Conclusion: test 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, sampling depth – 8 m.

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

The plasticity number is determined by formula (2.12):

I=41-23=18 – the soil is classified as clay (I>17) in accordance with Table B.11.

The porosity coefficient is determined using 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”, the method of double interpolation is used to find

Conclusion: the test 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, sampling depth – 12 m.

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

The plasticity number is determined by formula (2.12):

I=20-13=7 – the soil is classified as sandy loam (1I7) in accordance with Table B.11.

The porosity coefficient is determined using formula (2.10):

,

Determine the consistency coefficient using formula (2.13):

S= = 1

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

Determine the design resistance according to Appendix 3 R=300kPa.

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

3 Collection of loads acting on foundations

Loads are collected onto the cargo area, which is installed depending on the static layout of the structure. In this case, a structural scheme with transverse load-bearing walls located with 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 absorbs 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 planning level.

Figure 3.1 - Load area

When calculating temporary loads, we take the load reliability factor to be 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 topography with absolute elevations of 129.40 m, 130.40 m, 130.70 m.

The soil has a consistent bedding. Soils, being in their natural state, can serve as the basis for shallow foundations. For this type of foundation, the base will be layer No. 2 - 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 foundation is made on the basis of a technical and economic comparison of the options most often used in the practice of industrial construction of foundations:

1 shallow laying;

2 pile foundations.

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

5.1 Calculation of shallow foundations on a natural basis

We set the depth of the foundation base, depending on the depth of freezing, 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 melting (freezing) of the soil, found by the formula

, (5.12)

where z 0 is the specific heat of the water–ice phase transformation,

;

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 plasticity (rolling) boundary, fractions of unity.

Temperature at which soil begins to freeze, °C.

T f,m t f,m - accordingly, the average air temperature according to long-term data for the period of negative temperatures, °C and the duration of this period, h;

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

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

The estimated freezing depth is determined by the formula

where k h is a coefficient taking 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 calculated freezing depth, the laying depth is taken for design 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 using the method of successive approximations in the following order:

The area of ​​the foundation base is preliminarily determined using the formula

, (5.15)

R o – design resistance of the soil under the base of the foundation, R 0 = 150 kPa;

h – depth of the base, 1.52 m;

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

g - specific gravity of 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 FBS 12.4.3 brand blocks, having dimensions: 1.18 m, 0.4 m, 0.28 m.

According to Table 2 of Appendix 2 for silty sand of average 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 using the formula

where g c1 and g c2 are operating conditions coefficients adopted according to Table 5.3

g с1 = 1.25 and g с2 = 1.2;

k – 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 the table. 1.3;

k Z – coefficient accepted at b< 10 м равным 1;

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

g II – calculated value of the specific gravity of soils lying below the base

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

g 1 II – the same, lying above the base, kN/m 3;

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

d 1 – depth of laying of 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 on the basement side, m,

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

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

for concrete g cf = 25 kN/m 3.

The depth to the basement floor is determined by the formula

d b =d-d 1 , (5.18)

d b =1.52-0.67=0.85m

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

g II , (5.19)

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

h n – 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 – specific gravity of water, kN/m3;

γ 1 =1.83×9.8=17.93 kN/m 3

γ 2 =1.9×9.8=18.62 kN/m3

γ 3 =2×9.8=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 using the formula

, (5.21)

where N f is the weight of the foundation, kN;

N g - weight of soil on foundation edges, kN;

b – foundation width, m;

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

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

5.2 Calculation of pile foundation

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

a) Determine the length of the pile:

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

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

, (5.28)

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

h n – thickness of the nth layer of soil, m;.

Then find the width of the conditional foundation using the formula:

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

where, h – pile length, m;

b 0 – distance between the outer edges of the outer rows of piles, m.

The sand is fine, of average density with e 0 = 0.66, n = 1.8 kPa and φ n = 31.6 o;

1.3; M g =6.18; M s =8.43.

,

Therefore, the foundation is designed correctly.

Figure 5.6 – Design diagram of a pile foundation

5.3 Technical and economic comparison of options

For strip and pile foundations, their costs are compared based on aggregated indicators. Cost estimates and comparisons of the main types of work when constructing foundations are carried out for 1 linear meter.

