Calculation of heat loss of the floor on the ground in gv. Floors technical requirements and rules for design, installation, acceptance, operation and repair in development How to calculate the bearing capacity of floors on the ground

Despite the fact that heat losses through the floor of most one-story industrial, administrative and residential buildings rarely exceed 15% of the total heat loss, and sometimes do not even reach 5% with an increase in the number of storeys, the importance of correctly solving the problem ...

The definition of heat loss from the air of the first floor or basement to the ground does not lose its relevance.

This article discusses two options for solving the problem posed in the title. Conclusions are at the end of the article.

Considering heat losses, one should always distinguish between the concepts of "building" and "room".

When performing the calculation for the entire building, the goal is to find the power of the source and the entire heat supply system.

When calculating the heat losses of each individual room of the building, the problem of determining the power and number of thermal devices (batteries, convectors, etc.) required for installation in each specific room in order to maintain a given indoor air temperature is solved.

The air in the building is heated by receiving thermal energy from the Sun, external sources of heat supply through the heating system and from various internal sources - from people, animals, office equipment, household appliances, lighting lamps, hot water supply systems.

The air inside the premises cools due to the loss of thermal energy through the enclosing structures of the building, which are characterized by thermal resistances measured in m 2 ° C / W:

R = Σ (δ i /λ i )

δ i- the thickness of the material layer of the building envelope in meters;

λ i- coefficient of thermal conductivity of the material in W / (m ° C).

The ceiling (ceiling) of the upper floor, external walls, windows, doors, gates and the floor of the lower floor (possibly the basement) protect the house from the external environment.

The external environment is the outside air and soil.

Calculation of heat loss by the building is carried out at the estimated outdoor temperature for the coldest five-day period of the year in the area where the object is built (or will be built)!

But, of course, no one forbids you to make a calculation for any other time of the year.

Calculation inexcelheat loss through the floor and walls adjacent to the ground according to the generally accepted zonal method by V.D. Machinsky.

The temperature of the soil under the building depends primarily on the thermal conductivity and heat capacity of the soil itself and on the ambient air temperature in the area during the year. Since the temperature of the outside air varies significantly in different climatic zones, the soil also has different temperatures in different periods of the year at different depths in different areas.

To simplify the solution of the complex problem of determining heat loss through the floor and walls of the basement into the ground, for more than 80 years, the method of dividing the area of ​​enclosing structures into 4 zones has been successfully used.

Each of the four zones has its own fixed heat transfer resistance in m 2 °C / W:

R 1 \u003d 2.1 R 2 \u003d 4.3 R 3 \u003d 8.6 R 4 \u003d 14.2

Zone 1 is a strip on the floor (in the absence of soil penetration under the building) 2 meters wide, measured from the inner surface of the outer walls along the entire perimeter or (in the case of a subfloor or basement) a strip of the same width, measured down the inner surfaces of the outer walls from soil edges.

Zones 2 and 3 are also 2 meters wide and are located behind zone 1 closer to the center of the building.

Zone 4 occupies the entire remaining central area.

In the picture below, zone 1 is located entirely on the basement walls, zone 2 is partially on the walls and partially on the floor, zones 3 and 4 are completely on the basement floor.

If the building is narrow, then zones 4 and 3 (and sometimes 2) may simply not be.

Area gender zone 1 in the corners is counted twice in the calculation!

If the entire zone 1 is located on vertical walls, then the area is considered in fact without any additions.

If part of zone 1 is on the walls and part is on the floor, then only the corner parts of the floor are counted twice.

If the entire zone 1 is located on the floor, then the calculated area should be increased by 2 × 2x4 = 16 m 2 when calculating (for a rectangular house in plan, i.e. with four corners).

If there is no deepening of the structure into the ground, then this means that H =0.

Below is a screenshot of the Excel calculation program for heat loss through the floor and recessed walls. for rectangular buildings.

