made from beams from 10 to 24 cm thick. The length of the piles is determined by the depth of their immersion, but, as a rule, does not exceed 8 m.

During operation, it is necessary to protect the pit from filling with precipitation. To do this, it is necessary to plan the surface around the excavation and ensure the flow outside the construction site.

It is necessary to develop the soil of the pit and build the foundation in a short time, without leaving the bottom of the pit open for a long time (the longer the gap between the end of the earthworks and the foundation, the more the foundation soil and the slopes of the pit are destroyed).

After the foundation is erected, the sinuses between the walls of the foundation and the foundation pit are filled with soil, laid in layers with a rammer.

For a given volume of earthworks of the zero cycle, we select a scraper set of earthmoving machines: an E1252 single-bucket excavator (with a bucket capacity of 1.25 m3), several D-498 scrapers (with a bucket capacity of 7 m3), D3-18 bulldozers (based on a T-100 tractor), ZIL-MM3-555 dump trucks.

When developing a pit (see Figure 9.1), soil is developed for a residential building up to the mark by an EO 1621 excavator with a bucket capacity of 0.15 m3. For the removal of soil, a GAZ-93A dump truck is used.

The fertile soil layer at the base of the embankments and in the area occupied by various excavations, before the start of the main excavation work, must be removed in the amount established by the construction organization project and transferred to dumps for subsequent use in reclamation or increasing the fertility of unproductive lands.

It is forbidden to use the fertile soil layer for the installation of lintels, bedding and other permanent and temporary earthworks.

Conclusion

In this project, the most rational foundation was developed for a 4-storey residential building - a shallow strip foundation. The choice of a rational type of foundation was carried out on the basis of a technical and economic comparison of two options for the foundations most commonly used in the construction of foundations: shallow and pile. A comparison of the options was made on the basis of their cost, established according to aggregated indicators for one meter of foundation, the cost was 791.03 rubles for a strip foundation, and 848.46 rubles for a pile foundation.

The strip foundation is installed at the level of 128.6 m, that is, it is located in dusty sand, of medium density with R = 150 kPa.

As a result of calculations, slabs of the brand FL 20.12, FL 14.12 and FL 12.12, and wall blocks FBS 12.4.6 and FBS 12.4.3 were accepted.

For the selected type of foundation in three characteristic sections of buildings, the foundations were calculated according to the limit state of the 2nd group and the obtained values ​​were compared with the limit values ​​\u200b\u200bof 10 cm: for section 1-1, the settlement is 1.61 cm, for section 2-2 - 2.61 cm, for section 3-3 - 2.54 cm.

The foundation was constructed; the scheme of production of works of the zero cycle is calculated, and brief information about the structure of the pit is also given.

List sources used

1. Berlinov, M.V. Examples of calculation of bases and foundations: Proc. for technical schools / M.V. Berlinov, B.A. Yagupov. – M.: Stroyizdat, 1986. – 173p.

2. Veselov, V.A. Design of foundations and foundations: Proc. manual for universities / V.A. Veselov. - M .: Stroyizdat, 1990. - 304 p.

3. GOST 25100-82. Soils. Classification. - M .: Standards, 1982.-9s.

4. Dalmatov, B.I. Soil mechanics, foundations and foundations / B.I. Dalmatov. - L .: Stroyizdat, Leningrad. department, 1988.-415s.

5. Kulikov, O.V. Calculation of the foundations of industrial and civil buildings and structures: Method. instructions for the implementation of the course project / O.V. Kulikov. - Bratsk: BrII, 1988. - 20p.

6. Soil mechanics / B.I. Dalmatov [i dr.]. - M .: Publishing house ASV; St. Petersburg: SPbGA-SU, 2000. - 204p.

7. Soil mechanics, foundations and foundations: Textbook for building. specialist. Universities / S.B. Ukhov [i dr.]. – M.: Vyssh.shk., 2004. – 566p.

8. Foundations, foundations and underground structures (Designer's Handbook) / ed. E.N. Sorochan, Yu.G., Trofimov. – M.: Stroyizdat, 1985. – 480s.

9. Designing the foundations of buildings and underground structures / B.I. Dalmatov [i dr.]. - M .: Publishing house ASV; St. Petersburg: SPbGA-SU, 2006. - 428p.