The volume of the pit is found using the formula

(5.30)

where a,b is the width of the pit at the bottom and, accordingly, at the top of the pit, m;

u – pit depth, m;

l – pit length, m;

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

For a pile foundation it will be equal to:

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

Table 5.1 - Technical and economic comparison of options

Conclusion: according to a preliminary estimate of the cost of the main types of work when constructing foundations, of the 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 internal wall. The depth of the base is taken similar to the depth of the wall in section 1-1 (see section 5.1). We determine the approximate dimensions of the foundation in plan using formula (5.15)

According to the table 6.5 and 6.6 we select a slab of the FL 14.12 brand, 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 appendix 2 for silty sand of medium plasticity with 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 according to φ II, we find the values ​​of the coefficients:

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

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

d b =1.32-0.47=0.85m

Using formula (5.16), we determine the calculated value of resistance R:

Checking the average pressure value under the base of the foundation

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

Because two-way filtering using case 0-1.

1) The total stabilized settlement 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 using 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 base, and the rigidity of the foundation, taken according to the table. 6.10

A wm =2.4 (for silty-clayey soils);

h e = 2.4 × 2 = 4.8 m

N = 2 h e = 2 ×4.8 = 9.6 m

Figure 7.4

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

, (7.13)

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

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

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

4) Using formula (7.11.) we find the draft

5) Determine the consolidation coefficient using the formula

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

K ft – average filtration coefficient, determined by the formula

where N is the thickness of the compressible thickness, m;

k f i - filtration coefficient of the i-th soil layer, 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

We set 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 interpolation coefficients according to table V.5.

We summarize the data in table 7.4.

Table 7.4

7.5 Calculation of the decay decay over time for section 2-2

The calculation is carried out using the equivalent layer method for layered soils in the following sequence:

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

h e = 2.4×1.4 = 3.36 m

H = 2 h e = 2 × 3.36 = 6.72 m

Figure 7.5

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

3) Using formula (7.11.) we find the draft

4) Find the average filtration coefficient using the formula (7.15.)

,

5) Determine the consolidation coefficient using the formula (7.14.):

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

year =0.9N days =21.6N hours,

The calculation of settlement S t is summarized in Table 7.5.

Table 7.5 - Calculation of settlement attenuation

Conclusion: since precipitation in all sections does not exceed the maximum values, the dimensions of the foundations and their laying depth are calculated correctly.

Figure 7.7– Graph of precipitation decay over time

8 Foundation design

After geodetic alignment of the building axes, 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. Reinforced concrete foundation 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, with the marking index K, and the lower one - C. The working reinforcement is a hot-rolled periodic profile rod made of class A-III steel and a periodic profile wire made of BP-1 class steel. Distribution fittings - smooth reinforcing wire made of class B-I steel.

To ensure the spatial rigidity of the prefabricated foundation, a connection is provided between the longitudinal and transverse walls by tying them with foundation wall blocks or laying meshes of reinforcement with a diameter of 8-10 mm into the horizontal seams. Walls are protected from surface and underground water by constructing blind areas 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 the preparation of the basement floor. The outer surface of the basement walls is protected with coating insulation in one or two layers.

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

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

9. Work flow diagram

Figure 9.1 - Pit dimensions

The dimensions of the bottom of the pit in 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 installed near the foundations from the outside and the minimum width of the gap (allowing the installation of underground parts of the structure) between the additional structure and the wall of the pit. The dimensions of the pit at the top are the sum of the dimensions of the bottom of the pit, the width of the slopes or wall fastening structures and the gap between the edges of the foundations and slopes. The depth of the pit is determined by the foundation level.

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

To deliver materials, parts and transport mechanisms into the pit, descents are provided. The stability of the pit walls is ensured by various types of fastenings or by giving them appropriate slopes. The method of fastening depends on the depth of the pit, the properties and stratification of the soil, the level and flow rate of groundwater, the conditions of work, and the distance to existing buildings.

The construction of foundations and underground elements, as well as backfilling of excavation pits, should be carried out immediately after the development of the soil

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

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

During work, it is necessary to protect the pit from filling with precipitation. To do this, it is necessary to level the surface around the pit and ensure drainage beyond the construction site.

