House, design, repair, decor. Yard and garden. Do it yourself

III. Calculation of stone structures

Load for simpleness (Fig. 30) in the level of regeneration of the first floor overlap, KN:

snow for the II Snow Area

rolled roofing carpet - 100 N / m 2

asphalt screed with n / m 3 Thickness 15 mm

insulation - Wood-fibrous plates with a thickness of 80 mm with a density of n / m 3

parosolation - 50 N / m 2

precast concrete coating plates - 1750 N / m 2

weight of reinforced concrete farm

the weight of the cornice on the brick masonry wall at N / m 3

weight brick masonry Above marks +3.03

focused on the beelels of overlaps (conditionally excluding the continues of the riglels)

window filling weight with N / m 2

the total calculation load on the simpleness in the level of Options. +3.03

According to clause 6.7.5 and 8.2.6, it is allowed to be considered a wall dismantled in altitude to single-span elements with the location of the support joints in the level of the riglels. In this case, the load from the upper floors is taken applied in the center of severity section of the wall of the overlying floor, and all the loads of the KN within this floor are considered applied with the actual eccentricity relative to the center of severity of the wall section.

According to clause 6.9, paragraph 8.2.2, the distance from the point of application of the reaction reactions of Rigel P. Prior to the inner edge of the wall in the absence of supports, fixing the position of the reference pressure, no more than one third of the depth of the reigleel seal and not more than 7 cm (Fig. 31) is taken.

With the depth of sealing the riglel into the wall but z \u003d 380 mm, but Z: 3 \u003d 380: 3 \u003d

127 mm\u003e 70 mm We accept the point of reference pressure

R \u003d 346.5 kN at a distance of 70 mm from the inner edge of the wall.

The estimated height of the simplest in the lower floor

For the design scheme for simpleness of the lower floor of the building we take a rack with a pinch in the level of the edge of the foundation and with a hinewing in the overlap level.

The flexibility of the selence made of silicate brick brand 100 on the brand 25 solution, R. \u003d 1.3 MPa in Table. 2 is determined according to a note 1 to the table. 15 with an elastic masonry characteristic A \u003d 1000;

coefficient longitudinal bend Table. 18 j \u003d 0.96. In accordance with clause 4.14 in the walls with a rigid upper support, the longitudinal deflection in the reference sections may not be taken into account (J \u003d 1.0). In the middle third of the height of the seal, the coefficient of longitudinal bending is equal to the calculated value of J \u003d 0.96. In the pre-arm of the height J, it varies linearly from J \u003d 1.0 to the calculated value of J \u003d 0.96 (Fig. 32). The values \u200b\u200bof the coefficient of longitudinal bending in the estimated sections of the simpleness, in the levels of the vertex and the bottom of the window opening

Fig. 31.

the magnitudes of bending moments in the level of restructuring and in the estimated sections of simpleness at the level of the top and bottom of the window opening

knm;

knm;

Fig.32

The magnitude of normal forces in the same sections

Eccentricity longitudinal forces e. 0 = M.: N.:

MM.< 0,45 y. \u003d 0.45 × 250 \u003d 115 mm;

MM.< 0,45 y. \u003d 115 mm;

MM.< 0,45 y. \u003d 115 mm;

The bearing ability of an extracentrately compressed simpleness of rectangular cross section according to claim 4.7 is determined by the formula

where (J- The coefficient of longitudinal deflection for the entire cross section of the rectangular element; ); m G. - coefficient taking into account the effect of long-term operation (with h. \u003d 510 mm\u003e 300 mm take M G. = 1,0); BUT - Sewage area of \u200b\u200bsimpleness.

We check the strength of the brick simpleness of the bearing wall of the residential building of the variable floor in Vologda.