Zone areas F 1 , F 2 , F 3 , F 4 calculated according to the rules of ordinary geometry. The task is cumbersome and often requires sketching. The program greatly facilitates the solution of this problem.

The total heat loss to the surrounding soil is determined by the formula in kW:

Q Σ =((F 1 + F1y )/ R 1 + F 2 / R 2 + F 3 / R 3 + F 4 / R 4 )*(t vr -t nr)/1000

The user only needs to fill in the first 5 lines in the Excel table with values ​​​​and read the result below.

To determine heat losses to the ground premises zone areas will have to be calculated manually. and then substitute in the above formula.

The following screenshot shows, as an example, the calculation in Excel of heat loss through the floor and recessed walls. for the lower right (according to the figure) basement room.

The sum of heat losses to the ground by each room is equal to the total heat losses to the ground of the entire building!

The figure below shows simplified diagrams of typical floor and wall structures.

The floor and walls are considered non-insulated if the coefficients of thermal conductivity of materials ( λ i), of which they are composed, is more than 1.2 W / (m ° C).

If the floor and / or walls are insulated, that is, they contain layers with λ <1,2 W / (m ° C), then the resistance is calculated for each zone separately according to the formula:

Rinsulationi = Rnon-insulatedi + Σ (δ j /λ j )

Here δ j- the thickness of the insulation layer in meters.

For floors on logs, heat transfer resistance is also calculated for each zone, but using a different formula:

Ron the logsi =1,18*(Rnon-insulatedi + Σ (δ j /λ j ) )

Calculation of heat losses inMS excelthrough the floor and walls adjacent to the ground according to the method of Professor A.G. Sotnikov.

A very interesting technique for buildings buried in the ground is described in the article “Thermophysical calculation of heat losses in the underground part of buildings”. The article was published in 2010 in №8 of the ABOK magazine under the heading "Discussion Club".

Those who want to understand the meaning of what is written below should first study the above.

A.G. Sotnikov, relying mainly on the findings and experience of other predecessor scientists, is one of the few who, for almost 100 years, has tried to move the topic that worries many heat engineers. I am very impressed with his approach from the point of view of fundamental heat engineering. But the difficulty of correctly assessing the temperature of the soil and its thermal conductivity in the absence of appropriate survey work somewhat shifts the methodology of A.G. Sotnikov into a theoretical plane, moving away from practical calculations. Although at the same time, continuing to rely on the zonal method of V.D. Machinsky, everyone just blindly believes the results and, understanding the general physical meaning of their occurrence, cannot definitely be sure of the obtained numerical values.

What is the meaning of the methodology of Professor A.G. Sotnikov? He proposes to assume that all heat losses through the floor of a buried building “go” into the depths of the planet, and all heat losses through walls in contact with the ground are eventually transferred to the surface and “dissolve” in the ambient air.

This seems to be partly true (without mathematical justification) in the presence of sufficient deepening of the floor of the lower floor, but with a deepening of less than 1.5 ... 2.0 meters, doubts arise about the correctness of the postulates ...

Despite all the criticisms made in the previous paragraphs, it is the development of the algorithm of Professor A.G. Sotnikova seems to be very promising.

Let's calculate in Excel the heat loss through the floor and walls into the ground for the same building as in the previous example.

We write down the dimensions of the basement of the building and the estimated air temperatures in the block of initial data.

Next, you need to fill in the characteristics of the soil. As an example, let's take sandy soil and enter its thermal conductivity coefficient and temperature at a depth of 2.5 meters in January into the initial data. The temperature and thermal conductivity of the soil for your area can be found on the Internet.

The walls and floor will be made of reinforced concrete ( λ=1.7 W/(m °C)) 300mm thick ( δ =0,3 m) with thermal resistance R = δ / λ=0.176 m 2 ° C / W.

And, finally, we add to the initial data the values ​​of the heat transfer coefficients on the inner surfaces of the floor and walls and on the outer surface of the soil in contact with the outside air.

The program performs the calculation in Excel using the formulas below.