10. SNiP 2.02.01-83*. Foundations of buildings and structures / Gosstroy of the USSR. - M.: Stroyizdat, 1985. - 40 p.

11. SNiP 2.02.03-85. Pile foundations / Gosstroy of the USSR. - M .: CITP Gosstroy of the USSR, 1986. - 48s.

12. SNiP 2.01.07-85. Loads and impacts / Gosstroy of the USSR. - M .: CITP Gosstroy of the USSR, 1986. - 36 p.

13. SNiP 3.02.01-83. Bases and foundations / Gosstroy of the USSR. - M .: CITP Gosstroy of the USSR, 1983. - 39p.

14. Tsytovich, N.A. Soil mechanics / N.A. Tsytovich. - M.: Vyssh.shk., 1979. - 272p.

The main task of the foundation is to transfer the load from the structure to the soil. Therefore, the collection of loads on the foundation is one of the most important tasks that must be solved even before the construction of the building begins.

What to consider when calculating the load

The correctness of the calculation is one of the key steps in construction that must be solved. When making incorrect calculations, most likely, under the pressure of the loads, the foundation will simply settle and "go underground." When calculating and collecting loads on the foundation, it must be taken into account that there are two categories - temporary and permanent loads.

• The first is, of course, the weight of the building itself. The total weight of the structure is made up of several components. The first component is the total weight of the building's floors for the floor, roof, interfloor, etc. The second component is the weight of all its walls, both load-bearing and internal. The third component is the weight of communications that are laid inside the house (sewerage, heating, plumbing). The fourth and last component is the weight of the finishing elements of the house.
• Also, when collecting loads on the foundation, you need to take into account the weight, which is called the payload of the structure. This paragraph refers to the entire interior arrangement (furniture, appliances, residents, etc.) of the house.
• The third type of loads are temporary, which most often include additional loads that have appeared due to weather conditions. These include a layer of snow, loads in strong winds, etc.

An example of collecting loads on a foundation

In order to accurately calculate all the loads that will fall on the foundation, it is necessary to have an accurate building design plan, as well as to know what materials the building will be built from. In order to more clearly describe the process of collecting loads on the foundation, the option of building a house with a habitable attic, which will be located in the Ural region of the Russian Federation, will be considered.

• One-story house with a habitable attic.
• The size of the house will be 10 by 10 meters.
• The height between the floors (floor and ceiling) will be 2.5 meters.
• for the house will be built from aerated concrete blocks, the thickness of which is 38 cm. Also, from the outside of the building, these blocks will be covered with facing hollow bricks with a thickness of 12 cm.
• Inside the house there will be one load-bearing wall, the width of which will be 38 cm.
• Above the basement of the house there will be an empty floor made of reinforced concrete material. From the same material, the ceiling for the attic will also be equipped.
• The roof will be rafter type, and the roof will be made of corrugated board.

Calculation of loads on the foundation

After the load on the foundation of the house has been collected, you can proceed to the calculation.

• The first thing to calculate is the total area of ​​​​all floors. The size of the house is 10 by 10 meters, which means that the total area will be 100 square meters. m (10*10).
• Next, you can proceed to the calculation of the total area of ​​\u200b\u200bthe walls. This value also includes places for openings for doors and windows. For the first floor, the calculation formula will look like this - 2.5 * 4 * 10 \u003d 100 square meters. m. Since the house is with an inhabited attic, the loads on the foundation were collected taking into account this building. For this floor, the area will be equal to 65 square meters. m. After calculations, both values ​​\u200b\u200bare added up and it turns out that the total area of ​​\u200b\u200bthe walls for the building is 165 square meters. m.
• Next, you need to calculate the total area for the roof of the building. It will be 130 sq. m. - 1.3 * 10 * 10.

After carrying out these calculations, it is necessary to use the table for collecting loads on the foundation, which presents the average values ​​​​for those materials that will be used in the construction of the building.

Strip foundation

Since there are several types of foundation that can be used in the construction of a facility, several options will be considered. The first option is to collect loads on a strip foundation. The list of loads will include the mass of all elements used in the construction of the building.

1. Mass of external and internal walls. The total area is calculated excluding openings for windows and doors.
2. The area for the floors of the floor and the materials from which it will be built.
3. Ceiling and ceiling area.
4. The area of ​​the truss system for the roof and the weight of the materials for the roof.
5. The area of ​​stairs and other internal elements of the house, as well as the weight of the material from which they will be made.
6. It is also necessary to add the weight of materials that are used for fastening during construction, for arranging the plinth, thermal and air insulation, as well as for cladding the internal and / or external walls of the house.