Initial data:

The height of the floor is Net \u003d 2.8 m;

Number of floors - 8 fl;

Step of bearing walls - a \u003d 6.3 m;

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

Sizes of the severity section -1,53x0.68 m;

The thickness of the internal verst - 0.51 m;

Sequence area - a \u003d 1.04 m 2;

Length of the supporting platform plates of overlapping on the masonry

Materials: Silicate Brick Thickened Facial (250CH120CH88) GOST 379-95, Brand Sul-125/25, Silicate Porous Stone (250CH120CH138) GOST 379-95, Mark SRP -150/25 and Brick Silicate Hollow Thickened (250x120x88) GOST 379-95 Mark SURP-150/25. For masonry 1-5 floors, a cement-sandy solution M75 is used, for 6-8 floors, the density of the masonry \u003d 1800 kg / m 3, the laying of the multilayer, the insulation - polystyrene foam of the PSB-C-35 N \u003d 35 kg / m3 (GOST 15588 86). With a multilayer masonry, the load will be transferred to the inner vest of the outer wall, so when calculating the thickness of the outer verst and the insulation do not take into account.

The collection of loads from the coating and overlaps is presented in Tables 2.13, 2.14, 2.15. Estimated simpleness is presented in Fig. 2.5.

Figure 2.12. Estimated simpleness: a - plan; b - vertical section of the wall; B-calculation scheme; Mr. Moment

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

Load name	Regulatory value of kN / m2		CN / M2
Constant: 1. Layer of Locrome TKP, T \u003d 3.7 mm, weight 1m2 material 4.6 kg / m2, \u003d 1100 kg / m3 2. Layer of Locrome HPP, T \u003d 2.7 mm weight 1m2 material 3.6 kg / m2, \u003d 1100 kg / m3 3. Primer "Prica Bituminous"
4. Cement-sand screed, T \u003d 40 mm, \u003d 1800 kg / m3 5. Ceramzite gravel, T \u003d 180 mm, \u003d 600 kg / m3, 6. Insulation - PSB-C-35, T \u003d 200 mm, 35 kg / m3 7. Parosol 8. Reinforced concrete slab
Temporary: S0N \u003d 0.7CSQmacksT \u003d 0.7 h2.4 1ch1ch1

Table 2.14. Collection of loads for attic overlap, KN / m2

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

Table 2.16. Harvesting loads per 1 mp. from the outer wall t \u003d 680 mm, kn / m2

Determine the width of the cargo site by formula 2.12

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

a - the magnitude of the overlap plate, m.

The length of the freight space is determined by formula (2.13).

where L is the width of the simpleness;

l F - width of window openings, m.

The determination of a cargo area (respectively, Figure 2.6) is made by formula (2.14)

Figure 2.13. Scheme definition of freight space

Counting the efforts of N on the simplest floors at the level of the first floor overlap levels, which are based on the cargo space and the active loads on overlapping, coatings and roofing, load on the weight of the outer wall.

Table 2.17. Harvesting loads, kn / m

Load name	The calculated value of kN / m
1. Coating design
2. Attic overlap
3. Measuring overlap
4. Outdoor Wall T \u003d 680 mm

The calculation of the extracently compressed unarmed elements of stone structures should be made by formula 13

To perform the calculation of the walls for stability, it is necessary first of all to deal with their classification (see SNIP II -22-81 "Stone and Armocatament Designs", as well as a SNiP allowance) and understand what kind of walls are:

1. Bearing walls - These are the walls on which the slabs of overlapping, roof designs, etc. The thickness of these walls should be at least 250 mm (for brickwork). These are the most responsible walls in the house. They need to count on strength and stability.

2. Self-supporting walls - These are the walls that nothing rests, but they have a load from all overlapping floors. In essence, in a three-story house, for example, such a wall will be height in three floors; The load on it only on its own weight of the masonry is significant, but even the question of the stability of such a wall is very important - than the wall above, the greater the risk of its deformations.

3. Non-relaxing walls - these are outdoor walls that rest on the overlap (or to other constructive elements) And the load on them falls from the height of the floor only from its own weight of the wall. The height of non-vacant walls should be no more than 6 meters, otherwise they go to the category of self-supporting.

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

We will deal with the question of stable walls.

The first question arising from the "uninitiated" person: Well, where can the wall going? Find the answer by an analogy. Take a book in hardcover and put it on the edge. The more book format, the less its stability will be; On the other hand, than the book will thicker, the better it will stand on the edge. The same situation with the walls. The stability of the wall depends on the height and thickness.