Floor area:

F pl \u003dB*A

Wall area:

F st \u003d 2 *h *(B + A )

Conditional thickness of the soil layer behind the walls:

δ conv. = f(h / H )

Thermal resistance of the soil under the floor:

R 17 =(1/(4*λ gr )*(π / Fpl ) 0,5

Heat loss through the floor:

Qpl = Fpl *(tin — tgr )/(R 17 + Rpl +1/α in )

Thermal resistance of the soil behind the walls:

R 27 = δ conv. /λ gr

Heat loss through walls:

Qst = Fst *(tin — tn )/(1/α n +R 27 + Rst +1/α in )

General heat loss to the ground:

Q Σ = Qpl + Qst

Remarks and conclusions.

The heat loss of the building through the floor and walls into the ground, obtained by two different methods, differ significantly. According to the algorithm of A.G. Sotnikov value Q Σ =16,146 kW, which is almost 5 times more than the value according to the generally accepted "zonal" algorithm - Q Σ =3,353 kW!

The fact is that the reduced thermal resistance of the soil between the buried walls and the outside air R 27 =0,122 m 2 °C / W is clearly small and hardly true. And this means that the conditional thickness of the soil δ conv. not defined correctly!

In addition, the “bare” reinforced concrete of the walls, which I chose in the example, is also a completely unrealistic option for our time.

An attentive reader of the article by A.G. Sotnikova will find a number of errors, rather than those of the author, but those that arose when typing. Then in formula (3) a factor 2 appears in λ , then disappears later. In the example, when calculating R 17 no division sign after unit. In the same example, when calculating heat loss through the walls of the underground part of the building, for some reason the area is divided by 2 in the formula, but then it is not divided when recording the values ​​... What kind of non-insulated walls and floor are these in the example with Rst = Rpl =2 m 2 ° C / W? In this case, their thickness must be at least 2.4 m! And if the walls and floor are insulated, then, it seems, it is incorrect to compare these heat losses with the calculation option for zones for an uninsulated floor.

R 27 = δ conv. /(2*λ gr)=K(cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

As for the question, regarding the presence of a factor of 2 in λ gr has already been said above.

I divided the complete elliptic integrals by each other. As a result, it turned out that the graph in the article shows a function for λ gr =1:

δ conv. = (½) *TO(cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

But mathematically it should be:

δ conv. = 2 *TO(cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

or, if the factor is 2 λ gr not needed:

δ conv. = 1 *TO(cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

This means that the schedule for determining δ conv. gives erroneous underestimated values ​​by 2 or 4 times ...

It turns out that until everyone has nothing else to do, how to continue to either “count”, or “determine” heat losses through the floor and walls into the ground by zones? No other worthy method has been invented in 80 years. Or invented, but not finalized?!

I invite blog readers to test both calculation options in real projects and present the results in the comments for comparison and analysis.

Everything that is said in the last part of this article is solely the opinion of the author and does not claim to be the ultimate truth. I would be glad to hear the opinion of experts on this topic in the comments. I would like to understand to the end with the algorithm of A.G. Sotnikov, because it really has a more rigorous thermophysical justification than the generally accepted method.

ask respecting the work of the author to download a file with calculation programs after subscribing to article announcements!

P.S. (02/25/2016)

Almost a year after writing the article, we managed to deal with the questions raised a little higher.

Firstly, the program for calculating heat losses in Excel according to the method of A.G. Sotnikova thinks everything is correct - exactly according to the formulas of A.I. Pehovich!

Secondly, the formula (3) from the article by A.G. Sotnikova should not look like this:

R 27 = δ conv. /(2*λ gr)=K(cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

In the article by A.G. Sotnikova is not a correct entry! But then the graph is built, and the example is calculated according to the correct formulas!!!

