Calculation of the wall for stability. Calculation of brickwork for stability. Determining the stability of a brick column

When independent design brick house there is an urgent need to calculate whether the brickwork can withstand the loads that are included in the project. A particularly serious situation develops in areas of masonry weakened by window and doorways. In case of heavy load, these areas may not withstand and be destroyed.

The exact calculation of the resistance of the pier to compression by the overlying floors is quite complex and is determined by the formulas included in regulatory document SNiP-2-22-81 (hereinafter referred to as<1>). Engineering calculations of a wall's compressive strength take into account many factors, including wall configuration, compressive strength, strength of this type materials and much more. However, approximately, “by eye,” you can estimate the wall’s resistance to compression, using indicative tables in which the strength (in tons) is linked to the width of the wall, as well as brands of brick and mortar. The table is compiled for a wall height of 2.8 m.

Strength table brick wall, tons (example)

Stamps		Area width, cm
brick	solution	25	51	77	100	116	168	194	220	246	272	298
50	25	4	7	11	14	17	31	36	41	45	50	55
100	50	6	13	19	25	29	52	60	68	76	84	92

If the value of the wall width is in the range between those indicated, it is necessary to focus on the minimum number. At the same time, it should be remembered that the tables do not take into account all factors that can adjust the stability, structural strength and resistance of a brick wall to compression in a fairly wide range.

In terms of time, loads can be temporary or permanent.

Permanent:

• weight of building elements (weight of fences, load-bearing and other structures);
• soil and rock pressure;
• hydrostatic pressure.

Temporary:

• weight of temporary structures;
• loads from stationary systems and equipment;
• pressure in pipelines;
• loads from stored products and materials;
• climatic loads (snow, ice, wind, etc.);
• and many others.

When analyzing the loading of structures, it is imperative to take into account the total effects. Below is an example of calculating the main loads on the walls of the first floor of a building.

Brickwork load

To take into account the force acting on the designed section of the wall, you need to sum up the loads:

In the case of low-rise construction, the problem is greatly simplified, and many factors of temporary load can be neglected by setting a certain safety margin at the design stage.

However, in the case of the construction of 3 or more storey structures, a thorough analysis is required using special formulas that take into account the addition of loads from each floor, the angle of application of force, and much more. In some cases, the strength of the wall is achieved by reinforcement.

Load calculation example

This example shows the analysis of the current loads on the piers of the 1st floor. Only permanently taken into account here effective load from various structural elements building, taking into account the uneven weight of the structure and the angle of application of forces.

Initial data for analysis:

• number of floors – 4 floors;
• brick wall thickness T=64cm (0.64 m);
• specific gravity of masonry (brick, mortar, plaster) M = 18 kN/m3 (indicator taken from reference data, table 19<1>);
• width window openings is: Ш1=1.5 m;
• height of window openings - B1=3 m;
• pier section 0.64*1.42 m (loaded area where the weight of the overlying structural elements is applied);
• floor height Wet=4.2 m (4200 mm):
• the pressure is distributed at an angle of 45 degrees.

1. An example of determining the load from a wall (plaster layer 2 cm)

Hst = (3-4Ш1В1)(h+0.02) Myf = (*3-4*3*1.5)* (0.02+0.64) *1.1 *18=0.447MN.

Width of the loaded area P=Wet*H1/2-W/2=3*4.2/2.0-0.64/2.0=6 m

Nn =(30+3*215)*6 = 4.072MN

ND=(30+1.26+215*3)*6 = 4.094MN

H2=215*6 = 1.290MN,

including H2l=(1.26+215*3)*6= 3.878MN

1. Own weight of the walls

Npr=(0.02+0.64)*(1.42+0.08)*3*1.1*18= 0.0588 MN

The total load will be the result of a combination of the indicated loads on the walls of the building; to calculate it, the summation of the loads from the wall, from the floors of the second floor and the weight of the designed area is performed).

Scheme of load and structural strength analysis

To calculate the pier of a brick wall you will need:

• length of the floor (aka height of the site) (Wet);
• number of floors (Chat);
• wall thickness (T);
• width of the brick wall (W);
• masonry parameters (type of brick, brand of brick, brand of mortar);

1. Wall area (P)

1. According to table 15<1>it is necessary to determine the coefficient a (elasticity characteristic). The coefficient depends on the type and brand of brick and mortar.
2. Flexibility index (G)

1. Depending on indicators a and G, according to table 18<1>you need to look at the bending coefficient f.
2. Finding the height of the compressed part

where e0 is an indicator of extraness.