Now we take the worst option: a thin notebook of a large format and put on the edge - it will not just lose stability, but also bends. So the wall, if the conditions in the ratio of thickness and height are not observed, will begin to bent from the plane, and over time - to crack and collapse.

What do you need to avoid such a phenomenon? Need to explore PP 6.16 ... 6.20 Snip II -22-81.

Consider the issues of determining the stability of the walls on the examples.

Example 1. The partition is given from the M25 brand of M25 in the M4 grade solution with a height of 3.5 m, a thickness of 200 mm, 6 m wide, not associated with overlapping. In the septum door opening 1x2.1 m. It is necessary to determine the stability of the partition.

From table 26 (p. 2), we determine the masonry group - III. Find from tables 28? \u003d 14. Because The partition is not fixed in the upper section, it is necessary to reduce the value of β by 30% (in accordance with clause 6.20), i.e. β \u003d 9.8.

k 1 \u003d 1.8 - for a partition, not carrier load with its thickness 10 cm, and k 1 \u003d 1.2 - for a septum with a thickness of 25 cm. In the interpolation, we find for our partition with a thickness of 20 cm k 1 \u003d 1.4;

k 3 \u003d 0.9 - for partitions with openings;

so k \u003d k 1 k 3 \u003d 1.4 * 0.9 \u003d 1.26.

Finally β \u003d 1.26 * 9.8 \u003d 12.3.

Find the ratio of the height of the partition to the thickness: H / H \u003d 3.5 / 0.2 \u003d 17.5\u003e 12.3 - the condition is not performed, the septum of such thickness at a given geometry cannot be done.

What way can you solve this problem? Let's try to increase the solution brand to M10, then the laying group will become II, respectively, β \u003d 17, and taking into account the coefficients β \u003d 1.26 * 17 * 70% \u003d 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 performed. Also, it was also possible without increasing the brand of aerated concrete, lay in the septum structural reinforcement in accordance with clause 6.19. Then β increases by 20% and the stability of the wall is provided.

Example 2.Dana Outdoor N. bearing wall Of the lightweight masonry made of brick M50 brand on the M25 brand solution. The height of the wall is 3 m, thickness is 0.38 m, the length of the wall is 6 m. The wall with two windows size is 1.2x1.2 m. It is necessary to determine the stability of the wall.

From table 26 (p. 7), we determine the masonry group - i. From the tables 28 we find β \u003d 22. Because The wall is not fixed in the upper section, it is necessary to reduce the value of β by 30% (according to paragraph 6.20), i.e. β \u003d 15.4.

We find the coefficients k from tables 29:

k 1 \u003d 1.2 - for a wall that does not carry load at its thickness 38 cm;

k 2 \u003d √a n / a b \u003d √1.37 / 2.28 \u003d 0.78 - for walls with openings, where a b \u003d 0.38 * 6 \u003d 2.28 m 2 is the area of \u200b\u200bthe horizontal section of the wall, taking into account windows, and n \u003d 0.38 * (6-1.2 * 2) \u003d 1.37 m 2;

so k \u003d k 1 k 2 \u003d 1.2 * 0.78 \u003d 0.94.

Finally β \u003d 0.94 * 15.4 \u003d 14.5.

Find the ratio of the height of the partition to the thickness: H / H \u003d 3 / 0.38 \u003d 7.89< 14,5 - условие выполняется.

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

H + L \u003d 3 + 6 \u003d 9 m< 3kβh = 3*0,94*14,5*0,38 = 15.5 м - условие выполняется, устойчивость стены обеспечена.

Attention! For the convenience of answers to your questions, a new section "Free Consultation" has been created.

class \u003d "Eliadunit"\u003e

Comments

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0 # 212 Alexey 21.02.2018 07:08

Quote Irina:

profiles reinforcement will not replace

Quote Irina:

as for the foundation: the voids are allowed in the body of concrete, but not from the bottom, so as not to reduce the area of \u200b\u200bthe support, which is responsible for the supporting ability. That is, there should be a thin layer of reinforced concrete.
And what is the foundation - tape or stove? What are the soils?