So it should be according to A.I. Pekhovich (p. 110, additional task to item 27):

R 27 = δ conv. /λ gr\u003d 1 / (2 * λ gr ) * K (cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

δ conv. =R27 *λ gr =(½)*K(cos((h / H )*(π/2)))/К(sin((h / H )*(π/2)))

The essence of thermal calculations of premises, to some extent located in the ground, is to determine the influence of atmospheric "cold" on their thermal regime, or rather, to what extent a certain soil isolates a given room from atmospheric temperature effects. Because Since the thermal insulation properties of the soil depend on too many factors, the so-called 4-zone technique was adopted. It is based on the simple assumption that the thicker the soil layer, the higher its thermal insulation properties (the more the influence of the atmosphere is reduced). The shortest distance (vertically or horizontally) to the atmosphere is divided into 4 zones, 3 of which have a width (if it is a floor on the ground) or a depth (if it is a wall on the ground) of 2 meters, and the fourth has these characteristics equal to infinity. Each of the 4 zones is assigned its own permanent heat-insulating properties according to the principle - the farther the zone (the larger its serial number), the less the influence of the atmosphere. Omitting the formalized approach, we can make a simple conclusion that the farther a certain point in the room is from the atmosphere (by a factor of 2 m), the more favorable conditions (from the point of view of the influence of the atmosphere) it will be.

Thus, the countdown of conditional zones starts along the wall from the ground level, provided that there are walls along the ground. If there are no ground walls, then the first zone will be the floor strip closest to the outer wall. Next, zones 2 and 3 are numbered, each 2 meters wide. The remaining zone is zone 4.

It is important to consider that the zone can start on the wall and end on the floor. In this case, you should be especially careful when making calculations.

If the floor is not insulated, then the values ​​of heat transfer resistance of the uninsulated floor by zones are equal to:

zone 1 - R n.p. \u003d 2.1 sq.m * C / W

zone 2 - R n.p. \u003d 4.3 sq.m * C / W

zone 3 - R n.p. \u003d 8.6 sq.m * C / W

zone 4 - R n.p. \u003d 14.2 sq. m * C / W

To calculate the heat transfer resistance for insulated floors, you can use the following formula:

- resistance to heat transfer of each zone of an uninsulated floor, sq.m * C / W;

— insulation thickness, m;

- coefficient of thermal conductivity of the insulation, W / (m * C);

Examples of calculating the strength of a floor with a concrete underlayment

Example 1

It is required to determine the thickness of the concrete underlay in the passage of the warehouse. Floor covering, concrete, thickness h 1 = 2.5 cm. Floor load - from MAZ-205 cars; base soil - loam. Ground water is absent.

For the MAZ-205 vehicle, which has two axles with a wheel load of 42 kN, the calculated wheel load according to the formula ( 6 ):

R p \u003d 1.2 42 \u003d 50.4 kN

The wheel track area of ​​the MAZ-205 is 700 cm 2

According to the formula ( 5 ) we calculate:

r = D/2 = 30/2 = 15 cm

According to the formula ( 3 ) r p \u003d 15 + 2.5 \u003d 17.5 cm

2. For loamy soil of the base in the absence of groundwater according to Table. 2.2

TO 0 \u003d 65 N / cm 3:

For the underlying layer, we will take concrete in terms of compressive strength B22.5. Then, in the passage area in the warehouse, where stationary technological equipment is not installed on the floors (according to clause 2.2 group I), when loaded from trackless vehicles according to table. 2.1 Rδt = 1.25 MPa, E b = 28500 MPa.