1. Finding the area of ​​the compressed part of the section

Pszh = P*(1-2 e0/T)

1. Determination of the flexibility of the compressed part of the pier

Gszh=Vet/Vszh

1. Determination according to table. 18<1>fszh coefficient, based on gszh and coefficient a.
2. Calculation of the average coefficient fsr

Fsr=(f+fszh)/2

1. Determination of coefficient ω (Table 19<1>)

ω =1+e/T<1,45

1. Calculation of the force acting on the section
2. Definition of sustainability

U=Kdv*fsr*R*Pszh* ω

Kdv – long-term exposure coefficient

R – masonry compression resistance, can be determined from Table 2<1>, in MPa

1. Reconciliation

An example of calculating the strength of masonry

— Wet — 3.3 m

— Chat — 2

— T — 640 mm

— W — 1300 mm

- masonry parameters (clay brick made by plastic pressing, cement-sand mortar, brick grade - 100, mortar grade - 50)

1. Area (P)

P=0.64*1.3=0.832

1. According to table 15<1>determine the coefficient a.

1. Flexibility (G)

G =3.3/0.64=5.156

1. Bending coefficient (Table 18<1>).

1. Height of compressed part

Vszh=0.64-2*0.045=0.55 m

1. Area of ​​the compressed part of the section

Pszh = 0.832*(1-2*0.045/0.64)=0.715

1. Flexibility of the compressed part

Gszh=3.3/0.55=6

1. fsj=0.96
2. FSR calculation

Fsr=(0.98+0.96)/2=0.97

1. According to the table 19<1>

ω =1+0.045/0.64=1.07<1,45

To determine the effective load, it is necessary to calculate the weight of all structural elements affecting the designed area of ​​the building.

1. Definition of sustainability

Y=1*0.97*1.5*0.715*1.07=1.113 MN

1. Reconciliation

The condition is met, the strength of the masonry and the strength of its elements are sufficient

Insufficient wall resistance

What to do if the calculated pressure resistance of the walls is insufficient? In this case, it is necessary to strengthen the wall with reinforcement. Below is an example of an analysis of the necessary modernization of a structure with insufficient compressive resistance.

For convenience, you can use tabular data.

The bottom line shows indicators for a wall reinforced with wire mesh with a diameter of 3 mm, with a cell of 3 cm, class B1. Reinforcement of every third row.

The increase in strength is about 40%. Typically this compression resistance is sufficient. It is better to make a detailed analysis, calculating the change in strength characteristics in accordance with the method of strengthening the structure used.

Below is an example of such a calculation

Example of calculation of pier reinforcement

Initial data - see previous example.

• floor height - 3.3 m;
• wall thickness – 0.640 m;
• masonry width 1,300 m;
• typical characteristics of masonry (type of bricks - clay bricks made by pressing, type of mortar - cement with sand, brand of bricks - 100, mortar - 50)

In this case, the condition У>=Н is not satisfied (1.113<1,5).

It is required to increase the compression resistance and structural strength.

Gain

k=U1/U=1.5/1.113=1.348,

those. it is necessary to increase the structural strength by 34.8%.

Reinforcement with reinforced concrete frame

Reinforcement is carried out with a cage made of B15 concrete with a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 with a step of 0.150 m.

Section dimensions of the reinforced structure:

Ш_1=1300+2*60=1.42

T_1=640+2*60=0.76

With such indicators, the condition У>=Н is satisfied. Compression resistance and structural strength are sufficient.

Necessity of calculation brickwork when building a private house, it is obvious to any developer. In the construction of residential buildings, clinker and red bricks are used; finishing bricks are used to create an attractive appearance of the outer surface of the walls. Each brand of brick has its own specific parameters and properties, but the difference in size between different brands is minimal.

The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick.

Clinker bricks are used for the construction of luxury houses. It has a high specific gravity, attractive appearance, and high strength. Limited use due to the high cost of the material.

The most popular and in demand material is red brick. It has sufficient strength with a relatively low specific gravity, is easy to process, and is little susceptible to environmental influences. Disadvantages - sloppy surfaces with high roughness, the ability to absorb water at high humidity. Under normal operating conditions, this ability does not manifest itself.