It is necessary to excavate 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 interval between the completion of the excavation work and the installation of the foundation, the more the foundation soil and slopes of the pit are destroyed).

After the foundation is erected, the cavity between the walls of the foundation and the pit is filled with soil, laid in layers with a tamper.

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

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

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 amounts established by the construction organization project and moved to dumps for its subsequent use in reclamation or increasing the fertility of unproductive lands.

It is prohibited to use the fertile soil layer for the construction of lintels, bedding and other permanent and temporary earthen structures

Conclusion

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

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

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

For the selected type of foundation in three characteristic sections of buildings, the foundations were calculated according to the limit state of group 2 and the obtained values ​​were compared with the limit values ​​equal to 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; a diagram of the zero-cycle work is calculated, and brief information about the construction of the pit is given.

List sources used

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

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

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

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

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

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

7. Soil mechanics, foundations and foundations: Textbook for construction. specialist. Universities/S.B. Ukhov [and others]. – M.: Higher school, 2004. – 566 p.

8. Foundations, foundations and underground structures (Designer’s Handbook) / ed. E.N. Sorochana, Yu.G., Trofimova. – M.: Stroyizdat, 1985. – 480 p.

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

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 USSR. – M.: CITP Gosstroy USSR, 1986. – 48 p.

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

13. SNiP 3.02.01-83. Foundations and foundations/Gosstroy USSR. – M.: CITP Gosstroy USSR, 1983. – 39 p.

14. Tsytovich, N.A. Soil mechanics/N.A. Tsytovich. – M.: Higher school, 1979. – 272 p.

We talked about collecting loads for the case when the main load-bearing structures are the walls of the house. Nowadays, it increasingly happens that private residential buildings are built of the frame type: when the load-bearing columns are supported on columnar foundations and take the load from floors, beams, walls, partitions, floors, roofs - in general, everything that is designed in the house. The approach to collecting loads in this case is somewhat different.

Suppose we have a two-story house (the second floor is semi-attic) of a frame type: columnar foundations with foundation beams (under the walls of the 1st floor), monolithic columns, monolithic floors (beamless, only around the perimeter - a strapping beam), longitudinal monolithic beams on the second floor – roof supporting structures; wooden roof, external walls - aerated concrete, partitions - brick.

Let's try to collect loads for calculation:

1) columnar foundation for the central column (axis 2/B);

2) columnar foundation for a corner column (axis 1/B);

3) a columnar foundation for the outermost column (axis 4/G);

4) foundation beam.

Let's choose a design city (for snow load) - let it be Nikolaev.

Attention!The sections of load-bearing elements (thickness of the floor, dimensions of rafters, columns, beams) are taken simply as an example; their dimensions are not confirmed by calculation and may differ significantly from the accepted ones.

1. Load from 1 m 2 of floor above the first floor.

4 . Load from 1 m 2 of brick partition.

5 . Self-weight load of reinforced concrete structures (per 1 linear meter).

Now we need to move on to collecting loads on the foundations. Unlike the load on a strip foundation, which is determined per linear meter, the load on a columnar foundation is collected in kilograms (tons), since it is essentially concentrated and transmitted in the form of force N from the column to the foundation.

How to move from a uniformly distributed load to a concentrated one? You need to multiply it by the area (for a load measured in kg/m2) or by the length (for a load measured in kg/m2). So, the load is transferred to the column located at the intersection of axes “2” and “B” from a rectangle, indicated in pink in the figure above, the dimensions of this rectangle are 2.75x3 m 2. How to determine these sizes? Horizontally, we have two spans between adjacent columns: one is 4.5 m, the second is 1.5 m. From each of these spans, half the load falls on one column, and half on the other. As a result, for our column the load collection length will be equal to:

4.5/2 + 1.5/2 = 2.25 + 0.75 = 3 m.

The load collection length in the perpendicular direction is determined in the same way:

3/2 + 2.5/2 = 1.5 + 1.25 = 2.75 m.

The load collection area for the column along axis 2/B is equal to: 3 * 2.75 = 8.25 m 2.

But for the same column, the area for collecting the load from the roof will be different, because there is no longer a column along axis “3” on the second floor (this can be seen in the section of the house), and the span to the right of the column increases to 4.5 m. This will be taken into account in the tabular calculation.