These few points are for any structure that will be erected on a belt-type support.

Calculation methods for strip foundation

There are two ways to calculate the strip foundation. The first method involves the calculation of the bearing capacity of the soil under the base of the foundation, and the second - according to the deformation of the same soil. Since it is recommended to use the first method for calculations, it will be considered. Everyone knows that direct construction begins with the foundation, but the design of this site is carried out last. This is due to the fact that the main purpose of this design is to transfer the load from the house to the soil. And the collection of loads on the foundation can be carried out only after a detailed plan of the future structure is known. The actual calculation of the foundation can be conditionally divided into 3 stages:

• The first step is to determine the load on the foundation.
• The second stage is the choice of characteristics for the tape.
• The third stage is the adjustment of parameters depending on the operating conditions.

Foundation for a column

During the construction of houses, columns can be used as supports. However, it is quite difficult to carry out the calculation for this type of supporting structure. The whole complexity of the calculation lies in the fact that it is rather difficult to carry out the collection of loads on the foundation of the column on your own. To do this, you must have a special construction education and certain skills. In order to resolve the issue of calculating the load on the foundation of the column, it is necessary to have the following data:

• The first parameter to consider is weather conditions. It is necessary to determine the climatic conditions in the region in which the construction is carried out. In addition, an important parameter will be the type and power of winds, as well as the frequency of rainfall and their strength.
• At the second stage, it is necessary to make a geodetic map. It is necessary to take into account the flow of groundwater, their seasonal shift, as well as the type, structure and thickness of underground rocks.
• At the third stage, of course, it is necessary to calculate the load on the columns coming from the building itself, that is, the weight of the future building.
• Based on previously obtained data, it is necessary to choose the right brand of concrete according to its characteristics, strength and composition.

How to calculate the foundation for the column

When calculating the foundation for a column, the calculation of the load per square centimeter of the area of ​​\u200b\u200bthis foundation is implied. In other words, in order to calculate the necessary foundation for a column, you need to know everything about the building, soil and groundwater that flows nearby. It is necessary to collect all this information, systematize it, and based on the results obtained, it will be possible to carry out a complete calculation of the loads on the foundation under the column. In order to have all the necessary information, you need to do the following:

1. It is necessary to have a complete project of the building with all communications that will take place inside the building, and also know what materials will be used to build the building.
2. It is necessary to calculate the total area of ​​​​one support for the building.
3. It is necessary to collect all the parameters of the building and, on their basis, calculate the pressure that the building will exert on the column-type support.

foundation edge

The edge of the foundation is the upper part of the supporting concrete structure, which bears the main pressure from the structure. There is a certain sequence in which it is necessary to collect loads on the edge of the foundation, as well as their further calculation. In order to determine the load on the edge, it is necessary to have a typical floor plan of the building if it is a multi-storey building, or a typical basement plan if the building has only one floor. In addition, it is necessary to have a plan of longitudinal and transverse sections of the building. For example, in order to calculate the load on the edge of the foundation in a ten-story building, you need to know the following:

• Weight, thickness and height of a brick wall.
• The weight of the hollow cores that are used as floors, and multiply this number by the number of floors.
• The weight of the partitions multiplied by the number of floors.
• It is also necessary to add the weight of the roof, the weight of the waterproofing and the vapor barrier.

conclusions

As you can see, in order to calculate the load on a foundation of any type, it is necessary to have all the data about the building, as well as to know many formulas for calculation.

However, at present this task is somewhat simplified by the fact that there are electronic calculators that do all the calculations instead of people. But for their correct and productive work, it is necessary to download into the device all the information about the building, about the material from which it will be built, etc.

The calculation of the load on the foundation is necessary for the correct choice of its geometric dimensions and the area of ​​​​the base of the foundation. Ultimately, the strength and durability of the entire building depends on the correct calculation of the foundation. The calculation comes down to determining the load per square meter of soil and comparing it with the allowable values.

To calculate, you need to know:

• The region in which the building is being built;
• Soil type and groundwater depth;
• The material from which the structural elements of the building will be made;
• The layout of the building, the number of storeys, the type of roof.

Based on the required data, the calculation of the foundation or its final verification is carried out after the design of the structure.