The Grids are not yet known, most likely there will be a clean field of Suglinka all sorts, originally thought the stove, but it will be low, I want to shoot higher, and also the upper fertile layer will have to shoot, so I tend to the ribbon or even a boxed foundation. I don't need a lot of carrying capacity of the soil - the house was still solved in the 1st floor, and the ceramzite concrete is not very heavy, the freezing there is no more than 20 cm (although in the old Soviet regulations 80).

I think to remove the upper layer of 20-30 cm, lay out geotextiles, pumped up with a sandy sandpaper and dissolve with a seal. Then a light preparatory screed - for aligning (in her, it seems to be not even the reinforcement, although not sure), on top of the waterproofing primer
and then there is already a dillem - even if you link the framework of reinforcement width 150-200mm x 400-600mm altitudes and put them in a meter step, then you need to form more voids among these frames and, ideally, these voids should be on top of the reinforcement (yes Also, with some distance from training, but at the same time they will also need to be turned on with a thin layer under 60-100mm a screed) - I think the PPS plates to deposit as voids - theoretically, it will be possible to pour into 1 by the vibration.

Those. As if with the form of a plate 400-600mm with a powerful reinforcement every 1000-1200mm, the bulk structure is united and easy in the rest of the places, while at about 50-70% of the volume will be the foam (in non-loaded places) - i.e. According to the consumption of concrete and reinforcement - quite comparable to a stove 200mm, but + a bunch of relatively cheap foam and more.

If I somehow replace the foam on a simple primer / sand - it will be even better, but instead of easy training it is wiser to make something more serious with the reinforcement and removal of reinforcement in the beams - in general there is not enough theory and practical experience.

0 # 214 Irin 22.02.2018 16:21

Quote:

it is a pity, generally they write that in light concrete (ceramzit concrete) bad connection with the reinforcement - how to deal with it? I understand the stronger concrete and what more square The surface of the reinforcement - the better the connection will be, i.e. It is necessary a ceramzite concrete with the addition of sand (and not only clamzit and cement) and the reinforcement is thin, but more

why deal with this? It is necessary to simply take into account in the calculation and when constructing. You see, the ceramzitobeton is good enough wall Material with your vicinity of advantages and disadvantages. Like any other materials. Now if you wanted to use it for monolithic overlap, I would dissuade you because
Quote:

Brick - enough durable building material, especially full, and during the construction of houses in 2-3 floors walls from ordinary ceramic bricks in additional calculations usually do not need. Nevertheless, there are different situations, for example, it is planned two-storey house With a terrace on the second floor. Metallic riglels, which will also rely on the metal beams of the terrace overlapping, it is planned to be leaked to brick columns from the front hollow brick 3 meters high, above will be the columns with a height of 3 m, to which the roof will be relying:

At the same time, a natural question arises: what minimal cross-section of the columns will provide the required strength and stability? Of course, the idea to lay out columns from clay brick, Moreover, the walls of the house, is far from the new and all possible aspects of the calculations of brick walls, commonens, pillars that are the essence of the columns are sufficiently detailed in SNiP II-22-81 (1995) "Stone and Armochement Constructions". That is what regulatory document and should be guided during calculations. The calculation below is no longer more than an example of using the specified SNIP.

To determine the strength and stability of the columns, you need to have sufficiently many source data, such as: brick brand for strength, the area of \u200b\u200breligioning rheel on columns, the load on the columns, the area of \u200b\u200bthe cross section of the column, and if it is not known any of this in the design stage, you can do in the following way:

under central compression

Designed: The terrace with dimensions of 5x8 m. Three columns (one in the middle and two along the edges) from the facial hollow brick cross section of 0.25x0.25 m. The distance between the axes of the column of 4 m. Brick brand for strength M75.