6. Let’s determine the load on the columnar foundation under the column in the center of the building (along the “2/B” axis).

Explanations:

1. The height of the column is calculated from the top of the foundation to the bottom of the floor plus from the top of the floor to the bottom of the beam under the ridge.

2. When calculating the load from roof structures, you need to pay attention to the area where the load is collected - for inclined elements the area is larger, for horizontal elements it is smaller. In this case, the rafters, metal tiles and sheathing are located obliquely and have a larger area than the horizontal wooden attic beams, insulation and drywall. For the other two columns the situation will be different.

3. The load from the weight of the partition is taken from that part of the partition that rests on the section of the floor from which the load is collected (shaded pink in the figure). Because Table 4 collected loads from 1 sq. meter of partition, then it must be multiplied by the height and length of the partition.

7. Let’s determine the load on the columnar foundation under the column along the outer wall (along the “1/B” axis).

Explanations:

1. The height of the strapping beam is calculated to the bottom of the floor, so as not to count the same concrete twice.

2. Insulation and drywall in this case are located obliquely, so their area is taken accordingly.

3. The height of the partition is not the same due to the sloping roof. We find the average height as the sum of the smallest and largest heights of the partition (in the area from which the load is collected), divided by two.

8. Let’s determine the load on the columnar foundation under the corner column (along the “4/G” axis).

Explanations:

1. The beam under the rafter is located only along the “4” axis, it is not along the “G” axis, so the length of the beam is taken to be 2.15 m. While the strapping beam goes along the perimeter of the building, and its length is found by adding sections of 2.15 m and 1.65 m, minus 0.3 m - the size of the side of the column (so as not to duplicate the same concrete twice).

2. The total height of the wall along the “G” axis is determined based on the following data: 2.8 m – the height of the masonry on the first floor; 1.57 m – the smallest wall height on the second floor in the area from which the load is collected; 2.32 m - the greatest height of the wall on the second floor in the area from which the load is collected.

9. Let’s determine the load per 1 linear meter of the foundation beam from the aerated concrete wall

From weight 1 linear. meter of ground floor wall (wall height 2.8 m)

Explanation:

Because The house is frame, then the load-bearing elements in it are columns that take the load from the roof and ceiling and transfer it to the columnar foundations. Therefore, the walls of the first and second floors serve only as filling and are perceived by the ceiling and foundation beams as a load, while they themselves do not carry anything.

So, the collection of the load on the foundation is completed, but not quite. If the columns are hinged to the foundations, then these (vertical) loads will be sufficient to calculate the foundations. If the connection of the columns with the foundations is rigid, then not only vertical force will be transferred to the foundation from the columns N (kg), but also bending moments in two planes Mx and Mu (kg*m) and transverse forces Qx and Qy (kg). To determine them, you need to count the columns of the first floor and find the moments and shear forces in the lower section. In this example, they will be small, but they will still be there; they cannot be ignored when calculating the foundations.

In continuation of this calculation, read the article “Collecting wind loads in a frame house” in which we will get closer to determining the moments and shear forces for calculating the foundation.

Attention! For the convenience of answering your questions, a new section “FREE CONSULTATION” has been created.

In the comments to this article, please ask questions only about the article.

At the planning stage, an important activity is the collection of loads on the foundation. The reliability and durability of both the foundation and the entire structure depend on the accuracy of the measurements taken. All mathematical calculations are performed in strict accordance with the requirements of governing documents and standards. To successfully implement this event, it would be useful to first study SNiPs and seek advice from specialists.

The need for it and its conditions

Calculation is necessary to identify the created load per 1 sq.m. soil in accordance with acceptable values.

The successful implementation of this activity requires the necessary consideration of the following parameters:

• climate conditions;
• soil type and its characteristics;
• groundwater boundaries;
• design features of the building and the amount of material used;
• layout of the structure and type of roofing system.

Taking into account all the listed characteristics, calculation of the foundation and verification of compliance is carried out after approval of the construction project.

Performing a calculation

To properly collect the load, the weight of each structural element must be calculated and the depth of the supporting structure must be established.

Placement depth

This indicator is based on the depth of soil freezing and its structural analysis. For each region, the value under study is individual and is based on many years of experience of meteorologists.