Let's try to calculate the load on the foundation for a one-story house made of solid solid brick masonry, with a wall thickness of 40 cm. The dimensions of the house are 10x8 meters. The ceiling of the basement is reinforced concrete slabs, the ceiling of the 1st floor is wooden on steel beams. The roof is gable, covered with metal tiles, with a slope of 25 degrees. Region - Moscow region, soil type - wet loams with a porosity coefficient of 0.5. The foundation is made of fine-grained concrete, the wall thickness of the foundation for calculation is equal to the thickness of the wall.

Determining the depth of the foundation

The depth of laying depends on the depth of freezing and the type of soil. The table shows the reference values ​​​​of the depth of soil freezing in various regions.

Table 1 - Reference data on the depth of soil freezing

Reference table for determining the depth of the foundation by region

The depth of the foundation in the general case should be greater than the freezing depth, but there are exceptions due to the type of soil, they are indicated in table 2.

Table 2 - Dependence of the depth of the foundation on the type of soil

The depth of the foundation is necessary for the subsequent calculation of the load on the soil and determining its size.

We determine the depth of soil freezing according to table 1. For Moscow, it is 140 cm. According to table 2, we find the type of soil - loam. The laying depth must not be less than the estimated freezing depth. Based on this, the depth of the foundation for the house is selected 1.4 meters.

Roof load calculation

The load of the roof is distributed between those sides of the foundation, on which the truss system rests through the walls. For an ordinary gable roof, these are usually two opposite sides of the foundation, for a four-pitched roof, all four sides. The distributed load of the roof is determined by the area of ​​the projection of the roof, referred to the area of ​​the loaded sides of the foundation, and multiplied by the specific gravity of the material.

Table 3 - Specific gravity of different types of roofing

Reference table - Specific gravity of different types of roofing

1. We determine the area of ​​​​the projection of the roof. The dimensions of the house are 10x8 meters, the projection area of ​​the gable roof is equal to the area of ​​the house: 10 8 = 80 m 2.
2. The length of the foundation is equal to the sum of its two long sides, since the gable roof rests on two long opposite sides. Therefore, the length of the loaded foundation is defined as 10 2 = 20 m.
3. The area of ​​the foundation loaded with a roof with a thickness of 0.4 m: 20 0.4 \u003d 8 m 2.
4. The type of coating is metal tiles, the slope angle is 25 degrees, which means that the calculated load according to table 3 is 30 kg / m 2.
5. The load of the roof on the foundation is 80/8 30 \u003d 300 kg / m 2.

Snow load calculation

The snow load is transferred to the foundation through the roof and walls, so the same sides of the foundation are loaded as in the calculation of the roof. The area of ​​snow cover is calculated equal to the area of ​​the roof. The resulting value is divided by the area of ​​the loaded sides of the foundation and multiplied by the specific snow load determined from the map.

Table - calculation of snow load on the foundation

1. The length of the slope for a roof with a slope of 25 degrees is (8/2) / cos25 ° = 4.4 m.
2. The roof area is equal to the length of the ridge multiplied by the length of the slope (4.4 10) 2 \u003d 88 m 2.
3. The snow load for the Moscow region on the map is 126 kg / m 2. We multiply it by the roof area and divide by the area of ​​the loaded part of the foundation 88 126 / 8 = 1386 kg / m 2.

Floor load calculation

Ceilings, like the roof, usually rest on two opposite sides of the foundation, so the calculation is carried out taking into account the area of ​​\u200b\u200bthese sides. The floor area is equal to the area of ​​the building. To calculate the load of floors, you need to take into account the number of floors and the basement floor, that is, the floor of the first floor.

The area of ​​each overlap is multiplied by the specific gravity of the material from table 4 and divided by the area of ​​the loaded part of the foundation.

Table 4 - Specific gravity of floors

1. The floor area is equal to the area of ​​\u200b\u200bthe house - 80 m 2. The house has two floors: one of reinforced concrete and one of wood on steel beams.
2. We multiply the area of ​​the reinforced concrete floor by the specific gravity from Table 4: 80 500 = 40000 kg.
3. We multiply the area of ​​\u200b\u200bthe wooden floor by the specific gravity from table 4: 80 200 \u003d 16000 kg.
4. We summarize them and find the load on 1 m 2 of the loaded part of the foundation: (40000 + 16000) / 8 = 7000 kg / m 2.