With this design scheme, the maximum load will be on the middle bottom column. It is precisely her and should count on strength. The load on the column depends on the set of factors, in particular from the construction area. For example, snow load The roof in St. Petersburg is 180 kg / m & sup2, and in Rostov-on-Don - 80 kg / m & sup2. Taking into account the weight of the roof of 50-75 kg / m & sup2, the load on the column from the roof for Pushkin of the Leningrad region can be:

N with roof \u003d (180 · 1.25 +75) · 5 · 8/4 \u003d 3000 kg or 3 tons

Since the current loads from the floor of overlapping and from people, squeezing on the terrace, furniture, etc. yet, but reinforced concrete plate It is precisely not planned, but it is assumed that the overlap will be wooden, from separately underlying cutting boards, then for calculations of the load from the terrace, you can take a uniformly distributed load of 600 kg / m & sup2, then the focused force from the terrace acting on the central column will be:

N from terrace \u003d 600 · 5 · 8/4 \u003d 6000 kg or 6 tons

Own column weight 3 m 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 bottom column in the cross 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 is possible to take into account that there is no very high probability that the temporary burden of snow, the maximum in winter time, and the temporary load on the overlap, the maximum in the summer, will be attached simultaneously. Those. The sum of these loads can be multiplied by the probability ratio of 0.9, then:

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

The estimated load on the extreme columns will be almost two times less:

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

2. Determination of brickwork strength.

The M75 brick brand means that the brick must withstand the load of 75 kgf / cm & sup2, however, the strength of the brick and the strength of brickwork are different things. Understand this will help the following table:

Table 1. Estimated compression resistance for brickwork

But that's not all. All of the same SNIP II-22-81 (1995) Claim 3.11 a) recommends less than 0.3 m & sup2 at the area of \u200b\u200bpillars and seasplets, multiply the value of the calculated resistance to the working conditions coefficient γ C \u003d 0.8. And since the area of \u200b\u200bthe cross section of our column is 0.25x0.25 \u003d 0.0625 m & SUP2, it will have to use this recommendation. As we can see, for the brick of the M75 brand, even when using a masonry solution M100, the strength of the masonry will not exceed 15 kgf / cm & sup2. As a result, the calculated resistance for our column will be 15 · 0.8 \u003d 12 kg / cm & sup2, then the maximum compressive voltage will be:

10300/625 \u003d 16.48 kg / cm & sup2\u003e r \u003d 12 kgf / cm & sup2

Thus, to ensure the necessary strength of the column, it is necessary or used by the brick of greater strength, for example, M150 (the calculated compression resistance during the M100 solution marque will be 22 · 0.8 \u003d 17.6 kg / cm & sup2) or increase the cross section of the column or use cross-reinforcement of masonry. While we will focus on using a more durable facial brick.

3. Determination of the stability of the brick column.

The strength of the brickwork and the stability of the brick column is also different things and the same SNiP II-22-81 (1995) recommends determining the stability of the brick column according to the following formula:

N ≤ m g φRF (1.1)

m G. - coefficient taking into account the effect of long-term load. In this case, we, conventionally speaking, was lucky, since with the height of the section h. ≤ 30 cm, the value of this coefficient can be taken equal to 1.

φ - the coefficient of 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 the design are not set out here, only we note that according to SNIP II-22-81 (1995) clause 4.3: "The calculated heights of walls and pillars l. O. When determining the coefficients of longitudinal bending φ Depending on the conditions of supporting them on horizontal supports should be taken:

a) with fixed hinge supports l. O \u003d N.;

b) with an elastic upper support and hard pinch in the lower support: for single-ranking buildings l. O \u003d 1.5h, for multipress buildings l. O \u003d 1.25h;

c) for free standing designs l. O \u003d 2N;

d) for structures with partially pinched reference sections - taking into account the actual degree of pinching, but not less l. O \u003d 0.8Nwhere N. - distance between overlaps 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 clause b). That is, you can take l. O \u003d 1.25h \u003d 1.25 · 3 \u003d 3.75 meters or 375 cm. However, we can confidently use this meaning only when the lower support is really tough. If the brick column is laid out on a layer of waterproofing from the rubberoid, laid on the foundation, then such a support should be treated as a hinge, and not rigidly pinched. And in this case, our design in the plane parallel to the wall plane is geometrically variable, since the design of the overlap (separately lying boards) does not provide sufficient rigidity in the specified plane. 4 outputs are possible from a similar situation:

1. Apply a fundamentally different constructive scheme, for example, metal columns that are rigidly sealed to the foundation, to which the beelel of the overlap will be welded, then from aesthetic considerations, metal columns can be chosen by the face brick of any brand, since the entire load will be taken metal. In this case, the truth needs to be calculated by metal columns, but the calculated length can be taken l. O \u003d 1.25h.