There are two methods for laying bricks:

• tychkovy;
• spoon

When laying using the butt method, the brick is laid across the wall. The wall thickness must be at least 250 mm. The outer surface of the wall will consist of the end surfaces of the material.

With the spoon method, the brick is laid lengthwise. The side surface appears outside. Using this method, you can lay out half-brick walls - 120 mm thick.

What you need to know to calculate

The maximum amount of material can be calculated by determining the total volume of the walls and dividing it by the volume of one brick. The result obtained will be approximate and overestimated. For a more accurate calculation, the following factors must be taken into account:

• masonry joint size;
• exact dimensions of the material;
• thickness of all walls.

Manufacturers quite often, for various reasons, do not maintain standard product sizes. According to GOST, red masonry bricks must have dimensions of 250x120x65 mm. To avoid mistakes and unnecessary material costs, it is advisable to check with suppliers about the dimensions of available bricks.

The optimal thickness of external walls for most regions is 500 mm, or 2 bricks. This size ensures high strength of the building and good thermal insulation. The disadvantage is the large weight of the structure and, as a result, pressure on the foundation and lower layers of masonry.

The size of the masonry joint will primarily depend on the quality of the mortar.

If you use coarse-grained sand to prepare the mixture, the width of the seam will increase; with fine-grained sand, the seam can be made thinner. The optimal thickness of masonry joints is 5-6 mm. If necessary, it is allowed to make seams with a thickness of 3 to 10 mm. Depending on the size of the seams and the method of laying the brick, you can save some of it.

For example, let's take a seam thickness of 6 mm and the spoon method of laying brick walls. If the wall thickness is 0.5 m, you need to lay 4 bricks wide.

The total width of the gaps will be 24 mm. Laying 10 rows of 4 bricks will give a total thickness of all gaps of 240 mm, which is almost equal to the length of a standard product. The total area of ​​the masonry will be approximately 1.25 m2. If the bricks are laid closely, without gaps, 240 pieces fit in 1 m2. Taking into account the gaps, the material consumption will be approximately 236 pieces.

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Calculation method for load-bearing walls

When planning the external dimensions of a building, it is advisable to choose values ​​that are multiples of 5. With such numbers it is easier to carry out calculations, then carry them out in reality. When planning the construction of 2 floors, you should calculate the amount of material in stages for each floor.

First, the calculation of the external walls on the first floor is performed. For example, you can take a building with dimensions:

• length = 15 m;
• width = 10 m;
• height = 3 m;
• The thickness of the walls is 2 bricks.

Using these dimensions you need to determine the perimeter of the building:

(15 + 10) x 2 = 50

3 x 50 = 150 m2

By calculating the total area, you can determine the maximum amount of bricks for building a wall. To do this, you need to multiply the previously determined number of bricks for 1 m2 by the total area:

236 x 150 = 35,400

The result is inconclusive, the walls must have openings for installing doors and windows. The number of entrance doors may vary. Small private houses usually have one door. For large buildings, it is advisable to plan two entrances. The number of windows, their sizes and location are determined by the internal layout of the building.

As an example, you can take 3 window openings per 10-meter wall, 4 per 15-meter walls. It is advisable to make one of the walls blank, without openings. The volume of doorways can be determined by standard dimensions. If the dimensions differ from the standard ones, the volume can be calculated using the overall dimensions, adding to them the width of the installation gap. To calculate, use the formula:

2 x (A x B) x 236 = C

where: A is the width of the doorway, B is the height, C is the volume in the number of bricks.

Substituting standard values, we get:

2 x (2 x 0.9) x 236 = 849 pcs.

The volume of window openings is calculated similarly. With window sizes of 1.4 x 2.05 m, the volume will be 7450 pieces. Determining the number of bricks per temperature gap is simple: you need to multiply the length of the perimeter by 4. The result is 200 pieces.

35400 — (200 + 7450 + 849) = 26 901.

You should purchase the required quantity with a small margin, because errors and other unforeseen situations are possible during operation.