Wall load calculation

The load of the walls is determined as the volume of the walls, multiplied by the specific gravity from table 5, the result is divided by the length of all sides of the foundation, multiplied by its thickness.

Table 5 - Specific gravity of wall materials

Table - Specific weight of walls

1. The wall area is equal to the height of the building multiplied by the perimeter of the house: 3 (10 2 + 8 2) = 108 m 2.
2. The volume of the walls is the area multiplied by the thickness, it is equal to 108 0.4 \u003d 43.2 m 3.
3. We find the weight of the walls by multiplying the volume by the specific gravity of the material from table 5: 43.2 1800 \u003d 77760 kg.
4. The area of ​​​​all sides of the foundation is equal to the perimeter multiplied by the thickness: (10 2 + 8 2) 0.4 \u003d 14.4 m 2.
5. The specific load of the walls on the foundation is 77760/14.4=5400 kg.

Preliminary calculation of the foundation load on the ground

The load of the foundation on the soil is calculated as the product of the volume of the foundation and the specific density of the material from which it is made, divided by 1 m 2 of its base area. The volume can be found as the product of the depth of the foundation and the thickness of the foundation. The thickness of the foundation is taken in the preliminary calculation equal to the thickness of the walls.

Table 6 - Specific density of foundation materials

Table - specific gravity of the soil material

1. The foundation area is 14.4 m 2, the laying depth is 1.4 m. The volume of the foundation is 14.4 1.4 \u003d 20.2 m 3.
2. The mass of the foundation made of fine-grained concrete is: 20.2 1800 = 36360 kg.
3. Ground load: 36360 / 14.4 = 2525 kg / m 2.

Calculation of the total load per 1 m 2 of soil

The results of previous calculations are summarized, and the maximum load on the foundation is calculated, which will be greater for those sides on which the roof rests.

The conditional design soil resistance R 0 is determined according to the tables of SNiP 2.02.01-83 "Foundations of buildings and structures".

1. We sum up the weight of the roof, the snow load, the weight of the floors and walls, as well as the foundation on the ground: 300 + 1386 + 7000 + 5400 + 2525 \u003d 16 611 kg / m 2 \u003d 17 t / m 2.
2. We determine the conditional design soil resistance according to the tables of SNiP 2.02.01-83. For wet loams with a porosity coefficient of 0.5, R 0 is 2.5 kg/cm 2 or 25 t/m 2 .

It can be seen from the calculation that the load on the ground is within the permissible range.

Activities prior to determining the load on the foundation

When building a house, the foundation is first laid, through which the loads from the entire structure are transferred to the ground. The load on the foundation determines the stability, reliability and durability of the entire building, therefore, when starting the foundation, all technological processes must be observed. The correct calculation of the load on the foundation allows you to avoid cracks and destruction and ensure a uniform settlement of buildings.

At the heart of all houses is the foundation. The stability, reliability and durability of the entire building as a whole depends on the quality of its construction.

Before starting the construction of a house, it is impossible to do without geological work at the site of the planned construction, soil research. Important indicators include the indicator of the depth of groundwater and seasonal freezing of the soil. These indicators vary depending on the regions of construction. In the Moscow region, the soil freezes to a depth of 1.6 meters, in the South of Russia it can be less than 1 meter.

Soil indicators important for the strength of the foundation

The foundation is the soil. Rocks are considered the strongest.

For the soil, which is the base for the structure, two indicators are most important - strength and incompressibility.

The most durable are semi-rocky and rocky rocks. Therefore, when constructing the foundation of wooden houses, they do not make pits, but limit themselves to removing the subsiding topsoil.

If the house is laid in the conditions of non-rocky soils when they freeze up to 2 meters or more, the depth of the foundation pit should be calculated in accordance with the estimated depth of soil freezing. An exception is houses that are constantly in use, but at the same time they dig a foundation pit with a depth of at least 0.5 meters under the foundation.

From the standpoint of freezing, it should be taken into account that the soil is non-porous and heaving (dispersed).

Table of soil types and loads on them.

Dispersed soil freezes during the frosty winter months, which causes deformations and changes in the base of the foundation due to loads.

Therefore, the foundation of the pit is made at a level that is lower than the freezing depth.

If the soil is non-porous, deformation does not occur, but nevertheless, settlement can be somewhat reduced with the help of a practically incompressible material (coarse building sand, into which loose rock - gravel interferes) under the base.