3. σ R. The load from the car, according to par. 2.4 , is a load of a simple type and is transmitted along a round-shaped trace. Therefore, the calculated bending moment is determined by the formula ( 11 ). According to par. 2.13 let's ask approximately h\u003d 10 cm. Then according to p. 2.10 accept l= 44.2 cm. For ρ = r R / l\u003d 17.5 / 44.2 \u003d 0.395 according to the table. 2.6 find K 3 = 103.12. According to the formula ( 11 ): M p = TO 3 · R p \u003d 103.12 50.4 \u003d 5197 N cm / cm. According to the formula ( 7 ) calculate the stresses in the plate:

Tension in slab thickness h= 10 cm exceeds design resistance Rδt = 1.25 MPa. In accordance with par. 2.13 we repeat the calculation, setting a large value h= 12 cm, then l= 50.7 cm; p = r R / l = 17,5/50,7 = 0,345; TO 3 = 105,2; M R= 105.2 50.4 = 5302 N cm / cm

Received σ R= 1.29 MPa differs from the design resistance Rδt = 1.25 MPa (see tab. 2.1 ) by less than 5%, therefore, we accept the underlying layer of concrete in terms of compressive strength class B22.5 with a thickness of 12 cm.

Example 2

For mechanical workshops, it is required to determine the thickness of the concrete sub-base used as a floor without covering ( h 1 = 0 cm). Floor load - from the machine weighing P p= 180 kN, standing directly on the underlying layer, is evenly distributed along the track in the form of a rectangle measuring 220 × 120 cm. There are no special requirements for the deformation of the base. The base soil is fine sand, located in the zone of capillary rise of groundwater.

1. Let's determine the design parameters.

Estimated track length according to par. 2.5 and according to the formula ( 1 ) a p \u003d a \u003d 220 cm. Estimated track width according to the formula ( 2 ) b p = b = 120 cm. For the base soil of fine sand located in the zone of capillary rise of groundwater, according to Table. 2.2 K 0 \u003d 45 N / cm 3. For the underlying layer, we will take concrete in terms of compressive strength class B22.5. Then in mechanical workshops, where stationary technological equipment is installed on the floors without special requirements for the deformation of the base (according to paragraph 1 of Art. 2.2 group II), with a fixed load according to table. 2.1 Rδt = 1.5 MPa, E b = 28500 MPa.

2. Determine the tensile stress in the concrete of the slab during bending σ R. The load is transferred along a rectangular track and, according to par. 2.5 , is a load of simple form.

Therefore, the calculated bending moment is determined by the formula ( 9 ). According to par. 2.13 let's ask approximately h\u003d 10 cm. Then according to p. 2.10 accept l= 48.5 cm.

Taking into account α = a p / l= 220/48.5 = 4.53 and β = b p / l\u003d 120 / 48.5 \u003d 2.47 according to the table. 2.4 find TO 1 = 20,92.

According to the formula ( 9 ): M p = TO one · R p \u003d 20.92 5180 \u003d 3765.6 N cm / cm.

According to the formula ( 7 ) calculate the stress in the plate:

Tension in slab thickness h= 10 cm significantly smaller Rδt = 1.5 MPa. In accordance with par. 2.13 Let's recalculate and keep h\u003d 10 cm, we find a lower brand of concrete of the underlying layer slab, at which σ R » Rδt. Let's take concrete of class B15 for compressive strength, for which Rδt = 1.2 MPa, E b = 23000 MPa.

Then l= 46.2 cm; α = a p / l= 220/46.2 = 4.76 and β = b p / l= 120/46.2 = 2.60; according to the table 2.4 TO 1 = 18,63;. M R\u003d 18.63 180 \u003d 3353.4 N cm / cm.

The resulting tensile stress in a slab of concrete of compressive strength class B15 is less Rδt = 1.2 MPa. Let's take the underlying layer of concrete of the compressive strength class B15 with a thickness h= 10 cm.

Example 3

It is required to determine the thickness of the concrete underlayment of the floor in the machine-building shop under loads from automated line machines and ZIL-164 vehicles. The layout of the loads is shown in fig. 1 in", 1 in"", 1 at """. The center of the vehicle wheel track is 50 cm from the edge of the machine track. Weight of the machine in working condition R R= 150 kN is distributed evenly over the area of ​​a rectangular track 260 cm long and 140 cm wide.

The floor covering is the hardened surface of the underlying layer. The base soil is sandy loam. The base is located in the zone of capillary rise of groundwater

Let's define the calculated parameters.