According to the general principle, the foundation should be, with a reserve, deeper than the soil freezing limits, however, there are some exceptions to any rule. The required indicator will be required later to establish the permissible load and determine the base area.

To increase clarity, an example based on the tape type should be given. We will determine the depth of the foundation for a site located in the city of Smolensk and having a soil type of sandy loam. Using the first table, we find the city we are interested in and compare the indicator.

For the named settlement, it is 120 cm. Using the second table, we set the placement depth for the required type of soil, this figure is equal to at least ¾ of the calculated soil freezing depth, but not less than 0.7 m, thus we get a value of 80 cm, which satisfies all stated conditions.

The presented type of load through the walls of the structure on which the roofing system is placed is evenly distributed between the sides of the base. For a classic roof with two slopes, these are two opposite side walls. In the hipped roof version, the weight is distributed on all four sides.

The required indicator is established by the area of ​​the projection lines of the roof, related to the area of ​​the sides of the base subject to load, and multiplied by the total mass of the building material, which can be calculated according to the attached table.

Example:

1. The area of ​​the projection lines for a building size of 10×10 is 100 sq.m.
2. With a gable roof, the length of the sides of the base is calculated by the number of supporting walls, in our case there are 2 of them, thus we get 10 × 2 = 20 m.
3. The area of ​​the sides of the base subject to load, with a foundation thickness of 0.5 m, is equal to 0.5x20 = 10 sq.m.
4. Roof type – ceramic or cement-sand tiles with a slope of 45º, therefore, the load according to the attached table is 80 kg/sq.m.
5. The total load of the roof on the base is 100/10×80 = 800 kg/sq.m.

Snow load calculation

Snow creates pressure on the foundation through the roof and supporting walls, therefore the calculation of the load created by snow includes the forces of the roof on the foundation. The only thing that needs to be additionally established is the snow pressure area. The required indicator is equal to the area of ​​the equipped roof.

To obtain the final value, the roof area should be divided by the area of ​​the supporting walls of the base and multiplied by the average snow load, according to the table.

Example:

1. The length of the roof slope at 45º is 10/2/0.525 = 9.52 m
2. The roof area is equal to the length of the ridge part multiplied by the length of the slope (9.52x10) x 2 = 190.4 sq.m.
3. The snow load for Smolensk is 126 kg/sq.m. We multiply this value by the area of ​​the roof and divide by the area of ​​the loaded base walls (190.4x126/10 = 2399.04 kg/sq.m.).

Determination of loads generated by floors

The pressure of the floors is carried out in the same way as that of the roof on the supporting walls of the foundation; therefore, the calculation of the load is carried out in direct relation to their area. To determine the load, the first step is to calculate the area of ​​the intermediate elements of all floors, taking into account the floor slab.

The area of ​​one floor is multiplied by the total mass of the material embedded in its base, the value of which can be determined from the table, and the resulting value is divided by the area of ​​the loaded walls of the base.

Example:

The floor area of ​​each floor is equal to the area of ​​the structure - 100 sq.m. The building, for example, has a pair of floors: one is reinforced concrete, the second is wooden on metal (steel) guides.

1. We multiply the area of ​​each of the floors by their specific gravity. We get: 100 x 200 = 20,000 kg and 100 x 500 = 50,000 kg.
2. Let's summarize the presented indicators. calculate the load per square meter: (20000 + 50000) / 10 = 7000 kg/sq.m.

Calculation of loads created by walls

The presented indicator for the tape type is calculated as the product of the total volume of the wall elements and their total weight, which must be divided by the product of the length of the sides of the base and its thickness.

1. The area of ​​each wall is equal to the product of the height of the structure and the perimeter of the house: 3 x (10 x 2 + 10 x 2) = 120 sq.m.
2. We calculate their volume: the product of area and thickness (120 x 0.5 = 60 cubic meters).
3. We determine the total weight by finding the product of the volume and mass of the material indicated in the table: 60 x 1400 = 84,000 kg.
4. We set the area of ​​the supporting sides, which is equal to the product of the perimeter of the base and its thickness: (10 x 2 + 10 x 2) x 0.5 = 20 sq.m.
5. Load created by walls: 84,000/20 = 4,200 kg/sq.m.