2. Make another overlap, for example, from sheet materials, which will allow you to consider both the upper and lower support of the column, like hinged, in this case l. O \u003d H..

3. Make a diaphragm of stiffness In the plane parallel to the wall plane. For example, on the edges, lay out the columns, but rather a simple thing. It will also make it possible to consider both the upper and lower support of the column, as hinged, but in this case it is necessary to additionally calculate the rigidity diaphragm.

4. Do not pay attention to the above options and calculate columns, as separately standing with a rigid lower support, i.e. l. O \u003d 2N. In the end, the ancient Greeks put their columns (though not from the brick) without any knowledge of the resistance of materials, without the use of metal anchors, and so carefully written construction norms And the rules in those days were not, however, some columns stand still to this day.

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

λ H. \u003d L. O. / H. (1.2) or

λ I. \u003d L. O. (1.3)

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

It is not difficult to determine the radius of inertia in principle, it is necessary to divide the moment of inertia of the section on the cross section area, and then remove the square root from the result, but in this case there is no big necessity. In this way λ H \u003d 2 · 300/25 \u003d 24.

Now, knowing the value of the flexibility coefficient, you can finally determine the coefficient of longitudinal bending according to the table:

table 2. Longitudinal bending coefficients for stone and arm-change structures
(according to SNiP II-22-81 (1995))

At the same time, the elastic characteristic of 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 longitudinal bend coefficient will be about 0.6 (with the value of the elastic characteristic α \u003d 1200, according to claim 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 adopted section of 25x25 cm to ensure the stability of the lower central central-compressed column is not enough. To increase the stability, the most optimal will increase the cross section of the column. For example, if you lay out a column with emptiness inside a half of the brick, sizes 0.38x0.38 m, thus not only the area of \u200b\u200bthe cross section of the column up to 0.13 m & sup2 or 1300 cm & sup2 will increase, but the radius of the column inertia will increase i. \u003d 11.45 cm. Then λ i \u003d 600 / 11.45 \u003d 52.4, and the value of the coefficient φ \u003d 0.8.. In this case, the maximum load on the central column will be:

N p \u003d m g φγ from RF \u003d 1 · 0.8 · 0.8 · 22 · 1300 \u003d 18304 kg\u003e n with about \u003d 9400 kg

This means that the cross-section 38x38 cm to ensure the stability of the lower central central-compressed column is enough with a margin and can even reduce the brick brand. For example, with an initially accepted M75 brand, the limit 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 another detail. The foundation in this case is better to do with a ribbon (one for all three columns), and not a bit (separately for each column), otherwise even small foundations drawders will lead to additional voltages in the body of the column and it can cause destruction. Taking into account all the above, the most optimal cross section of the columns is 0.51x0.51 m, and from aesthetic point of view, such a cross section is optimal. The cross-sectional area of \u200b\u200bsuch columns will be 2601 cm & sup2.

Example of calculating the brick column for stability
With outcidentren compression

The extreme columns in the designed house will not be centrally compressed, as the Rigel will be based on them only on the one hand. And even if the riglels will be laid on the entire column, then the load from the overlap and the roof will be transmitted to the extreme column in the center of the cross section of the column. In what kind of place will be transmitted to the resultant of this load, depends on the angle of inclination of the riglels on the supports, modules of elasticity of rigels and columns and a number of other factors. This displacement is called the eccentricity of the load application E about. In this case, we are interested in the most unfavorable combination of factors, in which the load from overlapping on the columns will be transmitted as close as possible to the edge of the column. This means that the column besides the load itself will also act a bending moment equal to M \u003d NeAnd this moment should be taken into account when calculating. In the general case, the inspection for stability can be performed according to the following formula:

N \u003d φRF - MF / W (2.1)

W. - the moment of resistance to the section. In this case, the load for the lower extreme columns from the roof can be considered centrally applied, and the eccentricity will only create a load from overlapping. With an eccentricity 20 cm

N p \u003d φRF - MF / W \u003d1 · 0.8 · 0.8 · 12 · 2601 - 3000 · 20 · 2601· 6/51 3 \u003d 19975.68 - 7058,82 \u003d 12916.9 kg\u003eN cr \u003d 5800 kg

Thus, even with a very large eccentricity of the application of the load, we have more than double stock for strength.