For the ZIL-164 car, which has two axles with a wheel load of 30.8 kN, the calculated wheel load according to the formula ( 6 ):

R R= 1.2 30.8 = 36.96 kN

The wheel track area of ​​the ZIL-164 is 720 cm 2

According to par. 2.5

r R = r = D/2 = 30/2 = 15 cm

For sandy loam soil of the base, located in the zone of capillary rise of groundwater, according to Table. 2.2 TO 0 \u003d 30 N / cm 3. For the underlying layer, we will take concrete of the compressive strength class B22.5. Then for the machine-building shop, where an automated line is installed on the floors (according to paragraph 2.2 group IV), with the simultaneous action of fixed and dynamic loads according to Table. 2.1 Rδt = 0.675 MPa, E b= 28500 MPa.

Let's ask approximately h\u003d 10 cm, then according to p. 2.10 accept l= 53.6 cm. In this case, the distance from the center of gravity of the car wheel track to the edge of the machine track is 50 cm l = 321.6 cm, i.e. according to par. 2.4 The loads acting on the floor are complex loads.

In accordance with par. 2.17 set the position of the calculation centers in the centers of gravity of the trace of the machine (O 1) and the wheel of the car (O 2). From the load layout (Fig. 1 c") it follows that for the calculation center O 1 it is not clear which direction of the OS axis should be set. Therefore, we define the bending moment as with the direction of the OS axis parallel to the long side of the machine trace (Fig. 1 c"), and perpendicular to this side (Fig. 1 in""). For the calculation center O 2, we will take the direction of the OS through the centers of gravity of the traces of the machine and the wheel of the car (Fig. 1 in""").

Calculation 1 Determine the tensile stress in the concrete of the slab during bending σ R for the calculation center O 1 when the OS is directed parallel to the long side of the machine track (Fig. 1 c"). In this case, the load from the machine with a rectangular track refers to a load of a simple type. For the machine track according to p. 2.5 without floor covering h 1 \u003d 0 cm) a p \u003d a \u003d 260 cm; b p \u003d b \u003d 140 cm.

Taking into account the values ​​α = a р / l= 260/53.6 = 4.85 and β = b p / l\u003d 140 / 53.6 \u003d 2.61 according to the table. 2.4 find K 1 = 18,37.

For machine R 0 = R R= 150 kN in accordance with p. 2.14 determined by the formula ( 9 ):

M p = TO one · R p \u003d 18.37 150 \u003d 27555.5 N cm / cm.

Coordinates of the center of gravity of the car wheel track: x i= 120 cm and y i= 0 cm.

Taking into account the ratios x i /l= 120/53.6 = 2.24 and y i /l\u003d 0 / 53.6 \u003d 0 according to the table. 2.7 find TO 4 = -20,51.

Bending moment in calculation center O 1 from the car wheel according to the formula ( 14 ):

M i\u003d -20.51 36.96 \u003d -758.05 N cm / cm.

13 ):

M p I = M 0 + Σ M i= 2755.5 - 758.05 = 1997.45 N cm/cm

7 ):

Calculation 2 Determine the tensile stress in the concrete of the slab during bending σ R II for settlement center O 1 when the OS is directed perpendicular to the long side of the machine trace (Fig. 1 in""). We divide the area of ​​the machine footprint into elementary areas according to paragraph 1. 2.18 . Compatible with clearing house O 1 the center of gravity of an elementary square-shaped area with a side length a p = b p = 140 cm.

Let's define loads R i per elementary area according to the formula ( 15 ), for which we first determine the area of ​​the machine footprint F\u003d 260 140 \u003d 36400 cm 2;

To determine the bending moment M 0 from load R 0 is calculated for an elementary square-shaped platform with the center of gravity in the calculation center O 1 values ​​α = β = a p / l= b p / l\u003d 140 / 53.6 \u003d 2.61 and taking them into account according to table. 2.4 find K 1=36.0; according to the instructions of 2.14 and formula ( 9 ) we calculate:

M 0 = TO one · R 0 \u003d 36.0 80.8 \u003d 2908.8 N cm / cm.

