An example of calculating a brick wall for stability. Calculation of brickwork for strength. Cross-sectional dimensions of the reinforced structure

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

In this case, a natural question arises: what is the minimum section of the columns that will provide the required strength and stability? Of course, the idea of ​​laying clay brick columns, and even more so the walls of the house, is far from new, and all possible aspects of the calculations of brick walls, walls, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced masonry structures". It is this normative document that should be followed in the calculations. The calculation below is nothing more than an example of using the specified SNiP.

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

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

Designed:

Terrace with dimensions of 5x8 m. Three columns (one in the middle and two along 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 brick strength grade is M75.

Design assumptions:

With such a design scheme, the maximum load will be on the middle lower column. It is she who should be counted on strength. The load on the column depends on many factors, in particular on the area of ​​construction. For example, in St. Petersburg it is 180 kg / m 2, and in Rostov-on-Don - 80 kg / m 2. Taking into account the weight of the roof itself 50-75 kg / m 2, 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 actual loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but the reinforced concrete slab is not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculating the load from the terrace you can take a uniformly distributed load of 600 kg / m 2, then the concentrated force from the terrace, acting on the central column, will be:

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

The own weight of columns 3 m long will be:

N per 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 about \u003d 3000 + 6000 + 2 650 \u003d 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, which is maximum in winter, and the temporary load on the ceiling, which is maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:

N with about \u003d (3000 + 6000) 0.9 + 2 650 \u003d 9400 kg or 9.4 tons

The calculated 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 brand of brick M75 means that the brick must withstand a load of 75 kgf / cm 2, 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 strength for brickwork (according to SNiP II-22-81 (1995))

But that's not all. All the same SNiP II-22-81 (1995) p.3.11 a) recommends that if the area of ​​​​pillars and piers is less than 0.3 m 2, multiply the value of the design resistance by working conditions coefficient γ s =0.8. And since the cross-sectional area of ​​​​our column is 0.25x0.25 \u003d 0.0625 m 2, we will have to use this recommendation. As you can see, for a brick of the M75 brand, even when using the M100 masonry mortar, the masonry strength will not exceed 15 kgf / cm 2. As a result, the calculated resistance for our column will be 15 0.8 = 12 kg / cm 2, then the maximum compressive stress will be:

10300/625 \u003d 16.48 kg / cm 2\u003e R \u003d 12 kgf / cm 2

Thus, to ensure the necessary strength of the column, it is necessary either to use a brick of greater strength, for example, M150 (the calculated compressive strength with a brand of mortar M100 will be 22 0.8 = 17.6 kg / cm 2) or increase the section of the column or use transverse reinforcement of the masonry. For now, let's focus on using a more durable face brick.

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 all 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, relatively speaking, we are 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 coefficient m g, everything is not so simple, the details can be found in the comments to the article.

φ - coefficient of buckling, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l 0 , but it does not always coincide with the height of the column. The subtleties of determining the estimated length of the structure are set out separately, here we just note that according to SNiP II-22-81 (1995) p. 4.3: "The estimated heights of walls and pillars l 0 when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, one should take:

a) with fixed hinged supports l 0 = H;

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 = 2N;

d) for structures with partially pinched support sections - taking into account the actual degree of pinching, but not less than l 0 = 0.8N, where H- the distance between ceilings or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light.

At first glance, our calculation scheme can be considered as satisfying the conditions of paragraph b). i.e. you can take l 0 = 1.25H = 1.25 3 = 3.75 meters or 375 cm. However, we can confidently use this value only if the lower support is really rigid. If a brick column will be laid out on a roofing material waterproofing layer laid on a foundation, then such a support should rather be considered as hinged, and not rigidly clamped. And in this case, our structure 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 this plane. There are 4 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 crossbars will be welded, then, for aesthetic reasons, the metal columns can be overlaid with any brand of face brick, since the metal will carry the entire load. In this case, it is true that metal columns need to be calculated, but the estimated length can be taken l 0=1.25H.

2. Make another cover,

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

3. Make a hardness 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 column supports as hinged ones, but in this case it is necessary to additionally calculate the stiffness diaphragm.

