≡× MENU

• Interior
• Window
• Walls
• Projects
• The buildings

Calculation of brickwork. Calculation of brickwork for strength Calculation of brick walls for strength and stability

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 the configuration of the wall, its compressive strength, the strength of the type of material, and 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.

Table of brick wall strength, 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 interval 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 in some cases The strength of the pier 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)

Nst = (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 (also the 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 using a B15 concrete frame with a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 with a pitch 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. The compression resistance and structural strength are sufficient.

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 = 18000N/m 3	18000*((0,38+0,43)*0,5*0,51-0,13*0,25)* *6,74*1,1*0,001=23,2
weight brickwork above +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 window filling at V n =500N/m 2	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. At the same time, the load from 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

The longitudinal bending coefficient is φ=0.96; in walls with a rigid upper support, the 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 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 eccentrically compressed wall of rectangular cross-section

determined by the formula:

N=m g φ 1 RA*(1- )ω, whereω=1+ <=1.45,
, whereφ is the longitudinal bending coefficient for the entire section of the element rectangular shape h c =h-2e 0 ,m g - coefficient taking into account the influence of long-term load (for h = 510mm>300mm, take 1), A - 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

B e nding mo ments, 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

B e nding mo ments, 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 the 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 the 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

The need to calculate brickwork when building a private house 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:

• pinned;
• 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 sizes 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 a spoon laying method. 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.

Return to contents

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.

Picture 1. Calculation diagram for brick columns of the designed building.

A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea of ​​laying columns of clay bricks, and even more so the walls of a house, is far from new and all possible aspects of the calculations of brick walls, piers, pillars, which are the essence of the column, are described in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced stone structures". It is this regulatory document that should be used as a guide when making calculations. The calculation given below is nothing more than an example of using the specified SNiP.

To determine the strength and stability of columns, you need to have quite a lot of initial data, such as: brand of brick in terms of strength, area of ​​support of the crossbars on the columns, load on the columns, cross-sectional area of ​​the column, and if none of this is known at the design stage, then you can proceed in the following way:

An example of calculating a brick column for stability under central compression

Designed:

Terrace dimensions 5x8 m. Three columns (one in the middle and two at the edges) made of facing hollow brick with a section of 0.25x0.25 m. The distance between the axes of the columns is 4 m. The strength grade of the brick is M75.

Calculation prerequisites:

.

With this design scheme, the maximum load will be on the middle lower column. This is exactly what you should count on for strength. The load on the column depends on many factors, in particular the construction area. For example, in St. Petersburg it is 180 kg/m2, and in Rostov-on-Don - 80 kg/m2. Taking into account the weight of the roof itself 50-75 kg/m2, the load on the column from the roof for Pushkin, Leningrad region can be:

N from the roof = (180 1.25 + 75) 5 8/4 = 3000 kg or 3 tons

Since the current loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but a reinforced concrete slab is definitely not planned, and it is assumed that the floor will be wooden, from separately lying edged boards, then to calculate the load from the terrace you can accept a uniformly distributed load of 600 kg/m2, then the concentrated force from the terrace acting on the central column will be:

N from terrace = 600 5 8/4 = 6000 kg or 6 tons

The dead weight of columns 3 m long will be:

N from column = 1500 3 0.38 0.38 = 649.8 kg or 0.65 tons

Thus, the total load on the middle lower column in the section of the column near the foundation will be:

N with rev = 3000 + 6000 + 2 650 = 10300 kg or 10.3 tons

However, in this case, it can be taken into account that there is not a very high probability that the temporary load from snow, maximum in winter, and the temporary load on the floor, maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability coefficient of 0.9, then:

N with rev = (3000 + 6000) 0.9 + 2 650 = 9400 kg or 9.4 tons

The design load on the outer columns will be almost two times less:

N cr = 1500 + 3000 + 1300 = 5800 kg or 5.8 tons

2. Determination of the strength of brickwork.

The M75 brick grade means that the brick must withstand a load of 75 kgf/cm2, however, the strength of the brick and the strength of the brickwork are two different things. The following table will help you understand this:

Table 1. Calculated compressive resistance for brickwork (according to SNiP II-22-81 (1995))

But that's not all. All the same SNiP II-22-81 (1995) clause 3.11 a) recommends that for the area of ​​pillars and piers less than 0.3 m 2, multiply the value of the design resistance by working conditions factor γ s =0.8. And since the cross-sectional area of ​​our column is 0.25x0.25 = 0.0625 m2, we will have to use this recommendation. As you can see, for M75 grade brick, even when using M100 masonry mortar, the strength of the masonry will not exceed 15 kgf/cm2. As a result, the calculated resistance for our column will be 15·0.8 = 12 kg/cm2, then the maximum compressive stress will be:

10300/625 = 16.48 kg/cm 2 > R = 12 kgf/cm 2

Thus, to ensure the required strength of the column, it is necessary either to use a brick of greater strength, for example M150 (the calculated compressive resistance for the M100 grade of mortar will be 22·0.8 = 17.6 kg/cm2) or to increase the cross-section of the column or to use transverse reinforcement of the masonry. For now, let's focus on using more durable facing bricks.