M i, from loads located outside the calculation center O 1 . The calculated data are given in table. 2.10 .

Table 2.10

Calculated data with the calculation center O 1 and the direction of the y-axis perpendicular to the long side of the machine trace

I	x i	y i	x i /l	y i /l	TO 4 according to the table. 2.7	P i, kN	n i number of loads		M i = n i · TO 4 · P i
1	0	120	0	2,24	9,33	36,96	1		363,3
2	120	35	1,86	0,65	-17,22	17,31	4		-1192,3
								Σ M i= -829.0 Ncm/cm

Estimated bending moment from the wheel of the car and the machine according to the formula ( 13 ):

M p II = M 0 + Σ M i= 2908.8 - 829.0 = 2079.8 N cm / cm

Tensile stress in the plate during bending according to the formula ( 7 ):

Calculation 3 Determine the tensile stress in the concrete of the slab during bending σ R III for the settlement center O 2 (Fig. 1 in """). Divide the area of ​​the machine footprint into elementary areas according to p. 2.18 . Let's define loads R i per elementary area, according to the formula ( 15 ).

Let us determine the bending moment from the load created by the pressure of the car wheel, for which we find ρ = r R / l= 15/53.6 = 0.28; according to the table 2.6 find TO 3 = 112.1. According to the formula ( 11 ):M 0 = TO 3 · R p \u003d 112.1 36.96 \u003d 4143.22 N cm / cm.

Let us determine the total bending moment Σ M i from loads located outside the settlement center O 2 . The calculated data are given in table. 2.11 .

Table 2.11

Design data with settlement center O 2

I	x i	y i	x i /l	y i /l	TO 4 according to the table. 2.7	P i, kN	n i number of loads		M i = n i · TO 4 · P i
1	0	65	0	1,21	40,97	4,9	1		200,75
2	0	100	0	1,87	16,36	6,6	1		107,98
3	0	155	0	2,89	2,89	11,5	1		33,24
4	40	65	0,75	1,21	19,1	4,9	2		187,18
5	40	100	0,75	1,87	8,44	6,6	2		111,41
6	40	155	0,75	2,89	1,25	11,5	2		28,75
7	95	65	1,77	1,21	-10,78	8,7	2		-187,57
8	95	100	1,77	1,87	-5,89	11,5	2		-135,47
9	95	155	1,77	2,89	-2,39	20,2	2		-96,56
								Σ M i= 249.7 N cm/cm

Estimated bending moment from the wheel of the car and the machine according to the formula ( 13 ):

M p III = M 0 + Σ M i= 4143.22 + 249.7 = 4392.92 N cm/cm

Tensile stress in the plate during bending according to the formula ( 7 ):

more Rδt = 0.675 MPa, as a result of which we repeat the calculation, setting a large value h. We will carry out the calculation only according to the loading scheme with the calculation center O 2 , for which the value σ R III in the first calculation turned out to be the largest.

For the recalculation, we tentatively set h\u003d 19 cm, then according to p. 2.10 accept l= 86.8 cm; p = r R / l =15/86,8 = 0,1728; TO 3 = 124,7; M 0 = TO 3 · R p\u003d 124.7 36.96 \u003d 4608.9 N cm / cm.

Let us determine the total bending moment from loads located outside the calculation center O 2 . The calculated data are given in table. 2.12 .