Intermediate calculations of the load of the foundation on the ground

The general indicator of the load created by the strip support on the soil is calculated as follows: the volume of the foundation is multiplied by the density of the material embedded in its foundation and divided by the square meter of the base area. The volume should be calculated as the product of the placement depth and the thickness of the support layer.

As a rule, at the stage of preliminary calculations, the last indicator is taken as the thickness of the side walls.

1. Base area – 20 sq.m., placement depth – 80 cm, base volume 20 x 0.8 = 16 cubic meters.
2. The weight of the base, made of reinforced concrete, is: 16 x 2500 = 40,000 kg.
3. Total load on the ground: 40,000/20 = 2,000 kg/sq.m.

Determination of specific load per 1 sq.m. soil

Finally, we find the sum of all completed results, not forgetting to calculate the permissible load on the foundation. At the same time, it is worth considering that the pressure created by the walls with a roofing system on the support will be higher than their adjacent counterparts.

Watch the video on how to carry out a full calculation of the pressure on the foundation of a house.

The fixed indicator of soil resistance is calculated using the tables specified in SNiP 2.02.01-83 and describing the rules for the manufacture of foundations of buildings and structures.

1. We find the sum of the masses created by all elements of the structure, including the base: 800 + 2399.04 + 7,000 + 4,200 + 2,000 = 16,399.04 = 16.5 t/sq.m.
2. We determine the soil resistance index; for sandy loam with a porosity coefficient of 0.7 it is 17.5 t/sq.m.

From the calculations obtained, we can conclude that the pressure created by the building chosen for the example is located within the permissible limit.

Conclusion

As you can see from the example, performing load calculations is not such a difficult undertaking. To successfully implement it, it is necessary to strictly follow the requirements of regulatory documents and adhere to a certain number of rules.

The main task of the foundation is to transfer the load from the structure to the soil. Therefore, collecting loads on the foundation is one of the most important tasks that must be solved before the construction of a 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 resolved. If incorrect calculations are made, most likely, under the pressure of loads, the foundation will simply settle and “go underground.” When calculating and collecting loads on the foundation, you need to take 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 consists 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 (sewage, heating, water supply). The fourth and final 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 internal structure (furniture, appliances, residents, etc.) of the house.
• The third type of load is temporary, which most often includes additional loads resulting from weather conditions. These include a layer of snow, loads in strong winds, etc.

Example of collecting foundation loads

In order to accurately calculate all the loads that will fall on the foundation, it is necessary to have an accurate building design plan, and also to know what materials the building will be constructed from. In order to more clearly describe the process of collecting loads on the foundation, the option of building a house with an inhabited 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 ceilings (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, on the outside of the building, these blocks will be covered with facing hollow bricks 12 cm thick.
• There will be one load-bearing wall inside the house, the width of which will be 38 cm.
• Above the base of the house there will be an empty floor made of reinforced concrete material. The ceiling for the attic will also be constructed from the same material.
• The roof will be of rafter type, and the roof will be made of corrugated sheets.

Calculation of foundation loads

After the loads on the foundation of the house have been collected, you can begin the calculation.

• The first thing that needs to be calculated is the total area of ​​all floors. The size of the house is 10 by 10 meters, which means the total area will be 100 square meters. m (10*10).
• Next, you can begin to calculate the total area of ​​the walls. This value also includes space for openings for doors and windows. For the first floor, the calculation formula will look like this - 2.5*4*10=100 sq. m. Since the house has an inhabited attic, the loads on the foundation were collected taking into account this building. For this floor the area will be 65 square meters. m. After calculations, both values ​​are added up and it turns out that the total area of ​​the walls for the building is 165 sq. m. 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 foundation loads, which presents 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 when constructing 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. The mass of external and internal walls. The total area is calculated without taking into account openings for windows and doors.
2. The area for floor coverings and the materials from which it will be constructed.
3. The area of ​​the ceiling and ceiling.
4. The area of ​​the roof rafter system and the weight of roofing materials.
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 base, 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 strip-type support.