Determination of the bearing capacity of the soil

About 15-20% of the cost of building a house is the arrangement of the foundation.

The construction of the foundation of any house is from 15 to 20% of the cost of the total cost of construction. Moreover, the deeper the foundation is laid in the ground, the higher the cost of construction work. This reason quite often causes most developers to raise the sole of the foundation closer to the ground surface. In this case, it is necessary to correctly calculate the bearing capacity of soils. The calculation begins after the collection and analysis of information about the porosity of the soil, which is due to its resistance and degree of moisture.

An important indicator that should be taken into account is seismicity.

Simultaneously arising pressure due to static load and vibration reduces the strength of the soil, causes a pseudo-liquid state. The design resistance of soils in the seismic zone is usually increased by 1.5 times, which entails a corresponding increase in the area of ​​​​the foundation of the building.

Classification of loads on the building foundation

Foundation load classification scheme.

All loads are conditionally divided into permanent and temporary.

• the weight of the entire building, including the mass of the foundation of the house;
• operating loads (weight of people, furniture, equipment).

Temporary loads most often occur seasonally:

• snow load, determined by the angle of the roof slope and the region of construction of the house;
• wind load, depending on the location of the building: forest or open area, urban area.

The calculation of temporary loads is carried out taking into account the construction area.

Calculation of the load taking into account the area of ​​​​the base

Table of resistivity values ​​for various soils.

The main condition is that the load per 1 cm of soil should not be higher than the critical resistance value. The calculation of resistance is carried out depending on the type of soil:

• for gravel or coarse sands, it ranges from 3.5 to 4.5 kg / cm 2;
• for sands of medium size - from 2.5 to 3.5 kg / cm 2;
• hard clay soil - from 3.0 to 6.0 kg/cm 2 ;
• fine wet sands - within 2.0 - 3.0 kg / cm 2;
• plastic clay soil - from 1.0 to 3.0 kg/cm 2 ;
• fine very wet sands - 2.0 -2.5 kg / cm 2;
• pebbles, gravel, crushed stone - from 5.0 to 6.0 kg / cm 2.

The pressure on the soil under the influence of permanent and temporary loads leads to its compression. As a result, the foundation begins to sink, often unevenly, which leads to the appearance of cracks and deformations. Often this is a consequence of the fact that the load pressure of building structures was incorrectly calculated.

Therefore, already at the stage of construction planning and costing, it is necessary to correctly select materials, in particular, take into account the specific gravity of materials that determine the load.

For rubble stone, this indicator is determined in the range from 1600 to 1800 kg / m 3, rubble concrete and brick - in the range of 1800-2200 kg / m 3, reinforced concrete - kg / m 3.

Table of the specific gravity of walls made of different materials.

It is important to take into account the specific gravity of the walls, which is also determined by the materials. For wooden frame-panel walls, the specific gravity is from 30 to 50 kg / m 2, for block-beam, log walls - from 70 to kg / m 2.

When calculating attic floors, it is taken into account that they can exert pressure in the range of 150-200 kg / m 2. Basement ceilings can have different specific gravity, depending on the materials, it ranges from 100 to 300 kg / m 2. For reinforced concrete monolithic floors, the figure is even higher - up to 500 kg / m 2.

Permissible deformations of buildings and their excess

Table of indicators of deformation of buildings and structures.

Permissible deformations of a residential building are taken into account in advance when the total load on the base is calculated.

Precipitation, deformation of foundations is an inevitable phenomenon, the magnitude of which is regulated by the limit values ​​of deformations in Appendix 4 of SNiP 2.02.01-8.

Violation of building codes, along with uneven settlement of the house, leads to a change in the position of the house or deformation of the structure.

Common house deformities include:

Deflection and bulge arising from uneven settlement of the structure. With a deflection, the foundation is considered a dangerous zone, with a bend, the roof of the structure becomes the most dangerous.

The shift may be the result of a significant subsidence of the base on one side. The most dangerous area is the wall located in the middle zone.

Forms of structures deformations.

Roll occurs in buildings of sufficiently high height, which are characterized by a high degree of flexural rigidity. With the growth of the roll, the danger of the destruction of the building increases.

The skew occurs as a result of the uneven settlement of the house, which falls on a certain section of the long side of the building.