3. Determination of the stability of a brick column.

Brickwork strength and stability brick column- these are also different things and still the same SNiP II-22-81 (1995) recommends determining the stability of a brick column using the following formula:

N ≤ m g φRF (1.1)

Where m g- coefficient taking into account the influence of long-term load. In this case, we were, relatively speaking, lucky, since at the height of the section h≈ 30 cm, the value of this coefficient can be taken equal to 1.

Note: Actually, with the m g coefficient, everything is not so simple; details can be found in the comments to the article.

φ - buckling coefficient, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l 0 , and it does not always coincide with the height of the column. The subtleties of determining the design length of a structure are set out separately; here we only note that according to SNiP II-22-81 (1995) clause 4.3: “Calculated heights of walls and pillars l 0 when determining buckling coefficients φ depending on the conditions of supporting them on horizontal supports, the following should be taken:

a) with fixed hinged supports l 0 = N;

b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l 0 = 1.5H, for multi-span buildings l 0 = 1.25H;

c) for free-standing structures l 0 = 2H;

d) for structures with partially pinched supporting sections - taking into account the actual degree of pinching, but not less l 0 = 0.8N, Where N- the distance between floors or other horizontal supports, with reinforced concrete horizontal supports, the clear distance between them."

At first glance, our calculation scheme can be considered as satisfying the conditions of point b). i.e. you can take it l 0 = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only in the case when the lower support is really rigid. If a brick column is laid on a layer of roofing felt waterproofing laid on the foundation, then such a support should rather be considered as hinged rather than rigidly clamped. And in this case, our design in a plane parallel to the plane of the wall is geometrically variable, since the structure of the floor (separately lying boards) does not provide sufficient rigidity in the specified plane. There are 4 possible ways out of this situation:

1. Apply a fundamentally different design scheme

for example - metal columns, rigidly embedded in the foundation, to which the floor beams will be welded; then, for aesthetic reasons, the metal columns can be covered with facing bricks of any brand, since the entire load will be carried by the metal. In this case, it is true that the metal columns need to be calculated, but the calculated length can be taken l 0 = 1.25H.

2. Make another overlap,

for example, from sheet materials, which will allow us to consider both the upper and lower supports of the column as hinged, in this case l 0 = H.

3. Make a stiffening diaphragm

in a plane parallel to the plane of the wall. For example, along the edges, lay out not columns, but rather piers. This will also allow us to consider both the upper and lower supports of the column as hinged, but in this case it is necessary to additionally calculate the stiffness diaphragm.

4. Ignore the above options and calculate the columns as free-standing with a rigid bottom support, i.e. l 0 = 2H

In the end, the ancient Greeks erected their columns (though not made of brick) without any knowledge of the strength of materials, without the use of metal anchors, and there were no such carefully written building codes and regulations in those days, nevertheless, some columns stand and to this day.

Now, knowing the design length of the column, you can determine the flexibility coefficient:

λ h = l 0 /h (1.2) or

λ i = l 0 /i (1.3)

Where h- height or width of the column section, and i- radius of inertia.

Determining the radius of inertia is, in principle, not difficult; you need to divide the moment of inertia of the section by the cross-sectional area, and then take the square root from the result, but in this case there is no great need for this. Thus λ h = 2 300/25 = 24.

Now, knowing the value of the flexibility coefficient, you can finally determine the buckling coefficient from the table:

table 2. Buckling coefficients for stone and reinforced concrete stone structures(according to SNiP II-22-81 (1995))

In this case, the elastic characteristics of the masonry α determined by the table:

Table 3. Elastic characteristics of masonry α (according to SNiP II-22-81 (1995))

As a result, the value of the longitudinal bending coefficient will be about 0.6 (with the elastic characteristic value α = 1200, according to paragraph 6). Then the maximum load on the central column will be:

N р = m g φγ with RF = 1х0.6х0.8х22х625 = 6600 kg< N с об = 9400 кг

This means that the adopted cross-section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, it is most optimal to increase the cross-section of the column. For example, if you lay out a column with a void inside of one and a half bricks, measuring 0.38x0.38 m, then not only will the cross-sectional area of ​​the column increase to 0.13 m2 or 1300 cm2, but the radius of inertia of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the coefficient value φ = 0.8. In this case, the maximum load on the central column will be:

N r = m g φγ with RF = 1x0.8x0.8x22x1300 = 18304 kg > N with rev = 9400 kg

This means that a section of 38x38 cm is sufficient to ensure the stability of the lower central centrally compressed column and it is even possible to reduce the grade of brick. For example, with the initially adopted grade M75, the maximum load will be:

N r = m g φγ with RF = 1x0.8x0.8x12x1300 = 9984 kg > N with rev = 9400 kg

That seems to be all, but it is advisable to take into account one more detail. In this case, it is better to make the foundation strip (united for all three columns) rather than columnar (separately for each column), otherwise even small subsidence of the foundation will lead to additional stresses in the body of the column and this can lead to destruction. Taking into account all of the above, the most optimal section of the columns will be 0.51x0.51 m, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of ​​such columns will be 2601 cm2.