Calculation methods for strip foundations

There are two ways to calculate a strip foundation. The first method involves calculation based on the bearing capacity of the soil under the base of the foundation, and the second - based on the deformation of the same soil. Since it is recommended to use the first method for calculations, it will be considered. Everyone knows that the actual construction begins with the foundation, but the design of this section is carried out last. This is because the main purpose of this structure 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 the detailed plan of the future structure is known. The calculation of the foundation itself can be roughly divided into 3 stages:

• The first stage is determining the load on the foundation.
• The second stage is the selection of characteristics for the tape.
• The third stage is the adjustment of parameters depending on operating conditions.

Column foundation

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

• The first parameter that needs to be taken into account concerns weather conditions. It is necessary to determine the climatic conditions in the region in which construction is taking place. In addition, an important parameter will be the type and power of winds, as well as the frequency of rain 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, its seasonal movement, as well as the type, structure and thickness of underground rocks.
• At the third stage, of course, you need 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 select the correct brand of concrete according to its characteristics, strength and composition.

How to calculate the foundation for a column

When calculating a foundation for a column, it means calculating the load per square centimeter of the area of ​​this foundation. In other words, in order to calculate the required foundation for a column, you need to know everything about the building, the soil and the 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 full 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 design of the building with all communications that will take place inside the building, and also to know what materials will be used for the construction of 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, based on them, calculate the pressure that the building will exert on the column-type support.

Foundation cut

The foundation edge is the upper part of the load-bearing 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-story 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 the brick wall.
• The weight of hollow cores that are used as floors, and also multiply this amount 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 vapor barrier.

conclusions

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

However, nowadays this task is somewhat simplified by the fact that there are electronic calculators that perform all the calculations instead of people. But for their correct and productive operation, it is necessary to load into the device all the information about the building, the material from which it will be constructed, etc.

Before starting the calculation of any structure, we must collect all the loads on this structure. Let's find out what the loads are for calculating civil buildings:
1.) Permanent(the structure’s own weight and the weight of the overlying structures that rest on this one);
2.) Temporary;
- short-term(snow loads, wind loads, ice loads, weight of people);
- long-term(weight of temporary partitions, weight of the water layer);
3.) Special(seismic impacts, explosive impacts, impacts due to base deformation).
Now let's look at a couple of examples. For example, you have a 2-story frame-type cafe (reinforced concrete columns) in the city of Minsk and you need to find out what load is on the column. First, we must decide what loads will act on our column ( picture 1). In this case, it will be the self-weight of the column, the self-weight of the floor/covering, the snow load on the covering, the useful load on the 2nd floor and the useful load on the 1st floor. Next we must find the area on which the loads act (load area, figure 2).

Figure 1 – Diagram of application of loads on the column

Figure 2 – Load area per column

Standard value of snow load in Minsk – 1.2 kPa. We multiply the cargo area by our standard value and by the load reliability factor and get - 6 m * 4 m * 1.2 kPa * 1.4 = 43.2 kN. Those. The snow alone puts 4.32 tons of pressure on our column!
Standard payload value in dining halls (cafes) – 3 kPa. Just like with the snow load, we must multiply the cargo area by the value of the standard load, by the load safety factor and by two (because there are 2 floors). We get - 6 m * 4 m * 3 kPa * 1.2 * 2 floors = 172.8 kN.
The standard value of the self-weight of the floor will depend on the composition of the floor. Let the composition of the 1st floor floor, the 2nd floor floor and the roofing be the same and the standard load value is equal to 2.5 kPa. We also multiply it by the cargo area, by the load safety factor and by three floors. We have – 2.5 kPa*6 m*4 m*1.2*3 = 216 kN.
All that remains is the load from the column's own weight. Our column has a cross section of 300x300 mm and a height of 7.2 m. With a reinforced concrete density of 2500 kg/m3, the mass of the column will be equal to - 0.3 m*0.3 m* 7.2 m* 2500 kg/m3= 1620 kg. Then the calculated weight of the column will be equal to - 1620 kg * 9.81 * 1.2 = 19070 N = 19.07 kN.
If we sum up all the loads, we get the maximum possible load at the bottom of the column:

43.2 kN + 172.8 kN + 216 kN + 19.07 kN = 451.07 kN.

In the same way, for example, a crossbar is calculated. The loading area on the crossbar is shown in Figure 3.

Figure 3 - Load area on the crossbar