Load on the pier at the level of the bottom of the first floor floor beam, kN	Values, kN
snow for II snow region	1000*6,74*(23,0*0,5+0,51+0,25)*1,4*0,001=115,7
rolled roofing carpet-100N/m 2	100*6,74*(23,0*0,5+0,51+0,25)*1,1*0,001=9,1
asphalt screed at p=15000N/m 3 15 mm thick	15000*0,015*6,74*23,0*0,5*1,2*0,001=20,9
insulation - wood fiber boards 80 mm thick with a density p = 3000 N/m 3	3000*0,08*6,74*23,0*0,5*1,2*0,001=22,3
Vapor barrier - 50N/m 2	50*6,74*23,0*0,5*1,2*0,001=4,7
prefabricated reinforced concrete covering slabs – 1750N/m2	1750*6,74*23,0*0,5*1,1*0,001=149,2
reinforced concrete truss weight	6900*1,1*0,01=75,9
weight of the cornice on the brickwork of the wall at p = 18000 N/m 3	18000*((0,38+0,43)*0,5*0,51-0,13*0,25)* *6,74*1,1*0,001=23,2
brickwork weight above mark +3.17	18000*((18,03-3,17)*6,74 - 2,4*2,1*3)*0,51*1,1*0,001=857
concentrated from the floor crossbars (conditionally)	119750*5,69*0,5*3*0,001=1022
weight of window filling at V n =500N/m2	500*2,4*2,1*3*1,1*0,001=8,3

The total design load on the pier at the level of elevation. +3.17:

N=115.7+9.1+20.9+22.3+4.7+149.2+75.9+23.2+857.1+1022+8.3=2308.4.

It is permissible to consider the wall as divided in height into single-span elements with the location of the supporting hinges at the level of the support of the crossbars. In this case, the load from the upper floors is assumed to be applied at the center of gravity of the wall section of the overlying floor, and all loads P = 119750 * 5.69 * 0.5 * 0.001 = 340.7 kN within a given floor are considered to be applied with actual eccentricity relative to the center of gravity of the section .

The distance from the point of application of support reactions of the crossbar P to the inner edge of the wall in the absence of supports fixing the position of the support pressure is taken to be no more than a third of the depth of embedding of the crossbar and no more than 7 cm.

When the depth of embedding of the crossbar in the wall is a 3 = 380 mm, and 3: 3 = 380: 3 = 127 mm > 70 mm, we accept the point of application of the support pressure P = 340.7 kN at a distance of 70 mm from the inner edge of the wall.

Estimated height of the pier in the lower floor

l 0 =3170+50=3220 mm.

For the design diagram of the pier of the lower floor of the building we take a post with pinching at the level of the foundation edge and with hinged support at the floor level.

Flexibility of the wall made of sand-lime brick grade 100 on grade 25 mortar, at R=1.3 MPa with masonry characteristic α=1000

λ h =l 0:h=3220:510=6.31

Coefficient longitudinal bendingφ=0.96, in walls with a rigid upper support, longitudinal bending in the supporting sections may not be taken into account (φ=1) In the middle third of the pier height, the longitudinal bending coefficient is equal to the calculated value φ=0.96. In the support thirds of the height, φ changes linearly from φ=1 to the calculated value φ=0.96

Values ​​of the longitudinal bending coefficient in the design sections of the piers, at the levels of the top and bottom of the window opening:

φ 1 =0.96+(1-0.96)

φ 2 =0.96+(1-0.96)

The values ​​of bending moments at the level of support of the crossbar and in the design sections of the pier at the level of the top and bottom of the window opening, kNm:

M=Pe=340.7*(0.51*0.5-0.07)=63.0

M 1 =63.0

M 11 =63.0

Magnitude of normal forces in the same sections of the pier, kN:

N 1 =2308.4+0.51*6.74*0.2*1800*1.1*0.01=2322.0

N 11 =2322+(0.51*(6.74-2.4)*2.1*1800*1.1+50*2.1*2.4*1.1)*0.01=2416.8

N 111 =2416.8+0.51*0.8*6.74*1800*1.1*0.01=2471.2.

Eccentricities of longitudinal forces e 0 =M:N:

e 0 =(66.0:2308.4)*1000=27 mm<0.45y=0.45*255=115мм

e 01 =(56.3:2322)*1000=24 mm<0.45y=0.45*255=115мм

e 011 =(15.7:2416.8)*1000=6 mm<0.45y=0.45*255=115мм

e 0111 =0 mmy=0.5*h=0.5*510=255mm.

