Wall stability calculation online. Strength calculation of brickwork

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. The larger the book format, the less stable it will be; on the other hand, the thicker the book, the better it will stand on its edge. The situation is the same with walls. The stability of the wall depends on the height and thickness.

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

What is needed to avoid 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. An external non-load-bearing wall is made of lightweight masonry made of M50 grade brick with M25 grade mortar. 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 (item 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 paragraph 6.19:

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

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

I quote Irina:

profiles will not replace reinforcement

I quote Irina:

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

The soils are not yet known, most likely it will be an open field of all sorts of loam, initially I 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. I don’t need a lot of bearing capacity of the soil - after all, the house was built on the 1st floor, and expanded clay concrete is not very heavy, freezing there is no more than 20 cm (although according to old Soviet standards it is 80).

I’m thinking of removing the top layer of 20-30 cm, laying out geotextiles, covering it with river sand and leveling it with compaction. Then a light preparatory screed - for leveling (it seems like they don’t even make reinforcement into it, although I’m not sure), waterproofing 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 we somehow replaced the foam plastic 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 moving the reinforcement into the beams - in general, I lack both theory and practical experience here.

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? As I understand it, the stronger the concrete and the larger the surface area 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

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 of the supporting platform of the floor slabs per masonry

Materials: thickened facing silicate brick (250Х120Ч88) GOST 379-95, grade SUL-125/25, porous silicate stone (250Ч120Ч138) GOST 379-95, grade SRP -150/25 and thickened hollow silicate brick (250х120х88) GOST 379 -95 brand SURP-150/25. For masonry of 1-5 floors, cement-sand mortar M75 is used, 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

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

Table 2.15. Collection of loads on the interfloor ceiling, 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 current 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

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 wall configuration, compressive strength, strength of this type materials and much more. However, approximately, “by eye,” you can estimate the wall’s resistance to compression, using indicative tables in which the strength (in tons) is linked to the width of the wall, as well as brands of brick and mortar. The table is compiled for a wall height of 2.8 m.

Table of brick wall strength, tons (example)

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

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

Permanent:

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

Temporary:

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

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

Brickwork load

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

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.

Load calculation example

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

Initial data for analysis:

• number of floors – 4 floors;
• brick wall thickness T=64cm (0.64 m);
• specific gravity of masonry (brick, mortar, plaster) M = 18 kN/m3 (indicator taken from reference data, table 19<1>);
• the width of the window openings is: W1=1.5 m;
• height of window openings - B1=3 m;
• pier section 0.64*1.42 m (loaded area where the weight of the overlying structural elements is applied);
• floor height Wet=4.2 m (4200 mm):
• the pressure is distributed at an angle of 45 degrees.

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

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

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

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

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

H2=215*6 = 1.290MN,

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

1. Own weight of the walls

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

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

Scheme of load and structural strength analysis

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

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

1. Wall area (P)

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

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

where e0 is an indicator of extraness.

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

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

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

Gszh=Vet/Vszh

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

Fsr=(f+fszh)/2

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

ω =1+e/T<1,45

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

U=Kdv*fsr*R*Pszh* ω

Kdv – long-term exposure coefficient

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

1. Reconciliation

An example of calculating the strength of masonry

— Wet — 3.3 m

— Chat — 2

— T — 640 mm

— W — 1300 mm

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

1. Area (P)

P=0.64*1.3=0.832

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

1. Flexibility (G)

G =3.3/0.64=5.156

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

1. Height of compressed part

Vszh=0.64-2*0.045=0.55 m

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

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

1. Flexibility of the compressed part

Gszh=3.3/0.55=6

1. fsj=0.96
2. FSR calculation

Fsr=(0.98+0.96)/2=0.97

1. According to the table 19<1>

ω =1+0.045/0.64=1.07<1,45

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

1. Definition of sustainability

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

1. Reconciliation

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

Insufficient wall resistance

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

For convenience, you can use tabular data.

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

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

Below is an example of such a calculation

Example of calculation of pier reinforcement

Initial data - see previous example.

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

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

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

Gain

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

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

Reinforcement with reinforced concrete frame

Reinforcement is carried out using a B15 concrete frame with a thickness of 0.060 m. Vertical rods 0.340 m2, clamps 0.0283 m2 with a pitch of 0.150 m.

Section dimensions of the reinforced structure:

Ш_1=1300+2*60=1.42

T_1=640+2*60=0.76

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

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 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 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.

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 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 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.