An example of calculating a brick column for stability under eccentric compression

The outer columns in the designed house will not be centrally compressed, since the crossbars will rest on them only on one side. And even if the crossbars are laid on the entire column, then still, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the outer columns not in the center of the column section. Where exactly the resultant of this load will be transmitted depends on the angle of inclination of the crossbars on the supports, the modulus of elasticity of the crossbars and columns and a number of other factors, which are discussed in detail in the article "Calculation of the support section of a beam for bearing". This displacement is called the eccentricity of the load application e o. In this case, we are interested in the most unfavorable combination of factors, in which the load from the floor to the columns will be transferred as close as possible to the edge of the column. This means that in addition to the load itself, the columns will also be subject to a bending moment equal to M = Ne o, and this point must be taken into account in the calculations. IN general case The stability test can be performed using the following formula:

N = φRF - MF/W (2.1)

Where W- section moment of resistance. In this case, the load for the lower outermost columns from the roof can be conditionally considered centrally applied, and eccentricity will only be created by the load from the floor. At eccentricity 20 cm

N р = φRF - MF/W =1x0.8x0.8x12x2601- 3000 20 2601· 6/51 3 = 19975, 68 - 7058.82 = 12916.9 kg >N cr = 5800 kg

Thus, even with a very large eccentricity of load application, we have a more than double safety margin.

Note: SNiP II-22-81 (1995) “Stone and reinforced masonry structures” recommends using a different method for calculating the section, taking into account the features of stone structures, but the result will be approximately the same, therefore I do not present the calculation method recommended by SNiP here.

III. CALCULATION OF STONE STRUCTURES

Load on the pier (Fig. 30) at the level of the bottom of the first floor floor beam, kN:

snow for II snow region

rolled roofing carpet – 100 N/m2

asphalt screed at N/m 3, 15 mm thick

insulation – wood fiber boards 80 mm thick with a density of N/m 3

vapor barrier – 50 N/m 2

prefabricated reinforced concrete slabs coating – 1750 N/m 2

reinforced concrete truss weight

weight of the cornice on the brickwork of the wall at N/m 3

the weight of the brickwork is above +3.03

concentrated from the floor crossbars (conditionally without taking into account the continuity of the crossbars)

weight of window filling at N/m 2

total design load on the pier at the level of elevation. +3.03

According to clauses 6.7.5 and 8.2.6, it is permissible to consider the wall as divided in height into single-span elements with the support hinges located 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 kN loads within a given floor are considered to be applied with actual eccentricity relative to the center of gravity of the wall section.

According to clause 6.9, clause 8.2.2, the distance from the point of application of the crossbar support reactions P to the inner edge of the wall, in the absence of supports fixing the position of the support pressure, no more than one third of the embedding depth of the crossbar and no more than 7 cm is taken (Fig. 31).

At the depth of embedding of the crossbar into the wall A h = 380 mm, A h: 3 = 380: 3 =

127 mm > 70 mm accept the point of application of the reference pressure

R= 346.5 kN at a distance of 70 mm from the inner edge of the wall.

Estimated height of the pier in the lower floor

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 a wall made of sand-lime brick of grade 100 on mortar of grade 25, with R= 1.3 MPa according to table. 2, is determined according to Note 1 to Table. 15 with elastic characteristics of the masonry a= 1000;

buckling coefficient according to table. 18 j = 0.96. According to clause 4.14, in walls with a rigid upper support, the longitudinal deflection in the supporting sections may not be taken into account (j = 1.0). In the middle third of the pier height, the buckling coefficient is equal to the calculated value j = 0.96. In the supporting thirds of the height j varies linearly from j = 1.0 to the calculated value j = 0.96 (Fig. 32). Values ​​of the longitudinal bending coefficient in the design sections of the pier, at the levels of the top and bottom of the window opening

Rice. 31

the magnitude of bending moments at the level of the crossbar support and in the design sections of the pier at the level of the top and bottom of the window opening

kNm;

kNm;

Fig.32

The magnitude of normal forces in the same sections of the pier

Eccentricities of longitudinal forces e 0 = M:N:

Mm< 0,45 y= 0.45 × 250 = 115 mm;

Mm< 0,45 y= 115 mm;

Mm< 0,45 y= 115 mm;

Load-bearing capacity of an eccentrically compressed pier rectangular section according to clause 4.7 is determined by the formula

Where (j is the longitudinal deflection coefficient for the entire cross-section of a rectangular element; ); m g– coefficient taking into account the influence of long-term load action (with h= 510 mm > 300 mm accept m g = 1,0); A– cross-sectional area of ​​the pier.