Load-bearing capacity of an eccentrically compressed pier of rectangular cross-section

determined by the formula:

N=m g φ 1 RA*(1- )ω, whereω=1+ <=1.45,
, where φ is the longitudinal bending coefficient for the entire cross-section of a rectangular element h c = h-2e 0 , m g is a coefficient that takes into account the influence of long-term load (for h = 510 mm > 300 mm, take 1), A is the cross-sectional area of ​​the pier.

Bearing capacity (strength) of the pier at the level of support of the crossbar at φ=1.00, e 0 =27 mm, λ с =l 0:h с =l 0:(h-2е 0)=3220:(510-2*27 )=7.1,φ s =0.936,

φ 1 =0.5*(φ+φ s)=0.5*(1+0.936)=0.968,ω=1+
<1.45

N=1*0.968* 1.3*6740*510*(1-
)1.053=4073 kN >2308 kN

Bearing capacity (strength) of the wall in section 1-1 at φ=0.987, e 0 =24 mm, λ c =l 0:h c =l 0:(h-2e 0)=3220:(510-2*24) =6.97,φ s =0.940,

φ 1 =0.5*(φ+φ s)=0.5*(0.987+0.940)=0.964,ω=1+
<1.45

N 1 =1*0.964* 1.3*4340*510*(1-
)1.047=2631 kN >2322 kN

Bearing capacity (strength) of the pier in section II-IIatφ=0.970, e 0 =6 mm, λ c =l 0:h c =l 0:(h-2e 0)=3220:(510-2*6)=6 .47,φ s =0.950,

φ 1 =0.5*(φ+φ s)=0.5*(0.970+0.950)=0.960,ω=1+
<1.45

N 11 =1*0.960* 1.3*4340*510*(1- )1.012=2730 kN >2416.8 kN

Bearing capacity (strength) of the pier in section III-III at the foundation edge level under central compression at φ = 1, e 0 = 0 mm,

N 111 =1*1* 1.3*6740*510=4469 kN >2471 kN

That. The strength of the pier is ensured in all sections of the lower floor of the building.

Working fittings	Design cross section	Design force M, N mm	Design characteristics						Design reinforcement	Accepted fittings
			, mm		, mm			Reinforcement class
In the lower zone	In the extreme spans	123,80*10								, A s =760mm 2 in two flat frames
	On medium spans	94,83*10								, A s =628mm 2 in two flat frames
In the upper zone	In the second flight	52,80*10								, A s =308mm 2 in two frames
	In all medium spans	41,73*10								, A s =226mm 2 in two frames
	On a support	108,38*10								, A s =628mm 2 in one U-shaped mesh
	On a supportC	94,83*10								, A s =628mm 2 in one U-shaped mesh

Table 3

Loading scheme

Shear forces, kNm

M

In the extreme spans

M

On medium spans

M

M

M

M

M

Q

Q

Q

Q

Table 7

Arrangement of rods		Reinforcement cross-section, mm			Calculated characteristics
		Before rods A break	Breakable	After the breakage of rods A			mm x10			According to table 9
In the lower zone of the crossbar	At the end of the day: at support A
	at support B
	On average: at support B
In the upper zone of the crossbar	At support B: from the extreme span
	from the side of the middle span

Design cross section	Design force M, kN*m	Section dimensions, mm		Design characteristics		Longitudinal working reinforcement class AIII, mm		Actual load-bearing capacity, kN*m
				R b =7.65 MPa		R s =355 MPa	Actual accepted
In the lower zone of the extreme spans
In the upper zone above supports B at the edge of the column
In the lower zone of middle spans
In the upper zone above the supports C at the edge of the column

Ordinates

BENDING MOMENTS, k N m

In the extreme spans

M

On medium spans

M

M

M

M

M

Ordinates of the main diagram of moments when loading according to schemes 1+4

by the amount

M =145.2 kNm

Redistribution ordinates of diagram IIa

Ordinates of the main diagram of moments when loading according to schemes 1+5

Redistribution of forces by reducing the support moment M by the amount

Ordinates of the additional diagram at M =89.2 kNm

Redistribution ordinates of diagram IIIa

Loading scheme

BENDING MOMENTS, k N m

Shear forces, kNm

M

In the extreme spans

M

On medium spans

M

M

M

M

M

Q

Q

Q

Q

		Longitudinal reinforcement Breakable reinforcement	Transverse reinforcement step	Transverse force at the point where the rods break, kN		Length of launching breakable rods beyond the theoretical break point, mm	Minimum value ω=20d, mm	Accepted value ω,mm	Distance from support axis, mm
									To the place of theoretical break (scaled according to the diagram of materials)	To the actual location of the break
In the lower zone of the crossbar	At the end of the day: at support A
	at support B
	On average: at support B
In the upper zone of the crossbar	At support B: from the extreme span
	from the side of the middle span