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

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

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

λ h =l 0 /h (1.2) or

λ i =l 0 /i (1.3)

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

In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the area of ​​​​the section, and then extract the square root from the result, but in this case this is not very necessary. Thus λh = 2 300/25 = 24.

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

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

At the same time, the elastic characteristic of the masonry α determined by the table:

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

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

N p \u003d m g φγ with RF \u003d 1x0.6x0.8x22x625 \u003d 6600 kg< N с об = 9400 кг

This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then in this way not only the cross-sectional area of ​​\u200b\u200bthe column will increase to 0.13 m 2 or 1300 cm 2, but the radius of gyration of the column will also increase to i= 11.45 cm. Then λ i = 600/11.45 = 52.4, and the value of the coefficient φ = 0.8. In this case, the maximum load on the central column will be:

N p \u003d m g φγ with RF \u003d 1x0.8x0.8x22x1300 \u003d 18304 kg\u003e N with about \u003d 9400 kg

This means that a section of 38x38 cm is enough to ensure the stability of the lower central centrally compressed column with a margin, and even the brand of brick can be reduced. For example, with the originally adopted brand M75, the ultimate load will be:

N p \u003d m g φγ with RF \u003d 1x0.8x0.8x12x1300 \u003d 9984 kg\u003e N with about \u003d 9400 kg

It seems to be everything, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not 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 section of columns 0.51x0.51 m will be the most optimal, and from an aesthetic point of view, such a section is optimal. The cross-sectional area of ​​such columns will be 2601 cm 2.

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

The extreme 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 all the same, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the extreme 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, which are discussed in detail in the article " Calculation of the support section of the beam for collapse". This displacement is called the load application eccentricity e o. In this case, we are interested in the most unfavorable combination of factors, in which the floor load on 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 bending moment will also act on the columns, equal to M = Ne o, and this moment must be taken into account in the calculations. In general, stability testing can be performed using the following formula:

N = φRF - MF/W (2.1)

where W- section modulus. In this case, the load for the lower extreme columns from the roof can be conditionally considered to be centrally applied, and the eccentricity will be created only by the load from the ceiling. With an eccentricity of 20 cm

N p \u003d φRF - MF / W \u003d1x0.8x0.8x12x2601- 3000 20 2601· 6/51 3 = 19975, 68 - 7058.82 = 12916.9 kg >N cr = 5800 kg

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

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 give the calculation method recommended by SNiP here.

Brick is a fairly strong building material, especially solid, and when building houses of 2-3 floors, walls made of ordinary ceramic bricks usually do not need additional calculations. Nevertheless, the situations are different, for example, a two-story house with a terrace on the second floor is planned. The metal crossbars, on which the metal beams of the terrace floor will also rest, are planned to be supported on brick columns made of face hollow brick 3 meters high, there will be more columns 3 meters high, on which the roof will rest:

In this case, a natural question arises: what is the minimum section of the columns that will provide the required strength and stability? Of course, the idea of ​​laying clay brick columns, and even more so the walls of the house, is far from new, and all possible aspects of the calculations of brick walls, walls, pillars, which are the essence of the column, are set out in sufficient detail in SNiP II-22-81 (1995) "Stone and reinforced masonry structures". It is this normative document that should be followed in the calculations. The calculation below is nothing more than an example of using the specified SNiP.

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

with central compression

Designed: Terrace with dimensions of 5x8 m. Three columns (one in the middle and two along 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 brick strength grade is M75.

With such a design scheme, the maximum load will be on the middle lower column. It is she who should be counted on strength. The load on the column depends on many factors, in particular on the area of ​​construction. For example, the snow load on the roof 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 actual loads from the floor material and from people sitting on the terrace, furniture, etc. are not yet known, but the reinforced concrete slab is not exactly planned, but it is assumed that the floor will be wooden, from separately lying edged boards, then for calculating the load from the terrace it is possible to 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 the terrace = 600 5 8/4 = 6000 kg or 6 tons

The own weight of columns 3 m long will be:

N from the column \u003d 1500 3 0.38 0.38 \u003d 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 about \u003d 3000 + 6000 + 2 650 \u003d 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, which is maximum in winter, and the temporary load on the ceiling, which is maximum in summer, will be applied simultaneously. Those. the sum of these loads can be multiplied by a probability factor of 0.9, then:

N with about \u003d (3000 + 6000) 0.9 + 2 650 \u003d 9400 kg or 9.4 tons

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

N kr \u003d 1500 + 3000 + 1300 \u003d 5800 kg or 5.8 tons

2. Determination of the strength of brickwork.

The brand of brick M75 means that the brick must withstand a load of 75 kgf / cm & sup2, 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 strengths for masonry

But that's not all. All the same SNiP II-22-81 (1995) p. 3.11 a) recommends that if the area of ​​​​pillars and piers is less than 0.3 m2, multiply the value of the design resistance by the coefficient of working conditions γ c \u003d 0.8. And since the cross-sectional area of ​​​​our column is 0.25x0.25 \u003d 0.0625 m & sup2, we will have to use this recommendation. As you can see, for a brick of the M75 brand, even when using the M100 masonry mortar, the strength of the masonry will not exceed 15 kgf / cm². As a result, the design resistance for our column will be 15 0.8 = 12 kg / cm & sup2, then the maximum compressive stress will be:

10300/625 = 16.48 kg/cm² > R = 12 kgf/cm²

Thus, to ensure the necessary strength of the column, it is necessary either to use a brick of greater strength, for example, M150 (the calculated compressive strength with a brand of mortar M100 will be 22 0.8 = 17.6 kg / cm & sup2) or increase the section of the column or use transverse reinforcement of the masonry. For now, let's focus on using a more durable face brick.

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 all 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, relatively speaking, we are lucky, since at the height of the section h≤ 30 cm, the value of this coefficient can be taken equal to 1.

φ - coefficient of buckling, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column l o, but it does not always coincide with the height of the column. The subtleties of determining the estimated length of the structure are not set out here, we only note that according to SNiP II-22-81 (1995) p. 4.3: "The estimated heights of walls and pillars l o when determining the coefficients of buckling φ depending on the conditions of their support on horizontal supports, one should take:

a) with fixed hinged supports l o = H;

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 support sections - taking into account the actual degree of pinching, but not less than l o = 0.8N, where H- the distance between ceilings or other horizontal supports, with reinforced concrete horizontal supports, the distance between them in the light.

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

1. Apply a fundamentally different design scheme, for example - metal columns rigidly embedded in the foundation, to which the crossbars of the floor will be welded, then, for aesthetic reasons, the metal columns can be overlaid with a face brick of any brand, since the metal will carry the entire load. In this case, it is true that metal columns need to be calculated, but the estimated length can be taken l o = 1.25H.

2. Make another cover, for example, from sheet materials, which will allow us to consider both the upper and lower support of the column as hinged, in this case l o=H.

3. Make a hardness 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 column supports as hinged ones, but in this case it is necessary to additionally calculate the stiffness diaphragm.

4. Ignore the above options and count the columns as free-standing with a rigid bottom support, i.e. l o = 2H. In the end, the ancient Greeks put up their columns (though not of brick) without any knowledge of the resistance of materials, without the use of metal anchors, and there were no such carefully written building codes in those days, nevertheless, some columns stand and to this day.

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

λ h =l o /h (1.2) or

λ i =l o (1.3)

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

In principle, it is not difficult to determine the radius of gyration, you need to divide the moment of inertia of the section by the area of ​​​​the section, and then extract the square root from the result, but in this case this is not very necessary. Thus λh = 2 300/25 = 24.

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

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

At the same time, the elastic characteristic of the masonry α determined by the table:

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

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

N p \u003d m g φγ with RF \u003d 1 0.6 0.8 22 625 \u003d 6600 kg< N с об = 9400 кг

This means that the accepted section of 25x25 cm is not enough to ensure the stability of the lower central centrally compressed column. To increase stability, the most optimal would be to increase the section of the column. For example, if you lay out a column with a void inside one and a half bricks, with dimensions of 0.38x0.38 m, then in this way not only the cross-sectional area of ​​\u200b\u200bthe column will increase to 0.13 m2 or 1300 cm2, but the radius of gyration of the column will also increase to i= 11.45 cm. Then λi = 600/11.45 = 52.4, and the value of the coefficient φ = 0.8. In this case, the maximum load on the central column will be:

N p = m g φγ with RF = 1 0.8 0.8 22 1300 = 18304 kg > N with about = 9400 kg

This means that a section of 38x38 cm is enough to ensure the stability of the lower central centrally compressed column with a margin, and even the brand of brick can be reduced. For example, with the originally adopted brand M75, the ultimate load will be:

N p \u003d m g φγ with RF \u003d 1 0.8 0.8 12 1300 \u003d 9984 kg\u003e N with about \u003d 9400 kg

It seems to be everything, but it is desirable to take into account one more detail. In this case, it is better to make the foundation tape (single for all three columns), and not 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 section of columns 0.51x0.51 m will be the most optimal, 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 extreme 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 all the same, due to the deflection of the crossbars, the load from the floor and roof will be transferred to the extreme columns not in the center of the column section. Where exactly the resultant of this load will be transferred 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 load application eccentricity e o. In this case, we are interested in the most unfavorable combination of factors, in which the floor load on 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 bending moment will also act on the columns, equal to M = Ne o, and this moment must be taken into account in the calculations. In general, stability testing can be performed using the following formula:

N = φRF - MF/W (2.1)

W- section modulus. In this case, the load for the lower extreme columns from the roof can be conditionally considered to be centrally applied, and the eccentricity will be created only by the load from the ceiling. With an eccentricity of 20 cm

N p \u003d φRF - MF / W \u003d1 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 load application eccentricity, we have more than a double margin of safety.

Note: SNiP II-22-81 (1995) "Stone and reinforced stone 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.

Exterior load-bearing walls should, 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 consider the calculation of the bearing capacity of brickwork, and in the following articles - the rest of the 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 for this category.

carriers walls are called that perceive the load from floor slabs, coatings, beams, etc. resting on them.

You should also take into account the brand of brick for frost resistance. Since everyone builds a house for himself, at least for a hundred years, then with a dry and normal humidity regime of 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, with a thermal engineering calculation, the thickness of the insulation will turn out to be less, which will save money when buying it. Solid brick for external walls should be used only if it is necessary to ensure the strength of the masonry.

Reinforcement of masonry allowed only in the case when the increase in the grade of brick and mortar does not allow to provide the required bearing capacity.

An example of the calculation of a brick wall.

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

Choice of design section.

In blank walls, the section I-I at the level of the bottom of the floor with the longitudinal force N and the maximum bending moment M is taken as the calculated one. 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 section is taken at the level of the bottom of the lintels.

Let's look at the section I-I.

From a previous article Collection of loads on the wall of the first floor we take the obtained value of the total load, which includes the loads from the floor of the first floor P 1 \u003d 1.8t and the overlying floors G \u003d G P + P 2 +G 2 = 3.7t:

N \u003d G + P 1 \u003d 3.7t + 1.8t \u003d 5.5t

The floor slab rests on the wall at a distance a=150mm. The longitudinal force P 1 from the overlap 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 just 1/3 of the support length.

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

Since the load from the floor slab (P 1) is not applied in 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 on the shoulder.

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

Then the eccentricity of the longitudinal force N will be:

e 0 \u003d M / N \u003d 13.5 / 5.5 \u003d 2.5 cm

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

e 0 \u003d 2.5 + 2 \u003d 4.5 cm

y=h/2=12.5cm

When e 0 \u003d 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.

III. CALCULATION OF STONE STRUCTURES

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

snow for II snow region

rolled roof carpet - 100 N / m 2

asphalt screed at N/m 3 15 mm thick

insulation - wood fiber boards with a thickness of 80 mm at a density of N / m 3

vapor barrier - 50 N / m 2

prefabricated reinforced concrete floor slabs - 1750 N / m 2

concrete truss weight

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

brickwork weight above +3.03

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

window filling weight at N/m2

total calculated load on the partition at the level of elev. +3.03

According to paragraphs 6.7.5 and 8.2.6, it is allowed to consider the wall dissected in height into single-span elements with the location of the support hinges at the level of 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 the given floor are considered to be applied with the 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 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, no more than one third of the depth of the embedding of the crossbar and no more than 7 cm is taken (Fig. 31).

