Calculation of a brick column for strength and stability. Calculation of the strength of brickwork in the pier Calculation of the strength of a load-bearing brick wall

Let's check the strength of the brick pier of the load-bearing wall of a residential building of variable number of floors in Vologda.

Initial data:

Floor height - Net=2.8 m;

Number of floors - 8 floors;

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

The dimensions of the window opening are 1.5x1.8 m;

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

The thickness of the inner mile is 0.51 m;

Cross-sectional area of ​​the pier-A=1.04m2;

Length support platform floor slabs for masonry

Materials: thickened facing silicate brick (250Ch120Ch88) GOST 379-95, grade SUL-125/25, porous silicate stone (250Ch120Ch138) GOST 379-95, grade SRP -150/25 and thickened hollow silicate brick (250x120x88) GOST 379 -95 brand SURP-150/25. Used for laying 1-5 floors cement-sand mortar M75, for 6-8 floors, masonry density = 1800 kg/m3, multilayer masonry, insulation - polystyrene foam brand PSB-S-35 n = 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.

Load collection from pavement and floors is presented in tables 2.13, 2.14, 2.15. The calculated pier is shown in Fig. 2.5.

Figure 2.12. Design pier: a - plan; b - vertical section of the wall; c-calculation scheme; d - diagram of moments

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

Load name	Standard value kN/m2		Design value kN/m2
Constant: 1. Layer of linocrom TKP, t=3.7 mm, weight of 1 m2 of material 4.6 kg/m2, =1100 kg/m3 2. Layer of linocrom KhPP, t=2.7 mm weight of 1 m2 of material 3.6 kg/m2, =1100 kg/m3 3. Primer “Bitumen Primer”
4. Cement-sand screed, t=40 mm, =1800 kg/m3 5. Expanded clay gravel, t=180 mm, =600 kg/m3, 6. Insulation - polystyrene foam PSB-S-35, t=200 mm, =35 kg/m3 7. Paroisol 8. Reinforced concrete floor slab
Temporary: S0н =0.7ХSqмЧСeЧСt= 0.7Ч2.4 1Ч1Ч1

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

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

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

Let us determine the width of the cargo area using formula 2.12

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

a is the amount of support for the floor slab, m.

The length of the cargo area of ​​the partition is determined by formula (2.13).

where l is the width of the wall;

l f - width window openings, m.

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

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

We calculate the force N on the partition from the upper floors at the level of the bottom of the floors of the first floor based on the load area and the effective loads on the floors, coverings and roof, and the load from the weight of the outer wall.

Table 2.17. Load collection, kN/m

Load name	Design value kN/m
1. Cover design
2. Attic floor
3. Interfloor covering
4. Exterior wall t=680 mm

Calculation of eccentrically compressed unreinforced elements of masonry structures should be carried out according to formula 13

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 of bearing capacity brickwork, and in the following articles - the remaining calculations. So as not to miss the exit new article, subscribe to the newsletter and you will find out what the thickness of the wall should be after all the calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, we will consider all calculations specifically for this category.

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

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

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 cash when purchasing it. Solid brick For external walls it should be used only when it is necessary to ensure the strength of the masonry.

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

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 under the support section will be in the form of a triangle, and the center of gravity of the triangle is located at 1/3 of the length of the support.

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

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

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

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

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

Then the eccentricity of the longitudinal force N will be:

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

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

e 0 = 2.5 + 2 = 4.5 cm

y=h/2=12.5cm

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

Strength of masonry eccentrically compressed element determined by the formula:

N ≤ m g φ 1 R A c ω

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

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 calculating the load-bearing capacity of brickwork, and in subsequent articles we will look at other 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 bear the load from floor slabs, coverings, beams, etc. resting on them.

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

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.

An example of calculating a brick wall.

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

Selection of design section.

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

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

Let's look at section I-I.

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

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

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

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

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

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

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

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

Then the eccentricity of the longitudinal force N will be:

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

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

e 0 = 2.5 + 2 = 4.5 cm

y=h/2=12.5cm

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

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.

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

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

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

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

4. Partitions are interior walls less than 6 meters high, supporting only the load from its own weight.

Let's look at the issue of wall stability.

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

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

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

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

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

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

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

k 3 = 0.9 - for partitions with openings;

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

Finally β = 1.26*9.8 = 12.3.

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

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

Example 2. Dana external curtain wall made of lightweight brick masonry grade M50 on mortar grade M25. Wall height 3 m, thickness 0.38 m, wall length 6 m. Wall with two windows measuring 1.2x1.2 m. It is necessary to determine the stability of the wall.

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

We find the coefficients k from tables 29:

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

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

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

Finally β = 0.94*15.4 = 14.5.

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

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

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

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Comments

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

I quote Irina:

profiles will not replace reinforcement

I quote Irina:

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

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

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

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

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

0 #214 Irina 02.22.2018 16:21

Quote:

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

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

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:

A natural question arises: what is the minimum cross-section of columns that will provide the required strength and stability? Of course, the idea is to lay out columns from clay brick, and especially 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 masonry structures”. This is exactly what normative document and 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:

with 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.

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 load on the column from the roof for Pushkin Leningrad region may amount to:

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

Because effective loads from the flooring material and from people sitting on the terrace, furniture, etc. are not yet known, but reinforced concrete slab It’s not exactly planned, but it is assumed that the ceiling will be wooden, from separate edged boards, then for calculating the load from the terrace we can assume uniformly distributed load 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. Eventually design 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 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. Definition of sustainability 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 - coefficient longitudinal bending, depending on the flexibility of the column λ . To determine this coefficient, you need to know the estimated length of the column 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: “Calculated 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 is in the plane, parallel to the plane wall, is geometrically variable, since the design 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 completely different design diagram , 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 resistance of materials, without the use of metal anchors, and even so carefully written building codes and there were no rules in those days, however, some columns still stand 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 masonry and reinforced masonry 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.

An example of calculating a brick column for stability
with 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 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.