Вр1 with Rs=360 MPa, АIII with Rs=355 MPa

In the extreme areas between axes 1-2 and 6-7

In the extreme spans

In the middle spans

In the middle sections between axles 2-6

In the extreme spans

In the middle spans

Arrangement of rods		Reinforcement cross-section, mm 2			Design characteristics
		Before the rods break	torn off	After the rods break			b*h 0, mm 2 *10 -2				М=R b *b*h 0 *A 0 , kN*m
In the lower zone of the crossbar	In the extreme span: at support A
	at support B
	On the middle span: at support B
	at support C
In the upper zone of the crossbar	At support B: from the extreme span
	from the middle span
	At support C from both spans

Location of breakable rods		Longitudinal__ fittings__ breakable reinforcement	Transverse reinforcement _quantity_	Transverse force at the point of theoretical breakage of the rods, kN		Length of launching breakable rods beyond the theoretical break point, mm	Minimum value w=20d	Accepted value w, mm	Distance from support axis, mm
									To the point of theoretical break (according to the diagram of materials)	To the actual location of the break
In the lower zone of the crossbar	In the extreme span: at support A
	at support B
	On the middle span: at support B
	at support C
In the upper zone of the crossbar	At support B: from the extreme span
	from the middle span
	At support C from both spans

To perform a wall stability calculation, you first need to understand their classification (see SNiP II -22-81 “Stone and reinforced masonry structures”, as well as a manual for SNiP) and understand what types of walls there are:

1. Load-bearing walls- these are the walls on which floor slabs, roof structures, etc. rest. The thickness of these walls must be at least 250 mm (for brickwork). These are the most important walls in the house. They need to be designed for strength and stability.

2. Self-supporting walls- these are walls on which nothing rests, but they are subject to the load from all the floors above. In fact, in a three-story house, for example, such a wall will be three floors high; the load on it only from the own weight of the masonry is significant, but at the same time the question of the stability of such a wall is also very important - the higher the wall, the greater the risk of its deformation.

3. Curtain walls- these are external walls that rest on the ceiling (or other structural elements) and the load on them comes from the height of the floor only from the wall’s own weight. The height of non-load-bearing walls should be no more than 6 meters, otherwise they become self-supporting.

4. Partitions are internal walls less than 6 meters high that only support the load from their own weight.

Let's look at the issue of wall stability.

The first question that arises for an “uninitiated” person is: where can the wall go? Let's find the answer using an analogy. Let's take a hardcover book and place it on its edge. The larger the book format, the less stable it will be; on the other hand, the thicker the book, the better it will stand on its edge. The situation is the same with walls. The stability of the wall depends on the height and thickness.

Now let’s take the worst case scenario: a thin, large-format notebook and place it on its edge - it will not only lose stability, but will also bend. Likewise, the wall, if the conditions for the ratio of thickness and height are not met, will begin to bend out of plane, and over time, crack and collapse.

What is needed to avoid such a phenomenon? You need to study pp. 6.16...6.20 SNiP II -22-81.

Let's consider the issues of determining the stability of walls using examples.

Example 1. Given a partition made of aerated concrete grade M25 on mortar grade M4, 3.5 m high, 200 mm thick, 6 m wide, not connected to the ceiling. The partition has a doorway of 1x2.1 m. It is necessary to determine the stability of the partition.

From Table 26 (item 2) we determine the masonry group - III. From the tables do we find 28? = 14. Because the partition is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 9.8.

k 1 = 1.8 - for a partition that does not carry a load with a thickness of 10 cm, and k 1 = 1.2 - for a partition 25 cm thick. By interpolation, we find for our partition 20 cm thick k 1 = 1.4;

k 3 = 0.9 - for partitions with openings;

that means k = k 1 k 3 = 1.4*0.9 = 1.26.