Note: SNiP II-22-81 (1995) "Stone and Armocatament Designs" recommends using another method of calculating the cross section, which takes 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.

The external bearing walls should be at least calculated for strength, stability, local crumpled and heat transfer resistance. To find out what thickness should be brick wall , It is necessary to make its calculation. In this article, we will consider the calculation of the carrier ability of brickwork, and in the following articles - the remaining calculations. In order not to miss the output of the new article, subscribe to the newsletter and you will ounce which there should be a wall thickness after all calculations. Since our company is engaged in the construction of cottages, that is, low-rise construction, then we will consider all calculations for this category.

Carriers It is called walls that perceive the load from the slabs of overlapping, coatings, beams, etc.

You should also consider the brick stamp on frost resistance. Since everyone builds a house for itself, at least one hundred years, then with a dry and normal humidity mode of the premises, a brand (M RZ) is taken from 25 and higher.

During the construction of a house, cottage, garage, a host. Bruks and other systems with a dry and normal humidity regime, it is recommended to use hollow bricks for outer walls, as its thermal conductivity is lower than that of full-time. Accordingly, with a heat engineering calculation, the thickness of the insulation will be less, which will save money when purchasing it. Full-year bricks for external walls must be applied only if necessary to ensure the strength of the masonry.

Reinforcement of brick masonry It is allowed only if the increase in the brick and solution brand does not allow to provide the required carrying ability.

An example of calculating the brick wall.

The carrying ability of brickwork depends on many factors - from the brand of bricks, the grade of the solution, from the presence of openings and their size, from the flexibility of walls, etc. The calculation of the bearing capacity begins with the definition of the calculation scheme. When calculating walls on vertical loads, the wall is considered an operated on hinged-fixed supports. When calculating the walls on horizontal loads (wind), the wall is considered rigidly pinched. It is important not to confuse these schemes, since the moments will be different.

Selection of estimated section.

In the deaf walls for the calculated, the cross section of I-I at the level of overlapping with the longitudinal force N and the maximum bending moment M. is often dangerous section II-IISince the bending moment is slightly smaller than the maximum and equal to 2 / 3m, and the coefficients M G and φ are minimal.

In the walls with openings, the section is accepted at the bottom level of the jumpers.

Let's consider the cross section I-i.

From the past article Harvesting load on the wall of the first floor Take the resulting value of the full load, which includes loads from the overlap of the first floor P 1 \u003d 1.8T and the above 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 slab overlap relies on the wall at a distance A \u003d 150mm. The longitudinal force P 1 from the overlap will be at a distance of a / 3 \u003d 150/3 \u003d 50 mm. Why 1/3? Because the stress plot under the support area will be in the form of a triangle, and the center of gravity of the triangle is just 1/3 of the length of the support.

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

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

e \u003d H / 2 - A / 3 \u003d 250mm / 2 - 150mm / 3 \u003d 75 mm \u003d 7.5 cm,

it will create a bending moment (m) in section I-I. The moment is the work of strength on the shoulder.

M \u003d p 1 * e \u003d 1,800 * 7.5 cm \u003d 13.5 t * 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 carrier wall with a thickness of 25 cm, then calculated the value of the random eccentricity E ν \u003d 2 cm, then the total eccentricity is:

e 0 \u003d 2.5 + 2 \u003d 4.5 cm

y \u003d H / 2 \u003d 12.5 cm

At e 0 \u003d 4.5 cm< 0,7y=8,75 расчет по раскрытию трещин в швах кладки можно не производить.

The strength of the adki high-centranenially compressed element is determined by the formula:

N ≤ m g φ 1 r a c ω

Factors m G. and φ 1. In the considered section of the I - i are equal to 1.