Table 2.12

Calculated data for recalculation

I	x i	y i	x i /l	y i /l	TO 4 according to the table. 2.7	P i, kN	n i number of loads		M i = n i · TO 4 · P i
1	0	65	0	0,75	76,17	4,9	1		373,23
2	0	100	0	1,15	44,45	6,6	1		293,37
3	0	155	0	1,79	18,33	11,5	1		210,79
4	40	65	0,46	0,75	48,36	4,9	2		473,93
5	40	100	0,46	1,15	32,39	6,6	2		427,55
6	40	155	0,46	1,79	14,49	11,5	2		333,27
7	95	65	1,09	0,75	1,84	8,7	2		32,02
8	95	100	1,09	1,15	3,92	11,5	2		90,16
9	95	155	1,09	1,79	2,81	20,2	2		113,52
								Σ M i= 2347.84 N cm/cm.

M p= M 0 + Σ M i= 4608.9 + 2347.84 = 6956.82 Ncm/cm

Tensile stress in the plate during bending according to the formula ( 7 ):

Received value σ R= 0.67 MPa differs from Rδt = 0.675 MPa by less than 5%. We accept the underlying layer of concrete of the compressive strength class B22.5 with a thickness h= 19 cm.

Usually, floor heat losses in comparison with similar indicators of other building envelopes (external walls, window and door openings) are a priori assumed to be insignificant and are taken into account in the calculations of heating systems in a simplified form. Such calculations are based on a simplified system of accounting and correction coefficients for the resistance to heat transfer of various building materials.

Considering that the theoretical justification and methodology for calculating the heat loss of the ground floor was developed quite a long time ago (i.e. with a large design margin), we can safely talk about the practical applicability of these empirical approaches in modern conditions. The coefficients of thermal conductivity and heat transfer of various building materials, insulation and floor coverings are well known, and other physical characteristics are not required to calculate heat loss through the floor. According to their thermal characteristics, floors are usually divided into insulated and non-insulated, structurally - floors on the ground and logs.

The calculation of heat loss through an uninsulated floor on the ground is based on the general formula for estimating heat loss through the building envelope:

where Q are the main and additional heat losses, W;

BUT is the total area of ​​the enclosing structure, m2;

tv , tn- temperature inside the room and outside air, °C;

β - share of additional heat losses in total;

n- correction factor, the value of which is determined by the location of the enclosing structure;

Ro– resistance to heat transfer, m2 °С/W.

Note that in the case of a homogeneous single-layer floor slab, the heat transfer resistance Ro is inversely proportional to the heat transfer coefficient of the uninsulated floor material on the ground.

When calculating heat loss through an uninsulated floor, a simplified approach is used, in which the value (1+ β) n = 1. Heat loss through the floor is usually carried out by zoning the heat transfer area. This is due to the natural heterogeneity of the temperature fields of the soil under the floor.

The heat loss of an uninsulated floor is determined separately for each two-meter zone, the numbering of which starts from the outer wall of the building. In total, four such strips 2 m wide are taken into account, considering the soil temperature in each zone to be constant. The fourth zone includes the entire surface of the uninsulated floor within the boundaries of the first three strips. Heat transfer resistance is accepted: for the 1st zone R1=2.1; for the 2nd R2=4.3; respectively for the third and fourth R3=8.6, R4=14.2 m2*оС/W.

Fig.1. Zoning of the floor surface on the ground and adjacent recessed walls when calculating heat losses

In the case of recessed rooms with a soil base of the floor: the area of ​​the first zone adjacent to the wall surface is taken into account twice in the calculations. This is quite understandable, since the heat losses of the floor are added to the heat losses in the vertical enclosing structures of the building adjacent to it.

Calculation of heat loss through the floor is made for each zone separately, and the results obtained are summed up and used for the thermal engineering justification of the building design. The calculation for the temperature zones of the outer walls of recessed rooms is carried out according to formulas similar to those given above.

In calculations of heat loss through an insulated floor (and it is considered as such if its structure contains layers of material with a thermal conductivity of less than 1.2 W / (m ° C)) the value of the heat transfer resistance of an uninsulated floor on the ground increases in each case by the heat transfer resistance of the insulating layer:

Ru.s = δy.s / λy.s,

where δy.s– thickness of the insulating layer, m; λu.s- thermal conductivity of the material of the insulating layer, W / (m ° C).