Finally β = 1.26*9.8 = 12.3.

Let's find the ratio of the height of the partition to the thickness: H /h = 3.5/0.2 = 17.5 > 12.3 - the condition is not met, a partition of such thickness cannot be made with the given geometry.

How can this problem be solved? Let's try to increase the grade of mortar to M10, then the masonry group will become II, respectively β = 17, and taking into account the coefficients β = 1.26*17*70% = 15< 17,5 - этого оказалось недостаточно. Увеличим марку газобетона до М50, тогда группа кладки станет I , соответственно β = 20, а с учетом коэффициентов β = 1,26*20*70% = 17.6 >17.5 - the condition is met. It was also possible, without increasing the grade of aerated concrete, to lay structural reinforcement in the partition in accordance with clause 6.19. Then β increases by 20% and the stability of the wall is ensured.

Example 2. Dana external Not bearing wall made of lightweight brick masonry grade M50 on mortar grade M25. Wall height 3 m, thickness 0.38 m, wall length 6 m. Wall with two windows measuring 1.2x1.2 m. It is necessary to determine the stability of the wall.

From Table 26 (clause 7) we determine the masonry group - I. From Table 28 we find β = 22. Because the wall is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to clause 6.20), i.e. β = 15.4.

We find the coefficients k from tables 29:

k 1 = 1.2 - for a wall that does not bear a load with a thickness of 38 cm;

k 2 = √А n /A b = √1.37/2.28 = 0.78 - for a wall with openings, where A b = 0.38*6 = 2.28 m 2 - horizontal sectional area of ​​the wall, taking into account windows, A n = 0.38*(6-1.2*2) = 1.37 m2;

that means k = k 1 k 2 = 1.2*0.78 = 0.94.

Finally β = 0.94*15.4 = 14.5.

Let's find the ratio of the height of the partition to the thickness: H /h = 3/0.38 = 7.89< 14,5 - условие выполняется.

It is also necessary to check the condition stated in clause 6.19:

H + L = 3 + 6 = 9 m< 3kβh = 3*0,94*14,5*0,38 = 15.5 м - условие выполняется, устойчивость стены обеспечена.

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0 #212 Alexey 02/21/2018 07:08

I quote Irina:

profiles will not replace reinforcement

I quote Irina:

Regarding the foundation: voids in the concrete body are permissible, but not from below, so as not to reduce the bearing area, which is responsible for the load-bearing capacity. That is, there should be a thin layer of reinforced concrete underneath.
What kind of foundation - strip or slab? What soils?

The soils are not yet known, most likely it will be an open field of all sorts of loam, initially I was thinking of a slab, but it will turn out a little low, I want it higher, but I will also have to have the top one fertile layer remove, so I’m leaning towards a ribbed or even box-shaped foundation. I don’t need a lot of bearing capacity of the soil - after all, the house was built on the 1st floor, and expanded clay concrete is not very heavy, freezing there is no more than 20 cm (although according to old Soviet standards it is 80).

I’m thinking of removing the top layer of 20-30 cm, laying out geotextiles, covering it with river sand and leveling it with compaction. Then easy preparatory screed- for leveling (it seems like they don’t even put reinforcement in it, although I’m not sure), waterproof it with a primer on top
and then there’s a dilemma - even if you tie reinforcement frames with a width of 150-200mm x 400-600mm in height and lay them in steps of a meter, then you still need to form voids with something between these frames and ideally these voids should be on top of the reinforcement (yes also with some distance from the preparation, but at the same time they will also need to be reinforced on top thin layer under a 60-100mm screed) - I’m thinking of monolithing the PPS slabs as voids - theoretically it would be possible to fill this in one go with vibration.

Those. It looks like a slab of 400-600mm with powerful reinforcement every 1000-1200mm, the volumetric structure is uniform and light in other places, while inside about 50-70% of the volume there will be foam plastic (in unloaded places) - i.e. in terms of consumption of concrete and reinforcement - quite comparable to a 200mm slab, but + a lot of relatively cheap polystyrene foam and more work.

If you somehow replaced the foam with simple soil/sand, it would be even better, but then, instead of light preparation, it would be wiser to do something more serious with reinforcement and removal of the reinforcement into the beams - in in general here I lack both theory and practical experience.

