Brick is quite durable construction material, especially solid, and when building houses of 2-3 floors, walls made of ordinary ceramic bricks As a rule, additional calculations are not needed. Nevertheless, situations are different, for example, it is planned two-storey house with a terrace on the second floor. Metal crossbars on which they will also rest metal beams terrace ceilings, it is planned to rest on brick columns made of facing hollow bricks 3 meters high; above there will be more columns 3 m high, on which the roof will rest:
This raises a natural question: what minimum section will the columns 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 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: the brand of brick in terms of strength, the area of support of the crossbars on the columns, the load on the columns, the 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:
Designed: Terrace dimensions 5x8 m. Three columns (one in the middle and two at the edges) made of facing hollow brick with a cross-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.
With such a calculation scheme maximum load will be on the middle bottom 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, snow load for roofing in St. Petersburg 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, but it is assumed that the floor will be wooden, from separately lying edged boards, then to calculate the load from the terrace, you can take 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 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 time, and the temporary load on the floor, maximum in summer time, 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. Design compressive strengths for brickwork
But that's not 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 m², multiply the value of the design resistance by the operating conditions coefficient 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 brand brick, even when using masonry mortar M100, 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/cm² > R = 12 kgf/cm²
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 mortar grade 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.
The strength of brickwork and the stability of a brick column 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)
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.
φ - longitudinal bending coefficient, depending on the flexibility of the column λ . To determine this coefficient, you need to know effective length columns l o, and it does not always coincide with the height of the column. The subtleties of determining the design length of a structure are not outlined here, we only note that according to SNiP II-22-81 (1995) clause 4.3: “Calculation heights of walls and pillars l o 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 o = N;
b) with an elastic upper support and rigid pinching in the lower support: for single-span buildings l o = 1.5H, for multi-span buildings l o = 1.25H;
c) for free-standing structures l o = 2H;
d) for structures with partially pinched supporting sections - taking into account the actual degree of pinching, but not less l o = 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 o = 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 floor structure (separately lying boards) does not provide sufficient rigidity in the specified plane. From similar situation 4 outputs are possible:
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 brick 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 o = 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 o = 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 o = 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 o /h (1.2) or
λ i = l o (1.3)
h- height or width of the column section, and i- radius of inertia.
Determining the radius of gyration is not difficult in principle; you need to divide the moment of inertia of the section by the cross-sectional area, and then extract from the result Square root, however, 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 = 1 0.6 0.8 22 625 = 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 0.38 x 0.38 m, then not only will the cross-sectional area of the column increase to 0.13 m or 1300 cm, 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 р = m g φγ with RF = 1 0.8 0.8 22 1300 = 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 р = m g φγ with RF = 1 0.8 0.8 12 1300 = 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.
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 elastic moduli of the crossbars and columns and a number of other factors. 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 when calculating. In general, stability testing can be performed using the following formula:
N = φRF - MF/W (2.1)
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 =1 0.8 0.8 12 2601- 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 the calculation method recommended by SNiP is not given 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 the wall made of sand-lime brick grade 100 on a solution 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
 |
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;
 |
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- longitudinal deflection coefficient for the entire section of the element rectangular shape; 
); 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.
External load-bearing walls must, at a minimum, be designed for strength, stability, local collapse and resistance to heat transfer. To find out how thick should a brick wall be? , 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. In order not to miss the release of a 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 take 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, at least for a hundred years, then with dry and normal humidity conditions in the premises, a grade (M rz) of 25 and above is accepted.
When building 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 money when purchasing it. Solid bricks for external walls 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.
Calculation example 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 is under support area will be in the form of a triangle, and the center of gravity of the triangle is exactly 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
Because bearing wall 25 cm thick, then 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.
The strength of the masonry of an eccentrically compressed element is 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.
When independent design brick house there is an urgent need to calculate whether it can withstand brickwork those 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 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:
Temporary:
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.
To take into account the force acting on the designed section of the wall, you need to sum up the loads:
When low-rise construction the task is greatly simplified, and many factors of live 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.
This example shows the analysis of the current loads on the piers of the 1st floor. Here only permanent loads 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:
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
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).
To calculate the pier of a brick wall you will need:
where e0 is an indicator of extraness.
Pszh = P*(1-2 e0/T)
Gszh=Vet/Vszh
Fsr=(f+fszh)/2
ω =1+e/T<1,45
U=Kdv*fsr*R*Pszh* ω
Kdv – long-term exposure coefficient
R – masonry compression resistance, can be determined from Table 2<1>, in MPa
— 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)
P=0.64*1.3=0.832
G =3.3/0.64=5.156
Vszh=0.64-2*0.045=0.55 m
Pszh = 0.832*(1-2*0.045/0.64)=0.715
Gszh=3.3/0.55=6
Fsr=(0.98+0.96)/2=0.97
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.
Y=1*0.97*1.5*0.715*1.07=1.113 MN
The condition is met, the strength of the masonry and the strength of its elements are sufficient
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.
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.
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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 |
|
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 mortar grade 25, 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 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 |
|||||
|
|
|
|
Reinforcement class |
|||||||
|
In the lower zone |
In the extreme spans |
123,80*10 |
in two flat frames |
|||||||
|
On medium spans |
94,83*10 |
in two flat frames |
||||||||
|
In the upper zone |
In the second flight |
52,80*10 |
in two frames |
|||||||
|
In all medium spans |
41,73*10 |
in two frames |
||||||||
|
On a support |
108,38*10 |
in one U-shaped mesh |
||||||||
|
On a supportC |
94,83*10 |
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 |
|
|
A 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 |
|
|||||||||||||
|
| ||||||||||||||
|
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 |
|||||||||||||
|
Ordinates of the additional diagram at | ||||||||||||||
|
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 |
|||||||||
|
Until 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 |
|
| ||||||||
, mm
, mm
, A s =760mm 2
, A s =628mm 2
, A s =308mm 2
, A s =226mm 2
, A s =628mm 2
, A s =628mm 2
x10 
by the amount
M 
=145.2 kNm
by the amount
M 
=89.2 kNm