Horizontal displacement occurs, as a rule, in the area of ​​\u200b\u200bthe foundation or basement walls with excessive horizontal load.

With the correct calculation of the permissible deformations of the base for civil multi-storey and one-storey buildings, the maximum settlement of the bases should be within 8-12 cm, depending on the frame materials.

Causes and ways to eliminate uneven settlement of foundations

Scheme of the causes of deformation of buildings.

Complex deformation of the structure can occur due to the following reasons:

• violation of heating systems, sewerage, including stormwater, water supply, causing soil to be washed out from under the base of the house;
• heterogeneous base, represented by layers of different thicknesses or densities;
• the presence of underground workings;
• change in the level of industrial or ground groundwater;
• increase in soil porosity due to the movement of its particles when washed away by water flows;
• excessive moisture in any part of the base;
• excessive removal of soil during the construction of the foundation (leveling bedding that replaces it has a lower strength);
• the presence of uneven load on the base;
• construction of separate parts of the building in different periods of time;
• the presence in the base of materials that are susceptible to decay (wood, tree roots);
• soil compaction associated with an increase in weight during the operation of the house;

Uneven precipitation can be prevented by the following changes:

• to give the house, individual parts of the building rise, corresponding to the magnitude of the possible draft;
• increase the flexural rigidity of the short house, as well as reduce the flexural rigidity of the long house;
• build a compensating foundation;
• carry out horizontal reinforcement of all walls;
• when designing a house, it is necessary to provide for the need for uniform transfer of the entire load of the building to the foundation;
• arrange sedimentary and expansion joints;
• carry out preventive maintenance of sewerage systems, water supply systems, as well as those that drain rainwater.

The measures taken will increase the strength and durability of the house.

Technological features of the columnar foundation

The columnar foundation is arranged according to the pile foundation principle. The columnar foundation gives the entire frame stability and increases the ability to withstand destructive influences.

Scheme of the device of the columnar foundation.

The main feature of the design is the installation of a system of pillars at the points of the structure that are most subject to load, for example, in its corners, at the junctions of wall panels. This allows you to stabilize the entire building.

Poles are located in places that require special attention due to possible problems with the stability and safety of the structure. When calculating the foundation of a columnar type, the installation of a grillage is included - a reinforced belt and mounting elements (strapping beams and end beams). Thanks to the grillage, the displacement of the foundation in the horizontal plane is prevented, the pressure is evenly distributed over the system of pillars.

Monolithic reinforced concrete is most often used as a material for a columnar foundation.

An example of calculating loads on a columnar foundation

When determining the bearing capacity of the foundation, the following indicators are taken into account:

• wall perimeter - 12.0x6.0 m;
• two floors;
• walls made of gas blocks with a density of D600 and a thickness of 40 cm;
• floors on the 1st floor on the ground, bulk;
• overlap between floors - reinforced concrete slabs;
• the roof is tiled, pitched (angle 45°), on wooden logs;
• soil - plastic clay;
• section of the upper part of the column - 40x40 centimeters;
• section of the lower part (sole) of the column - 80x80 centimeters;
• the gap between two pillars is 2.0 m.

In order for the foundation structure to resist frost heaving, it is given an outer inclined surface. In the upper part, the foundation will be 0.4 meters wide, and to the sole it will expand to 0.5 meters.

When calculating the total load per 1 m of the length of the base of the foundation, the normative load from snow, ceilings, walls of the house, roofing and building material is taken into account.

If we assume that the total load was 4380 kgf, we should multiply it by 2 meters (the distance between the installed pillars), we get 8760 kgf. To the result obtained, add the mass of one column.

The columnar foundation will have a volume for this design of 0.25 m 3. According to the table, the density for reinforced concrete is determined - 2500 kg / m 3. The weight of the foundation will be equal to 625 kgf (0.25 m 3 x2500 kg / m 3).

The supporting surface of each of the pillars is 80 cm x 80 cm = 6400 cm 2. Knowing the indicator that determines the bearing capacity of the soil (in our case, 1.5 gf / m 2), it is possible to calculate the maximum loads of the entire foundation on the underlying soil: 6400 cm 2 x 1.5 gf / m 2 \u003d 9600 kgf. This figure is more than 9385 kgf - design loads, so the calculated columnar foundation will be a reliable support for the entire structure of the house.

Calculation of the load on the foundation: soil indicators, specific gravity of materials