0 #214 Irina 02.22.2018 16:21

Quote:

It’s a pity, in general they just write that lightweight concrete (expanded clay concrete) has a poor connection with the reinforcement - how to deal with this? I understand what stronger than concrete and with what larger area surface of the reinforcement - the better the connection will be, i.e. you need expanded clay concrete with the addition of sand (and not just expanded clay and cement) and thin reinforcement, but more often

why fight it? you just need to take it into account in the calculations and design. You see, expanded clay concrete is quite good wall material with its own list of advantages and disadvantages. Just like any other materials. Now, if you wanted to use it for monolithic ceiling, I would dissuade you, because
Quote:

External load-bearing walls must, at a minimum, be designed for strength, stability, local collapse and resistance to heat transfer. To find out what thickness should it be Brick wall , you need to calculate it. In this article we will look at the calculation bearing capacity brickwork, and in the following articles - the remaining calculations. So as not to miss the exit new article, subscribe to the newsletter and you will find out what the thickness of the wall should be after all the calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, we will consider all calculations specifically for this category.

Bearing are called walls that bear the load from floor slabs, coverings, beams, etc. resting on them.

You should also take into account the brand of brick for frost resistance. Since everyone builds a house for themselves for at least a hundred years, in dry and normal humidity conditions of the premises, a grade (M rz) of 25 and above is accepted.

During the construction of a house, cottage, garage, outbuildings and other structures with dry and normal humidity conditions It is recommended to use hollow bricks for external walls, since its thermal conductivity is lower than that of solid bricks. Accordingly, during thermal engineering calculations, the thickness of the insulation will be less, which will save cash when purchasing it. Solid brick for external walls it should be used only when it is necessary to ensure the strength of the masonry.

Reinforcement of brickwork is allowed only if increasing the grade of brick and mortar does not provide the required load-bearing capacity.

An example of calculating a brick wall.

The load-bearing capacity of brickwork depends on many factors - the brand of brick, the brand of mortar, the presence of openings and their sizes, the flexibility of the walls, etc. The calculation of bearing capacity begins with determining the design scheme. When calculating walls for vertical loads, the wall is considered to be supported by hinged and fixed supports. When calculating walls for horizontal loads (wind), the wall is considered rigidly clamped. It is important not to confuse these diagrams, since the moment diagrams will be different.

Selection of design section.

In solid walls, the design section is taken to be section I-I at the level of the bottom of the floor with a longitudinal force N and a maximum bending moment M. It is often dangerous section II-II, since the bending moment is slightly less than the maximum and is equal to 2/3M, and the coefficients m g and φ are minimal.

In walls with openings, the cross-section is taken at the level of the bottom of the lintels.

Let's look at section I-I.

From the previous article Collection of loads on the first floor wall Let's take the resulting value of the total load, which includes the load from the floor of the first floor P 1 = 1.8 t and the overlying floors G = G p +P 2 +G 2 = 3.7t:

N = G + P 1 = 3.7t +1.8t = 5.5t

The floor slab rests on the wall at a distance of a=150mm. The longitudinal force P 1 from the ceiling will be at a distance a / 3 = 150 / 3 = 50 mm. Why 1/3? Because the stress diagram under the support section will be in the form of a triangle, and the center of gravity of the triangle is located at 1/3 of the length of the support.

The load from the overlying floors G is considered to be applied centrally.

Since the load from the floor slab (P 1) is not applied at the center of the section, but at a distance from it equal to:

e = h/2 - a/3 = 250mm/2 - 150mm/3 = 75 mm = 7.5 cm,

then it will create a bending moment (M) in section I-I. Moment is the product of force and arm.

M = P 1 * e = 1.8t * 7.5cm = 13.5t*cm

Then the eccentricity of the longitudinal force N will be:

e 0 = M / N = 13.5 / 5.5 = 2.5 cm

Since the load-bearing wall is 25 cm thick, the calculation should take into account the value of the random eccentricity e ν = 2 cm, then the total eccentricity is equal to:

e 0 = 2.5 + 2 = 4.5 cm

y=h/2=12.5cm

At e 0 =4.5 cm< 0,7y=8,75 расчет по раскрытию трещин в швах кладки можно не производить.

Strength of masonry eccentrically compressed element determined by the formula:

N ≤ m g φ 1 R A c ω

Odds m g And φ 1 in the section under consideration, I-I are equal to 1.