With a depth of embedding the crossbar into the wall a h = 380 mm, a h: 3 = 380: 3 =

127 mm > 70 mm accept the reference pressure application point

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

Estimated height of the wall in the lower floor

For the design scheme of the pier of the lower floor of the building, we take a rack with pinching at the level of the edge of the foundation and with hinged support at the level of the ceiling.

The flexibility of a wall made of silicate brick grade 100 on a mortar grade 25, at R= 1.3 MPa according to the table. 2 is determined according to note 1 to table. 15 with elastic characteristic of 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 wall height, the buckling coefficient is equal to the calculated value j = 0.96. In the reference thirds of the height, j changes linearly from j = 1.0 to the calculated value j = 0.96 (Fig. 32). The values ​​of the buckling coefficient in the design sections of the partition, in the levels of the top and bottom of the window opening

values ​​of bending moments in the level of support of the crossbar and in the design sections of the wall at the level of the top and bottom of the window opening

kNm;

kNm;

The value 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;

The bearing capacity of an eccentrically compressed rectangular section wall according to clause 4.7 is determined by the formula

where (j- coefficient of longitudinal deflection for the entire section of a rectangular element; ); m g- coefficient taking into account the effect of long-term load (at h= 510mm > 300mm accept m g = 1,0); BUT- cross-sectional area of ​​the wall.

Let's check the strength of the brick wall of the bearing wall of a residential building with a variable number of storeys in the city of Vologda.

Initial data:

Floor height - Net=2.8 m;

Number of floors - 8 floors;

The pitch of the bearing walls is a = 6.3 m;

Dimensions of the window opening - 1.5x1.8 m;

The cross-sectional dimensions of the pier are -1.53x0.68 m;

The thickness of the inner verst is 0.51 m;

Sectional area of ​​the wall-A=1.04m 2 ;

The length of the supporting platform of floor slabs per masonry

Materials: silicate brick thickened front (250CH120CH88) GOST 379-95, grade SUL-125/25, silicate porous stone (250CH120CH138) GOST 379-95, grade SRP -150/25 and hollow silicate brick thickened (250x120x88) GOST 379-95 brand SURP-150/25. For laying 1-5 floors, cement-sand mortar M75 is used, for 6-8 floors, masonry density \u003d 1800 kg / m 3, multilayer masonry, insulation - expanded polystyrene brand PSB-S-35 n \u003d 35 kg / m3 (GOST 15588- 86). With multi-layer masonry, the load will be transferred to the inner verst of the outer wall, therefore, when calculating the thickness of the outer verst and insulation, we do not take into account.

Collection of load from pavement and floors is presented in tables 2.13, 2.14, 2.15. The design wall is shown in fig. 2.5.

Figure 2.12. Settlement wall: a - plan; b - vertical section of the wall; c-calculation scheme; d - plot of moments

Table 2.13. Collection of loads on the coating, kN / m 2

Table 2.14. Collection of loads on the attic floor, kN/m2

Table 2.15. Collection of loads on the interfloor overlap, kN/m2

Table 2.16. Collection of loads per 1 r.m. from the outer wall t=680 mm, kN/m2

We determine the width of the cargo area according to the formula 2.12

where b is the distance between the center axes, m;

a - the value of the support of the floor slab, m.

The length of the loading area of ​​the pier is determined by formula (2.13).

where l is the width of the partition;

l f - width of window openings, m.

The determination of the cargo area (according to Figure 2.6) is carried out according to the formula (2.14)

Figure 2.13. Scheme for determining the cargo area of ​​the pier

The calculation of the force N on the wall from the higher floors at the level of the bottom of the floors of the first floor is based on the cargo area and the existing loads on the floors, coatings and roofs, the load from the weight of the outer wall.

Table 2.17. Collection of loads, kN/m

Load name	Design value kN/m
1. Coating design
2. Attic floor
3. Interfloor overlap
4. Outer wall t=680 mm

The calculation of eccentrically compressed unreinforced elements of stone structures should be made according to the formula 13